Arithmetic Progression

A sequence (finite or infinite) is called an arithmetic progression (abbreviated A.P.) iff the difference of any term from its preceding term is constant. This constant is usually denoted by d and is called common difference.

General term of an A.P.

   Tn = a +(n -1) d.Thus if a is the first term and d is the common difference of an A.P., then the A.P. is a, a +d, a +2 d, ..., a +(n -1) d or a, a +d, a +2 d, ... according as it is finite or infinite. Corollary. If the last term of an A.P. consisting of n terms is denoted by l, then      l = a +(n -1) d.Note. Three numbers a, b, c are in A.P. iff b -a = c -b i.e. iff a +c = 2b.

Illustrative Examples

Example

i. A sequence {an} is given by the formula an = 10 -3 n. Prove that it is an A.P. ii. Find the general term and 15th term of the sequence 3, 1, -1, -3, .... iii. If the 9th term of an A.P. is 99 and 99th term is 9, find its 108th term.

Solution

i. Given an = 10 -3 n => an +1 = 10 -3 (n +1)=> an +1 -an = 10 -3 (n +1) -(10 -3 n) = -3, which is a constant. Therefore, the given sequence {an} is an A.P.

ii. The given sequence is an A.P. with first term a = 3, and common difference d = -2.Hence general term, Tn = a +(n -1) d = 3 +(n -1)(-2) = 5 -2n.Putting n = 15, we get, T15 = 5 -2(15) = -25.

iii. Let a be the first term and d be the common difference of given A.P.Then Tn = a +(n -1) d.Putting n = 9, 99 we get T9 = a +8d = 99, T99 = a +98d = 9=> -90d = 90 => d = -1, a = 99 -8d = 107.Hence T108 = a +107d = 107 +107(-1) = 0.

Example

Show that the sequence {an} where an = 2n² +3 is not an A.P.

Solution

Given an = 2n² +3 =>  an+1 = 2(n +1)² +3.            = an+1 -an = (2(n +1)² +3) -(2n² +3)            = (2(n² +2n +1) +3) -(2n² +3) = 4n +2,which depends upon n and is not constant.Hence the given sequence is not an A.P.

Example

In a sequence, the sum of first n terms is Sn = 2n² +3n +1 for all values of n. Show that the series is an A.P. Find its 100th term.

Solution

We have Sn = 2n² +3n +1,Sn+1 = 2(n -1)² +3(n -1) +1 = 2n² -nTn = Sn -Sn-1 = (2n² +3n +1) -(2n² -n) = 4n +1Tn+1 -Tn = (4(n +1) +1) -(4n +1) = 4.Hence the given series is an A.P. with common difference d = 4.First term a = T1 = 4n +1 = 4(1) +1 = 5Now 100th term = T100 = a +99d = 5 +99(4) = 401.

Example

If a, b, c are in A.P., prove that the following are also in A.P.                (b +c)² -a², (c +a)² -b², (a +b)² -c²

Solution

(b +c)² -a², (c +a)² -b², (a +b)² -c² are in A.P.if (b +c +a)(b +c -a), (c +a +b)(c +a -b)(a +b +c),  and (a +b -c) are in A.P.i.e. if b +c -a, c +a -b, a +b -c are in A.P.                             (dividing each term by a +b +c)i.e. if (c +a -b) -(b +c -a) = (a +b -c) -(c +a -b)i.e. if 2 (a -b) = 2 (b -c)i.e. if a -b = b -ci.e. if b -a = c -bi.e. if a, b, c are in A.P. which is given to be true.Hence (b +c)² -a², (c +a)² -b², (a +b)² -c² are in A.P.

Exercise

1. Find the(i) n th term of the series 5 +1 -3 -7 ...(ii) r th term of the series (a +d) +(a +3d) +(a +5d) +...(iii) 100th term of the sequence 0, -3, -6, -9, ...(iv) m th term of the sequence n -1, n -2, n -3, ...(v) Eighteenth term of the sequence 9, 5, 1, ...

2. Find first four terms of the series(i) (6n +1)      (ii) whose nth term is 3 -2n(iii) given that series is A.P. with a = 14/2, d = 3/2.

3. If a, b are positive real numbers, show that the sequencelog a, log (ab), log (ab²), log (ab³), ... is an A.P.Also find its general term.

4. Which term of the series(i) 2 +5 +8 +11 +... is 53(ii) 20 +18 +16 +... is -2(iii) 4 +0 -4 -8 ... is -392?

5. Which term of the sequence 20, 19 is the first negative term? 6. Which term of the sequence 9 -8i, 8 -6i, 7 -4i, ... is

(i) real    (ii) purely imaginary? 7. (i) Find the arithmetic progression of 6 terms if first term is 2/3 and the last is 22/3.

(ii) Find the number of terms in the series 5 +8 +11 +14 +... if the last term is 95. 8. (i) Determine the 25th term of an A.P. whose 9th term is -6 and common difference is

5/4.(ii) The third term of an A.P. is 25 and tenth term is -3. Find the first term and the common difference.

9. (i) If 9th term of an A.P. is zero, prove that 29th term is double the 19th term.(ii) If 7 times the 7th term of an A.P. is equal to 11 times its 11th term show that 18th term of this A.P. is zero.

10. Second, 31st and last terms of an A.P. are 31/4, 1/2, -13/2 respectively. Find the number of terms in the A.P.

11. Determine k so that k +2, 4k -6 and 3k -2 are three consecutive terms of an A.P. 12. (i) If the nth term of a sequence is an expression of first degree in n, show that it is an

A.P.[Hint. Take Tn = an +b](ii) A sequence {an} is given by an = n² -1, n N. Show that it is not an A.P.

13. A person was drawing a monthly salary of Rs 2500 in 11th year of service and a salary of Rs 2900 in the 19th year. Given that pension is half the salary at retirement time, find his monthly pension if he had put in 25 years of service before retirement. Assume that annual increment is constant.

14. A person purchases a VCR for Rs 16000. Its life is estimated to be 20 years. If the yearly depreciation is assumed to be constant, find the rate of depreciation and the price after 8 years.

15. If a, b, c are in A.P. show that(i) (a -c)² = 4(b² -a c)      (ii) (a -c)² = 4(a -b)(b -c)

16. For all real a, b, prove that (a -b)², a² +b² and (a +b)² are in A.P. 17. How many numbers of two digits are divisible by 6?

[Hint: The first two digit number divisible by 6 is 12 and the last two digit number divisible by 6 is 96.]

Answers

1. (i) 9 -4n            (ii) a +(2r -1)d (iii) -297    (iv) n -m             (v) -592. (i) 7, 13, 19, 25  (ii) 1, -1, -3, -5  (iii) 7, 123. log (a bn-1)4. (i) 18 th              (ii) 12 th             (iii) 100 th5. 28th                  6. (i) 5th            (ii) 10th7. (i)2/3, 2, 10/3, 14/3, 6, 22/3         (ii) 318. (i) 14                 (ii) 33, -4           10. 5911. 3                     13. Rs 1600       14. Rs 800, Rs 960017. 15

Formulae for Sum of A.P.

Formulae

1. Sum of first n terms of an A.P. is     Sn = (n/2)[2a +(n -1)d]

2. If Sn Is sum of n terms of an A.P. whose first term is a and last term is l,then Sn = (n/2)(a + l)

3. If common difference is d, number of terms n and the last term l,then Sn = (n/2)[2l-(n -1)d]

Remark

Sometimes the question involves 3, 4 or 5 terms of an A.P.If the sum of the numbers is given, then in an A.P.,

i. three numbers are taken as a -d, a, a +d ii. four numbers are taken as a -3d, a -d, a +d, a +3d. iii. five numbers are taken as a -2d, a -d, a, a +d, a +2d.

This considerably simplifies the calculations of a and d.

Illustrative Examples

Example

i. Sum the series 2 +4 +6 +... upto 40 terms. ii. Find the sum of first 19 terms of the A.P. whose nth term is 2n +1. iii. Find the sum of the series 1 +3 +5 +... +99.

Solution

i. We see that given series is an A.P. with first term a = 2,common difference d = 2 and number of terms n = 40Hence S40 = n[2a +(n -1)d]/2 = 40[2.2 +(40 -1)2]/2                = 20(4 +78) = 1640

ii. Here first term a = T1 = 2n +1 = 2(1) +1 = 3,    and last term = T19 = 2(19) +1 = 39    S19 = n(a +l)/2 = 19(3 +39)/2 = 19.21 = 399

iii. Given series is an A.P. with a = 1, l = 99, d = 2To find the number of terms, we use l = a +(n -1)d=>   99 = 1 +(n -1)2 => n -1 = 49 => n = 50    Sn = n(a +l)/2 = 50(1 +99)/2 = 50.50 = 2500

Example

i. Sum up 0·7 +0·71 +0·72 +... upto 100 terms. ii. How many terms of the A.P. 17 +15 +13 +... must be taken so that sum is 72?

Explain the double answer.

Solution

i. We see that given series is an A.P. with first term     a = 0·7 and common difference d = 0·01     S100 = n[2a +(n -1)d]/2 = 100[2(0·7) +(100 -1)(0·01)]/2           = 50 (1· 4 +0·99) = 119·5

ii. We are given a = 17, d = -2, Sn = 72 and we have to find n.Using Sn = n[2a +(n -1)d]/2, we get     72 = n[2.17 +(n -1)(-2)]/2 = n(36 -2n)/2 = n(18 -n)=>  n² -18n +72 = 0  =>  (n -6)(n -12) = 0=>    n = 6, 12Both values of n, being positive integers are valid. We get double answer because sum of 7th to 12th terms is zero, as some terms are positive and some are negative.

Example

The interior angles of a polygon are in arithmetic progression. The smallest angle is 120° and the common difference is 5°. Find the number of sides in the polygon.

Solution

The sum of all exterior angles of a polygon = 360°. If the interior angles are in A.P., then exterior angles are also in A.P.; largest exterior angle is 180°-120° = 60° and common difference is -5°.Sum = 360° = n [2 a +(n -1) d]/2 = [2. 60° +(n -1)(-5°)]/2   => 720 = 120 n -5n² +5n  => 5n² -125n +720 = 0   =>  n² -25 n +144 = 0   =>  (n -9)(n -16) = 0  =>   n = 9 or n = 16.However if n = 16, then internal angles would vary from 120° to 120° +(16 -1)(5°) = 195°; so one of the internal angles will be 180° which is not possible in a polygon.Hence the only correct solution is n = 9.

Example

The sum of three numbers in A.P. is 18 and their product is 192. Find the numbers.

Solution

Let the three numbers in A.P. be a -d, a, a +d. By given conditions,      (a -d) +a +(a +d) = 18, (a -d)a(a +d) = 192=>  3a = 18, a(a² -d²) = 192   =>  a = 6, 6(36 -d²) = 192=>  d² = 36 -192/6 = 36 -32 = 4 => d = ±2.With a = 6, d = 2, we get three numbers as 4, 6, 8.With a = 6, d = -2, we get three numbers as 8, 6, 4.Hence the required numbers are 4, 6, 8.

Exercise

1. Find the sum of(i) 3 +7 +11 +15 ... to 30 terms(ii) 1 +2/3 +1/3 +0 +.. to 19 terms(iii) 5/2, 10/3, 25/6, 5 ... upto 24 terms(iv) 51 +50 +49 +... +21(v) 72 +70 +68 +... +40.

2. Find the sum of an A.P. of(i) 25 terms whose nth term is 2n+5(ii) 19 terms whose nth term is (2n+1)/3

3. If the sum of n terms of a series is an²+bn, where a, b are constants, show that it is an A.P. Find the first term and the common difference. Also find an A.P. whose sum of any number of terms is equal to the square of the number of terms.[Hint. an²+bn=n² for all n  => a = 1, b = 0.]

4. Of a, l, d, n, Sn determine the ones which are missing for the following arithmetic progressions:(i) a = -2, d = 5, Sn = 568(ii) l = 8, n = 8, S8 = -20(iii) d = 2/3, l = 10, n = 20.

5. (i) How many terms of series 13 +11 +9 +... make the sum 45? Explain the double answer.(ii) How many terms of the sequence 18, 16, 14,... should be taken so that their sum is zero?

6. (i) Determine the sum of first 35 terms of an A.P. if T2 = 2, T7 = 22.(ii) It the third term of an A.P. is 1 and 6th term is -11, find the sum of first 32 terms.

(iii) If the first term of an A.P. is 2 and the sum of first 4 terms is equal to one fourth of the sum of the next five terms, find the sum of first 30 terms.

7. If the sum of an n terms of two arithmetic series are in ratio(i) (14 -4n):(3n +5), find the ratio of their 8th terms(ii) (2 +3n):(3 +2n), find the ratio of their 7th terms.

8. (i) If the ratio of sum of m terms of an A.P. to the sum of n terms is m² : n², show that ratio of the pth and qth term is (2p -1) : (2q -1).(ii) The sums of m and n terms of an A.P. are in ratio (2m +1) : (3n +1). Find the ratio of its 7th and 10th terms.

9. (i) If in an A.P., S1 = 6 and S7 = 105, prove that    Sn : Sn -3 : : (n +3) : (n -3).(ii) If in an A.P., S3 = 6 and S6 = 3, prove that      2(2n +1)Sn +4 = (n +4)S2n +1.

10. (i) Find the sum of all two digit numbers.(ii) Find the sum of all natural numbers between 100 and 1000 which are divisible by 2 as well as by 5.(iii) Find the sum of all two digit numbers which leave 1 as remainder when divided by 3.(iv) Find the sum of all odd integers between 2 and 100 which are divisible by 3.[Hint. (iv) The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.]

11. (i) The sum of three numbers in A.P. is -3 and their product is 8. Find the numbers.(ii) The sum of three consecutive numbers in an A.P. is 24 and the sum of their squares is 194. Find the numbers.(iii) The sum of three numbers in an A.P. is 30, and the ratio of first to third is 3 : 7. Find the numbers.(iv) Find five numbers in an A.P. whose sum is 25 and the ratio of the first to the last is 2 : 3.(v) Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7 : 15.

Answers

1. (i) 1830       (ii) -38       (iii) 290    (iv) 1116    (v) 9522. (i) 775         (ii) 1333. a +b; 2a; 1, 3, 5, 7, ....4. (i) n = 16, l = 73            (ii) a = -13, d = 3    (iii) a = -8/3, s1= 220/35. (i) 5 or 9      (ii) 196. (i) 2310       (ii) -1696    (iii) 16557. (i) -23/25     (ii) 41/298. (ii) 513/75410. (i) 4905      (ii) 98450  (iii) 1605        (iv) 86711. (i) -4, -1, 2 (ii) 7, 8, 9  (iii) 6, 10, 14  (iv) 4, 9/2, 5, 11/2, 6      (v) 2, 6, 10, 14

Arithmetic Mean

The A.M. between two numbers a and b is        A =(a +b)/2

Similarly, if a, A1, A2, ..., An, b are in A.P., then A1, A2, ..., An are called n arithmetic means between a and b.

Illustrative Examples

Example

Insert six arithmetic means between 2 and 16. Also prove that their sum is 6 times the A.M. between 2 and 16.

Solution

Let A1, A2, ..., A6 be six A.M.s between 2 and 16. Then, by def., 2, A1, A2, ..., A6, 16 are in A.P.Let d be the common difference. Here 16 is the 8th term.16 = 2 +7d => d = 2.Hence six A.M.s are a +d, a +2d, ..., a +6d

      i.e. 4, 6, 8, 10, 12, 14.Now sum of these means      = 6(4 +14)/2 = 54 = 6.9 = 6(2 +16)/2     = 6 times the A.M. between 2 and 16.

Example

If A is the A.M. between a and b, show that            (A +2 a)/(A -b) + (A +2 b)/(A -a) = 4           

Solution

Since A is the A.M. between a and b, A = (a +b)/2Hence (A +2 a)/(A -b) + (A +2 b)/(A -a)

        = = (5a + b)/(a - b) + (5b + a)/(b - a) = (4a - 4b)/(a-b)= 4

Exercise

1. (i) Insert 5 arithmetic means between -2 and 10. Show that their sum is 5 times the arithmetic mean between -2 and 10.(ii) Insert 4 arithmetic means between 12 and -3.(iii) Insert 10 A.M.s between -5 and 17 and prove that their sum is 10 times the A.M. between -5 and 17.

2. If A is the A.M. between a and b, prove that(i) (A -a)² +(A -b)² = (a -b)²/2(ii) 4(a -A)(A -b) = (a -b)².

3. The ratio of the first to the last of n A.M.s between 5 and 35 is 1 : 4. Find n. 4. There are n arithmetic means between 3 and 17. The ratio of the first mean to the

last mean is 1 : 3. Find n. 5. If the A.M. between p th and q th items of an A.P. is equal to the A.M. between rth

and sth items, show that p +q = r +s.

Answers

1. (i) 0, 2, 4, 6, 8      (ii) 9, 6, 3, 0    (iii) -3, -1, 1, 3, 5, 7, 9, 11, 13, 153. n = 9                    4. n = 6

Geometric Progression (G.P.)

A sequence (finite or infinite) of non-zero numbers is called a geometric progression (abbreviated G.P.) iff the ratio of any term to its preceding term is constant. This non-zero constant is usually denoted by r and is called common ratio.

General term of a G. P.

   Tn = a rn-1

Thus, if a is the first term and r is the common ratio., then the G.P. is a, ar, ar², ..., arn -1 or a ,ar, ar², ar³, ... according as it is finite or infinite.

Remarks

1. If the last term of a G.P. consisting of n terms is denoted by l, then l = a. rn-1. 2. Three numbers a, b, c are in G.P. iff b/a = c/b i.e. iff b² = ac. 3. Sometimes the problem involves 3, 4 or 5 numbers in a G.P. If the product of the

numbers is given, then in a G.P.,(i) three numbers are taken as a/r, a, ar.(ii) four numbers are taken as a/r³, a/r, a r, ar³.(iii) five numbers are taken as a/r², a/r, a, a r, ar².

Illustrative Examples

Example

Find the n th and the 15th term of the series        1 - 1/2 + 1/4 - 1/8 ....

Solution

The given series is a G.P. with first term a = 1 and common ratio r = -1/2.Hence nth term,

     In particular, T15 = (-1)15-1. 21-15 = 2-14.

Example

Three numbers whose sum is 21 are in A.P. If 2, 2, 14 are added to them respectively, the resulting numbers are in G.P. Find the numbers.

Solution

Let the three given numbers in A.P. be a -d, a, a +d.Their sum = 3a = 21 => a = 7.Hence the numbers are 7 -d, 7, 7 +d.Adding 2, 2, 14 respectively, we get the numbers 9 -d, 9, 21 +d.Since these three are in G.P.,(9)² = (9 -d)(21 +d) => d² +12d -108 = 0  => (d +18)(d -6) = 0 => d = 6 or -18.Using d = 6, we get three numbers as 1, 7, 13, and using d = -18, we get three numbers as 25, 7, -11. Thus two sets of numbers are 1, 7, 13 or 25, 7, -11.

Example

i. If a, b, c, d are in G.P., prove that a +b, b +c, c +d are also in G.P. ii. Find all sequences which are simultaneously arithmetic and geometric progressions.

Solution

i. Given a, b, c, d are in G.P., let r be the common ratio.So b = ar, c = ar², d = ar³.Hence (b +c)² = (ar +ar²)² = a²r²(1 +r)²,   and (a +b)(c +d) = (a +ar)(ar² +ar²) = a²r²(1 +r)².Hence (b +c)² = (a +b)(c +d) => a +b, b +c, c +d are in G.P.

ii. Let a1, a2, ..., an, ... be a sequence which is A.P. as well as G.P. Since it is an A.P.,

      an +1 = , n 1Also, since it is a G.P., Let a1 0 and r be the common ratio,so an = a1. rn -1, an +1 = a1. rn, an+2 = a1. rn +1.Substituting these values in above equation, we get

     a1 rn = => r = (1 +r²)/2 => r² -2 r +1 = 0=> (r -1)² = 0 => r = 1Therefore, only a constant sequence a, a, a, ... (a 0) is both an A.P. and G.P.

Example

If a, b, c are the p th, q th and r th terms respectively of an A.P. and also the pth, qth and rth terms of a G.P., prove that ab -c. bc -a. ca -b = 1.

Solution

Let the A.P. be A, A +D, A +2 D, ... and G.P. be x, xR, xR², ... then      a = A +(p -1)D, b = A +(q -1)D, c = A +(r -1)D=>  a -b = (p -q)D, b -c = (q -r)D, c -a = (r -p)DAlso a = xRp -1, b = xRq-1, c = xRr -1

Hence ab -c.bc -a.ca-b = (xRp -1)(q -r)D.(xRq -1)(r -p)D. (x Rr -1)(p -q)D

         = x(q -r +r -p +p -q)D. R[(p -1)(q -r) +(q -1)(r -p) +(r -1)(p -q)]D

         = x°. R° = 1.1 = 1

Exercise

1. (i) Find the next term of the sequence 1/6, 1/3, 2/3...(ii) Find the 15th term of the series 3 +1/ 3 +1/3 3 +...

2. Which term of the series(i) 1 +1/3 +1/9 +1/(27)....is 1/(243)(ii) 3 -3 3 +9 -... is 729?

3. (i) Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.(ii) Find the number of terms of a G.P. whose first term is 3/4, common ratio is 2 and the last term is 384.(iii) Find the geometric series whose 4th term is 54 and the 7th term is 1458.(iv) The fourth term of a G.P. is the square of its second term and the first term is -3. Determine its seventh term.

4. Find the value of x such that(i) -2/7, x, -7/2 are three consecutive terms of a G.P.(ii) x +9, x -6 and 4 are three consecutive terms of a G.P.(iii) x, x +3, x +9 are first three terms of a G.P.

5. (i) Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.(ii) Three numbers whose sum is 70 are in G.P. If each of the extremes is multiplied by 4 and the mean by 5, the numbers will be in A.P. Find the numbers.

6. (i) The lengths of the sides of a triangle form a G.P. If the perimeter of the triangle is 37 cm and the shortest side is of length 9 cm, find the lengths of the other two sides.(ii) There are four numbers such that first three of them form an arithmetic sequence and the last three form a geometric sequence. The sum of the first and third terms is 2 and that of second and fourth is 26. What are these numbers?(iii) The sum of first three terms of a G.P. is 7 and the sum of their squares is 21. Determine the first five terms of the G.P.[Hint. Let the three terms be a, ar and ar².Use r4 +r² +1 = (r² +r +1)(r² -r +1)]

7. (i) If a, b, c are in A.P. as well in G.P., prove that a = b = c.(ii) Consider a G.P. of finitely many terms. Prove that product of n th term from the beginning and nth term from the end is a constant and equals to the product of first and the last terms.(iii) The terms of a G.P. with first term a and common ratio r are squared. Prove that resulting series is also a G.P. Find its first term, common ratio and the n th term.

8. (i) If a, b, c are in G.P. show that 1/a, 1/b, 1/c are also in G.P.(ii) If K is any positive real number and Ka, Kb, Kc are three consecutive terms of a G.P., prove that a, b, c are three consecutive terms of an A.P.(iii) If p, q, r are in A.P., show that p th, q th and r th terms of any G.P. are themselves in G.P.

9. If a, b, c, d are in G.P., show that(i) a² +b², b² +c², c² +d² are in G.P.(ii) a² -b², b² -c², c² -d² are in G.P.(iii) (a² +b² +c²)(b² +c² +d²) = (ab +bc +cd)²(iv) (b -c)² +(c -a)² +(d -b)² = (a -d)².

10. Does there exist a geometric progression containing 27, 8 and 12 as three of its terms? If it exists, how many such progressions are possible?[Hint. Take 27, 8, 12, as p th, q th, r th items. We get p +2 q = 3 r, which has infinite number of positive integral solutions.]

Answers

1. (i) 4/3      (ii) 3-27/2       2. (i) 6th          (ii) 11th3. (i) 3072   (ii) 10             (iii) 2 +6 +18 +54 +...  (iv) -21874. (i) ±1       (ii) 0 or 16      (iii) 35. (i) 2, 5, 8 or 26, 5, -16   (ii) 10, 20, 406. (i) 12 cm, 16 cm             (ii) -3, 1, 5, 25 or 7, 1, -5, 25    (iii) 1, 2, 4, 8, 16 or 4, 2, 1/2 , 1/4, 17. (iii) a², r², a² r2n -2

10. Yes, infinite many

Geometric Mean

When three numbers are in G.P., the middle one is said to be the geometric mean (briefly written as G.M.) between the other two.Let a, b be two (positive) numbers and G be the G.M. between them. Then, by definition, a, G, b are in G.P.=> G = b => G² = a b => G = ab(Conventionally, G is taken as positive).Thus the G.M. between two numbers a and b is G = abSimilarly, if a, G1, G2, ..., Gn, b are in G.P., then G1, G2, ..., Gn are called n geometric means between a and b.

Illustrative Example

Example

i. The A.M. between two numbers is 34 and G.M. is 16. Find the numbers. ii. Show that A.M. between two distinct positive numbers is always greater than G.M. iii. If x, y, z are distinct positive numbers, prove that

                     (x +y)(y +z)(z +x) > 8 x y z.Further if x +y +z = 1, show that (1 -x)(1 -y)(1 -z) > 8 x y z.

Solution

i. Let the two numbers be a and b. Then A.M. = (a +b)/2 = 34=>  a +b = 68    G.M. = ab = 16=>  a b = 256 => a (68 -a) = 256   =>   a² -68 a +256 = 0 => (a -64)(a -4) = 0=>  a = 4, 64Taking a = 64, we get b = 68 -a = 4Taking a = 4, we get b = 68 -a = 64Thus the two numbers are 4 and 64

ii. Let a, b be two given positive real numbers. Then A.M. = A = (a + b)/2     G.M. = G = abTherefore, A -G = (a +b)/2 - ab = (1/2)(a +b -2 ab) = (1/2)( a - b)²Now ( a - b)², being the square of a non-zero real number is positive.    A -G > 0  =>     A> G

iii. As the A.M. between two distinct positive numbers is greater than the G.M. between them, therefore      (x + y)/2 > xy   =>  x +y > 2 xy       ....(1)       (y + z)/2 > yz   =>  z + y > 2 yz       ...(2)       (z + x)/2 > zx  =>  x + z > 2 zx        ...(3)From (1), (2) and (3), we get        (x +y)(y +z)(z +x) > 2 x y 2 y z 2 z x     i.e.    (x +y)(y +z)(z +x) > 8 x y z             ...(4)When x +y +z = 1, then        y +z = 1 -x, z +x = 1 -y and x +y = 1 -z                (1 -x)(1 -y)(1 -z)> 8 x y z.             [using (4)]

Exercise

1. (i) Insert 3 geometric means between 1 and 256.(ii) Insert 4 geometric means between 4/9 and 27/8.(iii) Insert 5 geometric means between 576 and 9. Show that their product is equal to the fifth power of the single G.M. between the given numbers.

2. (i) The A.M. of two numbers a and b exceeds their G.M. by 2. The ratio of two numbers is 4. Find the numbers.(ii) Find two positive numbers whose difference is 12 and whose A.M. exceeds G.M. by 2.

3. Find the value of n so that may be the G.M. between a and b. 4. If one geometric mean G and two arithmetic means p and q are inserted between

two numbers, show that G² = (2 p -q)(2 q -p). 5. Show that if A and G are A.M. and G.M. between two positive numbers, then the

numbers are A ± (A² -G²).

6. If the A.M. and G.M. between two numbers are in the ratio m : n, then prove that the numbers are in the ratio m + (m² -n²) : m - (m² -n²)

7. If G is the geometric mean between a and b, show that   1/(G + a) + 1/(G+b) = 1/G

8. If G1 and G2 are two geometric means between a and b, show that   (i) G1 G2 = a b      (ii) G1² / G2 + G2²/G1 = a +b

9. If p, q, r are in A.P., a is G.M. between p and q, and b is G.M. between q and r,   then prove that a², q², b² are in A.P.

Answers

1. (i) 4, 16, 64               (ii) 2/3, 1, 3/2, 9/4                       (iii) 288, 144, 72, 36, 182. (i) 16, 4                      (ii) 16, 43. -1/2

Sum of first n Terms of a G.P.

Formulae

         This formula can also be written as

          Note that if r = 1, this formula can not be used. In that case Sn = a +a +a +... n times = na.Sum of an infinite G.P.When r < 1 numerically i.e. -1 < r < 1, then rn goes on decreasing numerically as n increases, and ultimately as n , rn 0

             as n The limiting sum of the G.P. a +a r +a r² +... to is a/(1 - r) when |r| < 1.Note that if |r| 1, the G.P. diverges, and sum can not be found.

Zeno's Paradox

Ancient mathematicians did not know the idea of limits, and they devised several paradoxes of endlessness. In Zeno (5th century B.C) paradox, Achilles and the tortoise had a race. Achilles could run 10 times as fast as the tortoise, but the tortoise had a hundred yard start. Achilles runs the hundred yards, but the tortoise is now 10 yards ahead. Achilles runs the 10 yards, but the tortoise is now 1 yard ahead. Achilles runs the 1 yard, but the tortoise is now 1/10 yard ahead, and so on. How can Achilles overtake the tortoise? The ancient Greeks did not know about limits, so in their logic the problem could not be solved. However, we know that 100 +10 +1 +1/10 + 1/100 ... has a limiting sum

= and so at this point, Achilles overtakes the tortoise.

Exercise

1. Find the sum of(i) 20 terms of the series 2 +6 +18 +...(ii) 10 terms of series 1 + 3 +3 +...(iii) 6 terms of 1, -2/3 , 4/9,...(iv) 100 terms of 0·7 +0· 07 +0· 007 +...(v) the series 81 -27 +9 ... -1/27

2. Sum the following series to infinity:7 -1 + 1/7 - 1/49 + ....

3. (i) How many terms of the sequence 1, 4, 16, ... must be taken to have their sum equal to 341?(ii) How many terms of the sequence 3, 3/2, 3/4 ... are needed to give the sum

3069/512?(iii) How many terms of the sequence 1, 2, 2, 2 2, ... are required to give a sum of 1023( 2 +1)?

4. (i) If the first term of a G.P. is 5 and the sum of first three terms is 31/5, find the common ratio.(ii) The sum of first three terms of a G.P. is to the sum of first six terms as 125 : 152. Find the common ratio of the G.P.(iii) In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.(iv) The sum of first three terms of a G.P. is 16 and sum of the next 3 terms is 128. Determine the first term, common ratio and sum to n terms of the G.P.

5. The first term of an infinite G.P. is 1 and any term is equal to the sum of all the succeeding terms. Find the series.

6. Find the sum of first n terms of the series(i) 3 +33 +333 +...(ii) 0· 4 +0· 44 +0· 444

7. A snail starts moving towards a point 3 cms away at a pace of 1 cm per hour. As he gets tired, he covers only half the distance compared to previous hour in each succeeding hour. In how much time will the snail reach his target?

8. If S be the sum, P the product and R the sum of reciprocals of n terms of a geometric

progression, find the value of

9. The sum of an infinite G.P. is 16 and the sums of the squares of its terms is . Find the common ratio and fourth term of the progression.

10. Find the sum of an infinite G.P. whose first term is 28 and the fourth term is 4/49. 11. Find the sum of divisors of 762048.

[Hint. Given number = 26. 35. 7²; sum of divisors is(1 +2 +2² +... +26)(1 +3 +3² +... +35)(1 +7 +7²) as the expansion contains all possible divisors]

Answers

1. (i) 320 -1     (ii) 121( 3 +1)      (iii) 133/243

    (iv)              (v)2. 49/83. (i) 5            (ii) 10            (iii) 204. (i) 1/5 or -6/5 (ii) 3/5  (iii) 2     (iv) 16/7, 2, 16(2n -1)/75. 1, 1/2, 1/4, 1/86. (i) (1/27)(10n+1 -9 n -10)

   (ii) 7. Never        8. 1        9. 1/4, 3/16

10.         11. 2634996

Harmonic Progression

A series of non-zero numbers is said to be harmonic progression (abbreviated H.P.) if the series obtained by taking reciprocals of the corresponding terms of the given series is an arithmetic progression.For example, the series 1 +1/4 +1/7 +1/10 +..... is an H.P. since the series obtained by taking reciprocals of its corresponding terms i.e. 1 +4 +7 +10 +... is an A.P.A general H.P. is 1/a + 1/(a + d) + 1(a + 2d) + ...nth term of an H.P. = 1/[a +(n -1)d]

Three numbers a, b, c are in H.P. iff 1/a, 1/b, 1/c are in A.P.i.e. iff 1/a + 1/c = 2/bi.e. iff b= 2ac/(a + c)Thus the H.M. between a and b is H = 2ac/(a + c)

Illustrative Examples

Example

The 7th term of an H.P. is 1/10 and 12th term is 1/25 Find the 20th term, and the nth term.

Solution

Let the H.P. be 1/a + 1/(a + d) + 1(a + 2d) + ...The 7th term = 1/(a + 6d) = 1/10 =>  a +6 d = 10The 12th term = 1/(a + 11d) = 1/25  => a +11 d = 25Solving these two equations, a = -8, d = 3Hence 20th term = 1/(a+19d) = 1/[-8 + 9(3)] = 1/49and nth term = 1/[a +(n -1)d] = 1/[-8 +(n -1) 3] = 1/[3n - 11]

Example

Prove that three quantities a, b, c are in A.P., G.P., or H.P. iff       (a-b)/(b-c) = a/a, a/b or a/c respectively.

Solution

a, b, c are in A.P. iff b -a = c -b i.e.iff (a-b)/(b-c) = 1 = a/aAlso a, b, c are in G.P.iff b/a = c/b i.e iff 1 - b/a = 1 - c/bi.e. iff (a-b)/a = (b-c)/bi.e. iff (a-b)/(b-c) = a/bSimilarly a, b, c are in H.P.iff 1/b - 1/a = 1/c - 1/bi.e. iff (a-b)/ab = (b-c)/bci.e. iff (a-b)/(b-c) = ab/bc = a/c

Example

If A, G, H are arithmetic, geometric and harmonic means between two distinct, positive real numbers a and b, show that

i. G² = AH i.e. A, G, H are in G.P. ii. A, G, H are in descending order of magnitude i.e. A > G > H.

Solution

Here A = A.M. between a and b = (a+b)/2,        G = G.M. between a and b = (ab),        H = H.M. between a and b = 2ab/(a+b)(i) A.H = [(a+b)/2].[2ab/(a+b)] = [ (ab)]² = G²(ii) A - G = (a+b)/2 - (ab) = (1/2)[a +b -2 (ab)] = (1/2)( a - b)²> 0as ( a - b)² is square of a non-zero real number.Also G - H = ab - 2ab/(a+b) = [( ab)/(a+b)](a + b - 2 ab)                  = [( ab)/(a+b)]( a - b)²> 0Hence A > G > H.

Exercise

1. Find the n th term of series 5/2 +20/13 + 10/9 + 20/23 +.... 2. Find the 10th term of series 3 +3/4 +3/7 +.... 3. If the 10th term of an H.P. is 1/20 and the 17th term is 1/41, find its n th term. 4. In an H.P. the p th term is qr and q th term is pr. Show that r th term is pq. 5. Find x such that 2 +x, 3 +x, 5 +x may be in H.P. 6. Which term of the series 12/17 + 2/3 + 12/19 +.... is 12/25? 7. The sum of three numbers in H.P. is 11/12 and sum of their reciprocals is 12. Find

the numbers. 8. The sum of the reciprocals of three numbers in H.P. is 12 and the product of the

numbers is 1/48. Find the numbers.

Answers

1. 20/(5n +3)       2. 3/283. 1/(3n -10)        5. 1             6. 9th7. 1/2, 1/4, 1/6 or 1/6, 1/4, 1/28. 1/2, 1/4, 1/6 or 1/6, 1/4, ½

Arithmetico-Geometric Series &Some Special Sequences

A series each term of which is formed by multiplying the corresponding terms of an A.P. and G.P. is called arithmetico-geometric series.The general or standard form of such a series is a, (a +d) r, (a +2 d) r², ...

The sum of first n natural numbers

Let Sn (or n²) denote the sum of first n natural numbers, then   Sn = 1 +2 +3 +... +nThis is an A.P. with a = 1, d = 1 and n terms.Using formula Sn = n(a +l)/2,Sn = n(1 +n)/2 = n(n +1)/2       or      n= n(n +1)/2

The sum of squares of first n natural numbers

Let Sn (or n²) denote the sum of squares of the first n natural numbers, then    Sn = 1² +2² +3² +... +n².Consider the identity x³ -(x -1)³ = 3 x² -3 x +1 ...(i)Putting x = 1, 2, 3, ..., n in (i), we get  1³ -0³ = 3. 1² -3. 1 +1  2³ -1³ = 3. 2² -3. 2 +1  3³ -2³ = 3. 3² -3. 3 +1  ..........................................  ..........................................  n³ -(n -1)³ = 3. n² -3. n +1Adding column wise, we get n³ -0³ = 3(1² +2² +... +n²) -3 (1 +2 +... +n) + n n³ = 3 n² -3 n +ntherefore n² = n(n +1)(2n +1)/6

The sum of cubes of the first n natural numbers

Let Sn or n³ denote the sum of the cubes of the first n natural numbers.Then Sn = 1³ +2³ +3³ +... +n³.Consider the identity x4 -(x -1)4 = 4 x³ -6 x² +4 x -1, we get

     

Method of differences

In some series, the differences of successive terms (Tn and Tn -1) may be in some particular sequence -A.P., G.P., A.G.P. etc. In these cases, the sum of series can be found by a technique known as "method of differences".

Illustrative Examples

Example

Sum to n terms the series   2 +5 x +8 x² +11 x³ +..., |x| < 1. Deduce the sum to infinity.

Solution

The given series is formed by multiplying corresponding terms of A.P. 2, 5, 8, 11, ... and G.P. 1, x, x², x³, ...Hence n th term of given arithmetico-geometric series is  [2 +(n -1) 3]. xn -1 = (3 n -1) xn -1

Let Sn = 2 +5 x +8 x² +11 x³ +... +(3 n -4) xn -2 +(3 n -1) xn -1

  x Sn =        2 x +5 x² +8 x³ +...                           +(3 n -4) xn -1 +(3 n -1) xn

(1 -x) Sn = 2 +(3 x +3 x² +3 x³ +... +3 xn -1) -(3n -1) xn

         = 2 +3 x (1 +xn -1)/(1-x) -(3 n -1) xn

Sn =   2/(1-x)   + 3 x  (1 +xn -1)/(1-x)² -(3 n -1) xn/(1-x)To find limiting sum S, we note that   xn -1, xn => 0, n xn =>  because |x| < 1Therefore S = 2/(1-x) + 3x/(1 -x)² = (2 + x)/(1 -x)²

Example

Sum to infinity the series 1² +2². x + 3². x² +..., |x| < 1.

Solution

Though the given series is not an arithmetico-geometric series, it can be summed up using a similar method.Let S = 1 +4 x +9 x² +16 x³ +... x       Sx =   x +4 x² + 9 x³ +...=>  (1 -x) S = 1 +3 x +5 x² +17 x³ +...,which is an arithmetico-geometric series      x (1 -x) S =     x +3 x²+ 5 x³ +...=>   (1 -x)² S = 1 +2 x +2 x² +2 x³ +...                      = 1 + 2x/(1-x) = (1 +x)/(1-x).=>      S = (1 +x)/(1 -x)³

Example

Find the sum of the series 2. 5 +5. 8 +8. 11 +... to n terms.

Solution

This series is formed by multiplying the corresponding terms of sequences 2, 5, 8, ... and 5, 8, 11, ... both of which are A.P.s.Now nth term of first A.P. = 2 +(n -1)3 = 3n -1and nth term of second A.P. = 5 +(n -1)3 = 3n +2Hence nth term of the given series,    Tn = (3 n -1)(3 n +2) = 9 n² +3 n -2Thus the required sum,    Sn = Tn = (9 n² +3 n -2) = 9 n² +3 n -2 n          = 9 . n (n +1)(2n +1)/6 +3n (n +1)/2 -2 n    = n (6 n² +12 n +2)/2 = n (3 n² +6 n +1)/2

Example

Find the sum of first n terms of the series 1 +(1 +2) +(1 +2 +3) +.........

Solution

nth term of the given series,Tn = 1 +2 +3 +... up to n = n (n +1)/2 = n²/2 +n/2Sum of first n terms of the given series

     =[ n (n +1)/12](2n +1 +3) = n(n +1)(n +2)/6

Example

Find the nth term of series 1 +3 +7 +13 +..., and hence find the sum of first n terms.

Solution

Here the differences of successive terms are 2, 4, 6, ... which form an A.P.Let Tn denote the nth term and Sn denote the sum of first n terms.Then Sn = 1 +3 +7 +13 +... +Tn -1 +Tn

Also Sn = 1 +1 +3 +17 +............ +Tn -1 +Tn

          0 = 1 +2 +4 +6 +... upto n terms -Tn

        Tn = 1 +(2 +4 +6 +... upto n-1 terms)             = 1 + (n -1)[2. 2 +(n -2)2]/2 = n² -n +1        Sn = n² - n +n              = n (n +1)(2 n +1)/6 n (n +1)/2 +n              = n [2 n² +3 n +1 -3 (n +1) +6]/6             = n (2n² +4)/6 = n (n²+2)/3

Example

Find the n th term, sum of n terms, and sum to infinity of the series                  1/2.5 +1/5.8 +1/8.11+ ...

Solution

Tn = 1/ [(n th terms of 2, 5, 8,....)(n th term of 5, 8, 11,...)]       = 1/[(2 +(n -1) 3)(5 +(n -1) 3)]    = 1/ [(3n -1)(3 n +2)]

      = Putting n = 1, 2, 3, ..., n, we get

    T1 =

     T2 = .....................

Tn =

Adding, Sn =

Also sum to infinity =

                   =

Exercise

1. Sum to infinity the following series:(i) 1 +3/2 +5/2² + 7/2³ +(ii) 3 +5/4 + 7/4² + ....(iii) 1 +4 x² +7 x4 + 10 x6 +..., |x| < 1

2. Sum to n terms the following series:(i) 2. 1 +3. 2 +4. 4 +5. 8 +...(ii) 1. 2 +2. 2² +3. 2³ +4. 24 +...

3. Sum to n terms the following series:(i) 1 +(1 +3) +(1 +3 +5) +...(ii) 1 + (1 +2)/2 +(1 +2 +3)/3 +...(iii) 2² +5² +8² +...(iv) 1. 3 +3. 5 +5. 7 +...

4. (i) Find the sum of series 2² +5² +8² +... upto 15 terms.(ii) Find the sum of squares of first 100 odd natural numbers.(iii) Find the sum of cubes of first 100 odd natural numbers.(iv) Find the sum of the products of first n natural numbers taken two at a time.

5. Sum to n terms the following series:(i) 1 +5 +14 +30 +55 +...

(ii) 3 +7 +14 +24 +37 +...(iii) 6 +9 +16 +27 +42 +...

(iv)

(v) (vi) 1³ +(1³+2³)/2 +(1³ +2³ +3³)/3 +...(vii) 4 +44 +444 +...(viii) 0·5 +0·55 +0·555 +...

6. Sum the following series to n terms 5 +7 +13 +31 +85 +... 7. Prove that sum of cubes of any number of consecutive natural numbers is always

divisible by the sum of these numbers.[Hint. Let the m consecutive natural numbers be n +1, n +2, ..., n +m]

Answers

1. (i) 6             (ii) 44/9                     (iii) 1/(1 -x²) +3x²/(1 -x²)²2. (i) n. 2n                                         (ii) 2 +(n -1) 2 n +1

3. (i) n (n +1)(2n +1)/6                      (ii) n (n +3)/4    (iii) n (6 n² +3 n -1)/2                    (iv) n (4 n² +6 n -1)/34. (i) 10455     (ii) 1333300               (iii) 199990000     (iv) (n -1)n(n +1)(3n +2)/24           [Hint. Consider (1 +2 +... + n)²]5. (i) n (n +1)² (n +2)/12                   (ii) n(n² +n +4)/2     (iii) n(4 n² -3 n +35)/6                   (iv) 2 n -2 + 1/2n-1

    (v)    (vi) n (n +1)(n +2)(3n +5)/48   (vii) 4(10n +1 -9n -10)/81               (viii) 5(9n -1 +1/10n)/816. (3n +8 n -1)/2

Solving Quadratic Equations by Factorisation

This method uses the fact that ab = 0 => a = 0 or b = 0.Thus x² -3x +2 = 0 => (x -1)(x -2) = 0 => x -1 = 0 or x -2 = 0=> x = 1 or x = 2. Thus x = 1, x = 2 are roots of the equation x² -3x +2 = 0.

Illustrative Examples

Example

Solve for x: (x +3)(x -3) = 40.

Solution

(x +3)(x -3) = 40 => x² -9 = 40=> x² -49 = 0 => (x -7)(x +7) = 0=> x -7 = 0 or x +7 = 0 => x = 7 or x = -7.Hence the roots of the given equation are 7, -7.

Example

Solve(x -3)/(x +3) + (x +3)/(x -3) = 5/2

Solution

Given (x -3)/(x +3) + (x +3)/(x -3) = 5/2To clear the fractions, multiply both sides by L.C.M. of fractions i.e. by 2(x +3)(x -3) to get2(x -3)² +2(x +3)² = 5(x -3)(x +3)=> 2(x² -6x +9) +2(x² +6x +9) = 5(x² -9)=> 2x² -12x +18 +2x² +12x +18 -5x² +45 = 0=> -x² +81 = 0 => x² -81 = 0 => (x -9)(x +9) = 0

=> x -9 = 0 or x +9 = 0Hence the values x = 9, x = -9 satisfy the given equation.

Exercise

Solve the following equations by factorisation:1. 4 x² = 3 x 2. x (2 x +1) = 6 3. 21 x² -8 x -4 = 0 4. 3 x² -5 x -12 = 0 5. (x² -5 x)/2 = 0 6. 2x²/3 -x/3 = 1 7. (x -4)² +5² = 13² 8. 3 (x -2)² = 147 9. x² -(p +q) x +p q = 0 10. x + 1/x = 41/20 11. (x +2)/(x + 3) = (2 x -3)/(3x -7) 12. 8/(x +3) - 3/(2 -x) = 2

Answers

1. 0, 3/4            2. -2, 3/2         3. 2/3, -2/74. 3, -4/3          5. 0, 5              6. 1, 3/27. -8, 16           8. -5, 9            9. p, q10. 4/5, 5/4      11. -1, 5          12. -1/2, 5

Solving Quadratic Equations by Formula

The roots of the equation a x² +b x +c = 0, a 0 are                     [-b + (b² -4ac)]/2a, [-b - (b² -4ac)]/2aThus a quadratic equation has exactly two roots: equal or distinct; real or imaginary.

Remarks

1. If a quadratic equation in a variable x is satisfied by more than two values of x, then it will be satisfied by all values of x and hence it is an identity.

2. If the coefficients of quadratic are rational, and one of the roots is irrational, then the other root is its conjugate.

3. If the quadratic equation has real coefficients, then if the roots are complex numbers, they occur in conjugate pair.

Illustrative Examples

Example

Solve: 22x +8 +1 = 32. 2x.

Solution

Given 22x +8+1 = 32. 2x =>  28.22x +1 = 32. 2x

=>  256. (2x)² -32. 2x +1 = 0Putting 2x = t, we get, 256 t² -32t +1 = 0=>   t = [-(-32) + (- 32)² -4 (256)(1)]/[2 (256)] =   (32 ± 0)/512 = 1/16, 1/16But t = 2x. Hence 2x = t = 1, 1 = 2-4, 2-4

=>   x = -4, -4Hence the solution of the given equation is x = -4

Example

Solve (x +1)(x +2)(x +3)(x +4) = 120.

Solution

If the problem has four factors (x +a)(x +b)(x +c)(x +d), and the sum of two of the numbers a, b, c, d is equal to the sum of the other two, then we make the substitution as under:

       (x +1)(x +2)(x +3)(x +4) = 120=> [(x +1)(x +4)] [(x +2)(x +3)] = 120=> (x² +5 x +4)(x² +5 x +6) = 120Putting y = x² +5 x, we get(y +4)(y +6) = 120       => y² +10 y +24 = 120=>  y² +10 y -96 = 0   =>   y = [- 10 ± [(10)² -4 (1)(-96)]]/[2(1)]=>  y = (-10 ± 484)/2 = (- 10 ± 22)/2 = 6, -16When y = 6, we get x² +5 x = 6=> x² +5 x -6 = 0 => (x -1)(x +6) = 0=> x = 1, -6.When y = -16, we get x² +5x = -16 => x² +5x +16 = 0=>   x = [-5 ± [(5)² -4 (1)(16)]]/[2 (1)] = (-5 ± -39)/2 = (-5 ± i 39)/2Hence the required solutions are 1, -6, (-5 ±i 39)/2

Example

i. Evaluate

ii. If 3 = , find x.

iii. Evaluate the continued fraction

Solution

i. Let x = Squaring both sides, we get

x² = 6 +    = 6 +x=> x² -x -6 = 0 => (x -3)(x +2) = 0 => x = 3, -2As the value of given expression should be positive, we ignore x = -2. Hence the value of the given expression is 3.

ii. We are given that 3 = Squaring both sides, we get

9 = x + = x +3=> x = 6

iii. Let x = Then x = 2 +1/x  =>  x² = 2 x +1   =>  x² -2 x -1 = 0=>  x  = [-(-2) ± (-2)² -4 (1)(-1)]/[2(1)] = (2 ± 8)/2            = (2 ± 2 2)/2 = 1 ± 2But the given expression is positive. Hence the value of the given continued fraction is 1 ± 2.

Example

Solve the equation (2x +7) + (3x -18) = (7x +1).

Solution

Squaring both sides of the given equation, we get2x +7 +3x -18 +2 (2x +7). (3x -18) = 7 x +1=> 2 (2x +7). (3x -18) = 2 x +12=>   (2x +7). (3x -18) = x +6Squaring, (2x +7)(3x -18) = (x +6)²=> 6x² -15x -126 = x² +12 x +36

=> 5x² - 27x -162 = 0=> (5x +18)(x -9) = 0 => x = -18/5, 9For x = -18/5, (2x +7) is meaningless. Hence the only solution is x = 9.

Example

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence find the three numbers.

Solution

Since the middle number of the three consecutive numbers is x, the other two numbers are x -1 and x +1.According to the given condition, we have    x² = [(x +1)² -(x -1)²] +60=> x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60=> x² -4x -60 = 0 => (x -10)(x +6) = 0=> x = 10 or -6Since x is a natural number, we get x = 10Hence the three numbers are 9, 10, 11

Example

Determine all values of x for which

Solution

We know that ab = 1 if b = 0, a 0 or if a = 1

Given that Now x² -2x -48 = 0 => (x -8)(x +6) = 0 => x = 8, -6Also for x = 8 or -6, base = x² -5 x +5 0x² -5x +5 = 1 => x² -5x +4 = 0=> (x -1)(x -4) = 0 => x = 1, 4Hence the solutions of the given equation are x = -6, 1, 4, 8

Exercise

Solve the following equations (1-9):1. (i) x4 -8x² -9 = 0

(ii) x4 -5x² +6 = 0 2. (i) 3x +3-x = 2

(ii) 2x +1 +4x = 8 3. (i) 5x +1 +52 +x = 5³ +1

(ii) 4x -3x+1/2 = 3x+1/2 -22x -1 4. x2/3 -x1/3 -2 = 0 5. (x² +3x +2)² -8(x² +3x) -4 = 0 6. 1/(x + 1) - 2/(x + 2) = 3(x + 3) - 4/(x + 4) 7. x(x +2)(x +3)(x +5) = 72

8.

9.

10. Evaluate (i)

              (ii)

11. Solve 5 =

12. Evaluate the continued fraction 13. Solve (i) x +2 = (2x +7)

         (ii) (x +15) = x +3 14. Solve (x +5) + (x +21) = (6x +40) 15. The hypotenuse of a right angled triangle is 17 cm and the difference between the

other two sides is 7 cm. Find the two unknown sides. 16. A trader bought a number of articles for Rs 1200. Ten were damaged and he sold

each of the rest at Rs 2 more than what he paid for, thus clearing a profit of Rs 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.

17. Two years ago, a mans age was three times the square of his sons age. In three years time, his age will be four times his sons age. Find their present ages.

18. The sum of two numbers is 9 and the sum of their squares is 41. Taking one number as x, form an equation in x and solve it to find the numbers.

19. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that area of the walk is 120 square meters, find its width.

20. An express train makes a run of 240 km at a certain speed. Another train, whose speed is 12 km/hr less takes an hour longer to make the same trip. Find the speed of the express train.

21. A certain positive integer exceeds its square root by 20. Find the number. 22. Divide 39 into 2 parts whose product is 338. 23. A two digit number is four times the sum and three times the product of its digits.

Find the number. 24. The sum of the squares of two consecutive positive even integers is 340. Find the

integers.

Answers

1. (i) ± 3, ±i     (ii) ± 3, ± 2       2. (i)  0     (ii) 13. (i) -1, 2       (ii) 3/2                 4. -1, 85 . -4, -3, 0, 1                             6. 0, -5/27. 1, -6, (-5 +i 23)/2              8. 9/13, 4/139. -5/16                                      10. (i) (1 + 13)/2    (ii) 211. 20                                        12.  2 + 513. (i) 1   (ii) 1                            14. 415. 8 cm, 15 cm

16. (x -10) = 1200 +60; x = 10017. 29, 5 yrs18. x² +(9 -x)² = 41; the numbers are 4, 519. 2 meters                                20. 60 km/hr21. 25                                         22. 26, 1323. 24                                         24. 12, 14

Nature of roots of a Quadratic Equation

Discriminant

      = b² -4acCase I. When a, b, c are real numbers, a 0:

1. If = b² -4 a c = 0, then roots are equal (and real). 2. If = b² -4 a c > 0, then roots are real and unequal. 3. If = b² -4 a c < 0, then roots are complex. It is easy to see that roots are a pair of

complex conjugates.

Case II. When a, b, c are rational numbers, a 0:1. If = b² -4 a c = 0, then roots are rational and equal. 2. If = b² -4 a c > 0, and is a perfect square of a rational number, then roots are

rational and unequal. 3. If = b² -4 a c > 0 but is not a square of rational number, then roots are irrational

and unequal. They form a pair of irrational conjugates p + q, p - q where p, q Q, q> 0.

4. If = b² -4 a c < 0, then roots are a pair of complex conjugates.

Illustrative Examples

Example

Discuss the nature of the roots of the following equations:(i) 4 x² -12 x +9 = 0(ii) 3 x² -10 x +3 = 0(iii) 9 x² -2 = 0(iv) x² +x +1 = 0

Solution

i. Here coefficients are rational, and discriminant    = b² -4 a c = (-12)² -4 (4)(9) = 144 -144 = 0.Hence the roots are rational and equal.

ii. Here coefficients are rational, and discriminant      = b² -4 a c = (-10)² -4 (3)(3) = 100 -36 = 64.Now = 64 > 0, and 64 is a perfect square of a rational number.Hence the roots are rational and unequal.

iii. Here coefficients are rational, and discriminant       = b² -4 a c = (0)² -4 (9)(-2) = 72.Now = 72 > 0 but is not a perfect square of a rational number.Hence the roots are irrational and unequal.

iv. Here coefficients are rational, and discriminant       = b² -4 a c = (1)² -4 (1)(1) = -3 < 0.Hence the roots are a pair of complex conjugates.

Example

Discuss the nature of the roots of the equation    (m +6) x² +(m +6) x +2 = 0

Solution

Discriminant = (m +6)² -4 (m +6)(2)       = m² +12 m +36 -8 m -48       = m² +4 m -12 = (m +6)(m -2)

i. Roots are real and equal if  = (m +6)(m -2) = 0 i.e. if m = 2(Ignoring m= -6, as then equation becomes 2=0)

ii. Roots are real and unequal when  = (m +6)(m -2) > 0 i.e. when m < -6 or when m > 2

iii. Roots are a pair of complex conjugates when  = (m +6)(m -2) < 0 i.e. when -6 < m < 2

Exercise

1. Find the nature of roots of the following equations without solving them:(i) x² +9 = 0(ii) 4 x² -24 x +35 = 0(iii) x² -2 2 x +1 = 0(iv) 2 x² -2 5 x +3 = 0

2. Show that roots of the equation (x -a)(x -b) = a b x² where a, b R are always real. When are they equal?[Hint. = (a -b)² +(2 a b)²].

3. Show that the roots of the equation (x -a)(x -b) +(x -b)(x -c) +(x -c)(x -a) = 0, where a, b, c R are always real. Find the condition that the roots may be equal. What are the roots when this condition is satisfied?[Hint. = 2 ((a -b)² +(b -c)² +(c -a)²)]

4. Discuss the nature of roots of the following equations:(i) 3 x² -2 x - 3 = 0        (ii) x² -(p +1) x +p = 0(iii) (x -a)(x -b) = a b.It is given that p Q, and a, b R.

5. Find m so that roots of the equation (4 +m) x² +(m +1) x +1 = 0 may be equal. 6. Show that the roots of the equation x² +2 (3 a +5) x +2 (9 a² +25) = 0 are complex

unless a = 5/3.

7. If a, b, c, d R show that the roots of the equation (a² +c²) x² +2 (a b +c d) x + (b² +d²) = 0 cannot be real unless they are equal.

8. Determine a positive real value of k such that both the equations x² +k x +64 = 0 and x² -8 x +k = 0 may have real roots.

Answers

1. (i) pair of complex conjugates  (ii) rational and unequal    (iii) real and unequal     (iv) pair of complex conjugates2. Roots are real and equal when a = b = 03. Roots are equal when a = b = c. Then roots are a, a4. (i) real and distinct   (ii) rational and distinct when p 1; when p = 1, roots are rational and equal   (iii) real and distinct when a +b 0; when a +b = 0, roots are both 0.5. 5, -38. k = 16

Relations between Roots and Coefficients

Formulae

Sum and product of rootsSum of roots = + = - b/a = - (coefficient of x)/(coefficient of x²)Product of roots = = c/a = (constant term)/(coefficient of x²)

Symmetric functions of roots

From + = -b/a and = c/a, values of other functions of and can be calculated:( - )² = ( + )² -4 ,

² + ² = ( + )² -2 ,³ + ³ = ( + )³ -3 ( + ),4 + 4 = [( + )² -2 ]² -2 ( )²,² - ² = ( + )( - ) etc.

Formation of a quadratic equation with given roots

x² -S x +P = 0, where S = sum of roots and P = product of roots.

Illustrative Examples

Example

If , are roots of the equation x² -4 x +2 = 0, find the values ofi. ² + ² ii. ² - ² iii. ³ + ³ iv. 1/ + 1/

Solution

As , are roots of the equation x² -4 x +2 = 0,      + = -(-4)/1 = 4, = 2/1  = 2

i. ² + ² = ( + )² -2 = (4)² -2 (2) = 12 ii. ² - ² = ( + )( - ).

    Now ( - )² = ( + )² -4 = (4)² -4 (2) = 8     - = ± 8 = ±2 2     ² - ² = ( + )( - ) = (4)(±2) = ±8

iii. ³ + ³ = ( + )³ -3 ( + )    = (4)³ -3 (2)(4) = 64 -24    = 40

iv. 1/ + 1/ = ( + )/ = 4/2 =2

Example

If , are roots of the quadratic equation ax² +bx +c = 0, form an equation whose roots arei. - , - ii. 1/ , 1/

Solution

Since , are roots of a x 2 +b x +c = 0,                + = -b/a, = c/a

i. Here, S = (- ) +(- ) = -( + ) = -(-b)/a = b/aP = (- )(- ) = = c/aHence the required equation is x² -S x +P = 0i.e. x² - (b/a) x + c/a = 0 i.e. a x² -b x +c = 0

ii. Here, S = 1/ + 1/ = ( + )/ = - (b / a)/(c / a) = - b/cP = (1/ ).(1/ ) = 1/( ) = 1/(c/a) = a/cHence the required equation is x² -S x +P = 0i.e. x² -(- b/c) x + a/c = 0 i.e. cx² +bx +a = 0

Example

Find p, q if p and q are roots of the equation x² +px +q = 0.

Solution

Since p, q are roots of x² +px +q = 0, p +q = -p and pq = q.Now pq = q => pq -q = 0 => q(p -1) = 0 => q = 0 or p = 1.When q = 0 then p +q = -p => 2p = -q = 0 => p = 0.When p = 1, then p +q = -p => q = -2 p = -2.Hence the required solutions are p = q = 0 or p = 1, q = -2.

Example

If the roots of the equation 2 x² +(k +1) x +(k² -5 k +6) = 0 are of opposite signs then show that 2 < k < 3.

Solution

Since roots are of opposite signs, roots are real and distinct,    discriminant> 0 and product of roots < 0(k +1)² -4. 2. (k² -5 k +6) > 0 and (k² -5 k +6)/2 < 0First condition is always true when second holds.  (because (k +1)² = 0)Hence (k² -5 k +6)/2 < 0    =>   k² -5 k +6 < 0    =>   (k -2)(k -3) < 0=>   2 < k < 3

Example

The coefficient of x in the equation x² +p x +q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation.

Solution

By given conditions, -2 and -15 are roots of the equation x² +17 x +q = 0Hence product of roots = (-2)(-15) = q/1 => q = 30.Therefore, the original equation is x² +13 x +30 = 0=>   (x +10)(x +3) = 0      =>   x = -3, -10Hence the roots of the original equation are -3, -10.

Exercise

1. Write the sum and product of the roots of the following equations:(i) x² +9 = 0(ii) 2 x² - 2 x +4 = 0.

2. If  , are the roots of the equation 3 x² -6 x +4 = 0, evaluate

3. If the roots of the equation x² +p x +7 = 0 are denoted by and , and ² + ² = 22, find the possible values of p.

4. If , be the roots of the equation a x² +b x +c = 0, find the value of(i) ² + ²(ii) / + /(iii) ³ + ³(iv) | - (v) ² - ²(vi) 6 + 6

5. If , are roots of x² +k x +12 = 0 and - = 1, find k. 6. The sum of the roots of the equation 1/(x+a) + 1/(x+b) = 1/c is zero.

Prove that the product of the roots is - (a² +b²)/2[Hint. Rewrite the equation as x² +x (a +b -2 c) +(ab -bc -ca) = 0.]

7. Two candidates attempt to solve a quadratic equation of the form x² +p x +q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and -9. Find the correct roots and the equation.

8. Given that and are the roots of the equation x² = 7 x +4,(i) Show that ³ = 53 +28(ii) Find the value of / + /

9. Form an equation whose roots are(i) 3, -1/3(ii) 3, 1/ 3(iii) 0, 0

10. Form an equation with rational coefficients one of whose roots is( 3 +1)/( 3 -1)

11. Form an equation with real coefficients one of whose roots is(i) -1 -2 i(ii) -2 - -3(iii) 1/(2 + -2)

12. If , are the roots of the equation a x² +b x +c = 0, then form an equation whose roots are:(i) 2 , 2 (ii) /2, /2(iii) / , /(iv) ( + )², ( - )²(v) ( / ) +1, ( / ) +1(vi) +k, +k

13. If , be the roots of the equation 2 x² -3 x +1 = 0, find an equation whose roots are /(2 +3), /(2 +3)

Answers

1. (i) 0, 9        (ii) 1, 2 2    2. 8                 3. ±64. (i) (b² -2ac)/a²     (ii) (b² -2ac)/ac      (iii) (3a b c -b³)/a³ (iv) [ (b² -2ac)]/|a|    (v) ± [b (b² -2ac)]/a²               (vi) [b6 +9 a² b² c² -6ab4 c -2c³ a³]/a6

5. ±7                 7. -3, -4; x² +7 x +12 = 0    8. (ii) -57/49.(i) 3 x² -8 x -3 = 0 (ii) 3x² -4x + 3 = 0  (iii) x² = 010. x² -4 x +1 = 011. (i) x² +2 x +5 = 0 (ii) x² +4 x +7 = 0 (iii) 6 x² -4 x +1 = 012. (i) a x² +2 b x +4 c = 0     (ii) 4 a x² +2 b x +c = 0     (iii) a c x² -(b² -2 a c) x +a c = 0     (iv) a 4 x² -2 a² (b² -2 a c) x +b²(b² -4 ac) = 0     (v) a c x² +b (a +c) x +(c + a)² = 0     (vi) a x² -(2 a k -b) x +(a k² -b k +c) = 013. 40 x² -14 x +1 = 0

Linear Inequations

In general a linear inequation can always be written as   a x +b < 0, a x +b 0, a x +b > 0 or a x +b 0,where a and b are real numbers, a 0.

Replacement set

The set from which values of the variable (involved in the inequation) are chosen is called the replacement set.

Solution set

A solution to an inequation is a number (chosen from replacement set) which, when substituted for the variable, makes the inequation true. The set of all solutions of an inequation is called the solution set of the inequation.For example, consider the inequation x < 4Replacement set                             Solution set(i) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}      {1, 2, 3}(ii) {-1, 0, 1, 2, 5, 8}                      {-1, 0, 1, 2}(iii) {-5, 10}                                   {-5}(iv) {5, 6, 7, 8, 9, 10}                     Note that the solution set depends upon the replacement set.

Remark

Always reverse the symbol of an inequation when multiplying or dividing by a negative number.

Illustrative Examples

Example

Find the solution set of 2 ;< 3 (x - 2) +5 < 8, x W. Also represent its solution on the number line.

Solution

Given 2 3 (x -2) +5 < 8=>  2 3 x -6 + 5 < 8=>  2 3 x -1 < 8=>  2 +1 3 x -1 +1 < 8 +1      [Add 1]=>  3 3 x < 9=>  1 x < 3                             [Divide by 3]But x W i.e. x {0, 1, 2, 3 , 4, ...},Hence the solution set is {1, 2}The solution set is shown by thick dots on the number line.

      

Example

John needs a minimum of 360 marks in four tests in a Mathematics course to obtain an A grade. On his first three tests, he scored 88, 96, 79 marks. What should his score be in the fourth test so that he can make an A grade?

Solution

Let John score x marks in the fourth test. Then the sum of Johns test scores should be greater than or equal to 360 i.e.88 +96 +79 +x 360   => 263 + x 360=> x 360 +(-263)    =>    x 97John should score 97 or greater than 97 in the fourth test to obtain A grade.

Exercise

1. Solve the inequation 3 x -11 < 3 where x {1, 2, 3, ..., 10}. 2. Solve 2 (x -3) < 1, x {1, 2, 3, ..., 10}. 3. Solve 5 -4 x > 2 -3x, x W. 4. List the solution set of 30 -4(2x -1) < 30, given that x is a positive integer. 5. Solve 2 (x -2) < 3x -2, x {-3, -2, -1, 0, 1, 2, 3}. 6. If x is a negative integer, find the solution set of

     2/3 + (x +1)/3 > 0.

7. Solve (2 x -3)/4, x { 0, 1, 2, ..., 8 }. 8. Solve x -3 (2 +x) > 2 (3 x -1), x {-3, -2, -1, 0, 1, 2 }. 9. Given x {1, 2, 3, 4, 5, 6, 7, 9}, solve x -3 < 2x -1. 10. Given A = {x ; x I, -4 x < 4}, solve 2x - 3 < 3 where x has the domain A. 11. List the solution set of the inequation

    (1/2) +8 x > 5x - 3/2, x Z 12. List the solution set of (11 -2x)/5 (9 -3x)/8 + 3/4, x N 13. Find the values of x, which satisfy the inequation

  -2 1/2 - 2 x/3 11/6, x N 14. If x W, find the solution set of

    3 x/5 - (2 x -1)/3 > 1Also graph the solution set on the number line, if possible.

15. Given x {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2 x -1 < x +4. 16. Solve 1 15 -7x > 2x -27, x N. 17. If x Z, solve 2 +4x < 2x -5 < 3x. 18. Solve the inequality 2 x -10 < 3 x -15. 19. Find the solution set of the inequation x +5 = 2 x + 3, x R.

Answers

1. {1, 2, 3,4}           2. {1, 2, 3}               3. {0, 1, 2}4. {1, 2, 3, 4, ...}      5. {-1, 0, 1, 2, 3}     6. {-2, -1}7. {3, 4, 5, 6, 7, 8}   8. {-3, -2, -1}          9. {1, 2, 3, 4, 5, 6, 7, 9}10. {-4, -3, -2, -1, 0, 1, 2}                       11. {0, 1, 2, 3, ...}12. {1, 2, 3, ..., 13}   13. {1, 2, 3}            14. ; not possible15. {1, 2, 3, 4}         16. {2, 3, 4}            17. {-5, -4}18. {x ; x R, x > 5}  19. {x ; x R, x 2}

Solution of Simultaneous Linear Equations

A statement of any one of the following types:(i) a x +b y +c < 0           (ii) a x +b y +c 0(iii) a x +b y +c> 0          (iv) a x +b y +c 0where a, b, c are real numbers and at least one of a and b is non-zero, is called a linear inequation (or inequality) in two variables x and y.The set of all ordered pairs of real numbers which satisfy a given inequation is called the solution set of the given inequation.

Illustrative Examples

Example

Solve the inequations 3 x +2 y > 5 and x +y 1 simultaneously.

               

Solution

The given inequations are                3 x +2 y > 5           ... (i)         and x +y   1                ...(ii)To draw the graph of 3x +2y > 5Draw the straight line 3 x +2 y = 5 which passes through the points (5/3,0) and (0,5/2)The line divides the plane into two parts.Further, as O (0, 0) does not satisfy the inequation 3 x +2 y > 5 (as 3.0 +2.0 = 0 < 5),

therefore, the graph of (i) consists of that part of the plane divided by the line 3 x +2 y = 5 which does not contain the origin.Similarly, draw the graph of the inequation x +y 1.Shade the common part of the graphs of both the given inequations (i) and (ii).The solution set of the given inequations consists of all points in the shaded part of the co-ordinate plane shown in fig. The points on the part EC of the line DC are included in the solution.

Example

Solve the following inequations simultaneously:          3 y -2 x < 4, x +3 y > 3 and x +y 5.

            

Solution

The given inequations are               3 y -2 x < 4 ...(i)                  x +3 y > 3 ...(ii)            and x +y 5   ...(iii)To draw the graph of 3 y -2 x < 4.Draw the straight line 3 y -2 x = 4 which passes through the points (-2, 0) and (0,4/3). The line divides the plane into two parts. Further as O (0, 0) satisfies the inequation 3 y -2 x < 4 (as 3.0 -2.0 = 0 < 4), therefore, the graph consists of that part of the plane divided by the line 3 y -2 x = 4 which contains the origin.Similarly, draw the graphs of other two inequations x +3y > 3 and x +y 5.Shade the common part of the graphs of all the three given inequations (i), (ii) and (iii).The solution set consists of all the points in the shaded part of the co-ordinate plane shown in fig. The points on the line segment BC are included in the solution.

Exercise

Solve the following systems of linear inequations simultaneously:1. 3 x +2 y > 5 and y > 2. 2. 3 x +2 y 6 and x +2 y > 4. 3. 2 x +3 y> 12, x = 2 and y 1. 4. x -2 y +11> 0, 2 x -3 y +18 0 and y 0. 5. x +2 y 8, x -y 2, x > 0 and y > 0. 6. x +2 y 0, x +y 4, x > 0 and y < 2.

[Hint. The line x +2 y = 0 passes through (0, 0). Take any other point, say (1, 1). Since 1 +2. 1 = 3 > 0, so it satisfies the inequality x +2 y = 0 and hence the graph of x +2 y = 0 consists of that part of the plane divided by the line x +2 y = 0 in which the point (1, 1) lies.]

Answers

In each problem, the solution set consists of all points in the shaded part of the co-ordinate plane in their respective diagrams.

1.             2.

3.                4.

5.                  6.

Quadratic Inequations

Inequations of the type a x² +b x +c > 0, a x² + b x +c 0, a x² +b x +c < 0, a x² +b x +c 0 (a0) are called quadratic inequalities.

Solving Quadratic Inequalities

First make the coefficient of x² unity by multiplying throughout by 1/a, and reversing the inequality sign if a is negative. Now, three cases arise:Case I. If the quadratic has real distinct roots, say , and < , then

1. (x - )(x - ) > 0     => x < or x > 2. (x - (x - ) 0 => x or x 3. (x - )(x - ) < 0   =>  < x < 4. (x - )(x - ) 0 => x

Case II. If the quadratic has two real, identical roots say , , then we know that x² 0 for real x. Therefore

1. (x - )²  0    =>    x R (all real numbers) 2. (x - )² > 0    =>    x  R except x = 3. (x - )² 0   =>    x = 4. (x - )² < 0      =>   there is no solution.

Case III. If the quadratic has complex roots, then either there is no solution or the entire set of real numbers is the solution.

Sign of Quadratic expressions

1. a x² +b x +c has same sign as a except when a x² +b x +c = 0 has two real and distinct roots, say , ( < ) and < x < .

2. a x² +b x +c is +ve for all real x, when a > 0 and < 0. 3. a x² +b x +c is -ve for all real x, when a < 0 and < 0.

Range (Maximum / Minimum Value) of Quadratic Expressions

Hence the maxima / minima occurs at x = -b/2a and is equal to - /4a. If a > 0, we have a minima, otherwise we have a maxima.

Illustrative Examples

Example

Solve algebraically:i. 2 x² +x -1 < 0 ii. x² +2x +3 0 iii. x² +x +1 < 0

Solution

i. 2 x² +x -1 < 0    =>    (x +1)(2 x -1) < 0=>  (x +1)(x -1/2) < 0         (Dividing by 2)=> -1 < x < 1/2, which is the required solution.We can also write it as x (-1, 1/2)

ii. For x² +2 x +3 0, the discriminant= 2² -4.1.3 = -8 < 0, so roots are complex.

Now x² +2 x +3 = (x +1)² +2 2 for all real x.Hence x² +2 x +3 0 is true for all real x.

iii. For x² +x +1 < 0, discriminant= 1² -4.1.1 = -3 < 0, so roots are complex.

Now for all real x.So x² +x +1 < 0 has no solution.

Exercise

1. Discuss the sign of x² -3 x +2. 2. Solve the following inequalities:

(i) -x² +3 x -2 > 0            (ii) 4 x² -9 0(iii) 2 x² +x -15  0.

3. Solve the following inequalities:(i) 2 x² < x                 (ii) 3 -2 x²> 5 x(iii) x² -6 x +19 0     (iv) x² -6 x +11 < 1.

4. Find all real values of x which satisfy x² -3 x +2 > 0 and x² -3 x -4 0. 5. Find the range of x for which 6 +x < 2 x². 6. If x is real, prove that 5 x² -8 x +6 is always positive and find its minimum value. 7. If x be real, find the maximum value of a² +2 a x - x². 8. Find the values of a so that expression x² -(a +2) x +4 is always positive. 9. Construct a quadratic polynomial which is zero when x = 3, 5 and whose minimum

value is -1. How many such polynomials are possible? 10. Can you construct a quadratic polynomial which is zero when x = 3, 5 and whose

maximum value is(i) 2            (ii) -2?

11. Solve: log2 (x² -1) = 3. 12. Write the most general quadratic expression whose

(i) minimum value is K(ii) maximum value is K.

Answers

1. -ve when 1 < x < 2, zero at x = 1, 2; +ve otherwise.2. (i) (1, 2)      (ii) x > 3/2 or x < -3/2

    (iii)

3. (i)     (ii) -3 < x < 1/2       (iii) all real values of x    (v) no real value of x.4. -1 x < 1 or 2 < x 4.5. x < -3/2 or x > 2.                6. 14/57. 2a² at x = a.                        8. -6 < a < 2.9. x² -8 x +15; only one.10. (i) Yes, -2 x² +16 x -30    (ii) No.11. -3 x < -1 or 1 < x 3.12. (i) (a x +b)² +K                  (ii) K -(a x +b)²  

Method of Intervals

Consider the sign of the expressionf(x) = (x -a1)(x -a2) ... (x -an) where a1 < a2 < ... < an.

           

On the real number line, mark points a1, a2, ..., a n. Start with a positive sign for x > an; -ve for an -1 < x < an; and then alternating +ve and -ve signs. The logic is that when x> an; all factors x -a1, x -a2, ..., x -a n are +ve and hence f(x) is +ve; when an-1 < x < an, only one factor x -an is -ve, and so f(x) is -ve, and so on.

Quadratic fractions and their range

Expressions of the kind (a x² +b x +c)/(p x² +q x +r), where x is any real number, are called quadratic fractions.Their range can easily be found by putting       (a x² +b x +c)/(p x² +q x +r) = y       => a x² +b x +c = p y x² +q x y +r y=> x² (a -p y) +x (b -q y) +(c -r y) = 0Since the value of x is real, discriminant of this equation in x must be non-negative=> (b -q y)² -4 (a -p y)(c -r y) = 0.This inequality yields range of y.

Illustrative Examples

Example

Solve: x (x -1)(x -2)(x -3) > 0.

 

Solution

Mark points 0, 1, 2, 3 on real line. By method of intervals, we see that given expression is +ve when x > 3 or 1 < x < 2 or x < 0.Thus required solutions set is {x < 0; 1 < x < 2; x > 3}.

Example

Solve 1/x < 1.

Solution

Do not fall into this trap: (1/x) < 1    =>   x > 1,because we cant multiply by negative quantity without  changing the sign of inequality.Correct solution is:Since x² > 0 for all real x, x 0.  1/x < 1    =>  (1/x) x² < x²   =>  x < x²   =>  x² -x > 0=>   x (x -1) > 0  =>  x < 0 or x > 1, which is the required solution.

Example

For what values of a is the inequality (x² + a x -2)/(x² -x +1) < 2 satisfied for all real values of x?

Solution

(x² +a x -2)/(x² -x + 1) < 2=> (x² +a x -2)/(x² -x +1) -2 < 0

=>  (x² +a x -2 -2 x² +2 x -2)/(x² -x + 1) < 0=>    [-x² +x (a +2) -4]/(x² -x +1) < 0

=> -x² -x (a +2) -4 > 0   (as for all real x)=>  x² -x (a +2) +4 > 0Now the expression x² -x (a +2) +4 is positive for all real x if   < 0 (as coeff of x² = 1 > 0)=> [-(a +2)]² -4. 1. 4 < 0   => a² +4 a +4 -16 < 0=>  a² +4 a -12 < 0   =>  (a +6)(a -2) < 0=> -6 < a < 2, which is the required range for a.

Example

Show that for all real values of x, the expression (x² -2x +4)/(x² +2x +4) has greatest value 3 and least value 1/3.

Solution

Let (x² -2 x +4)/(x² +2 x +4) = y=>  x² -2 x +4 = x² y +2 x y +4 y=>  x² (1 -y) -2 x (1 +y) +4 (1 -y) = 0.Since x is real, discriminant of this quadratic equation in x must be non-negative=>  = b² -4 a c = 0=>   [-2 (1 +y)]² -4. (1 -y). 4 (1 -y) 0=>   4 [1 +y² +2 y -4 (1 +y² -2 y)] 0=> -3 y² +10 y -3 = 0    =>  3 y² -10 y +3 = 0=> (3 y -1)(y -3) ;  0    =>  1  y 3=> 1/3 (x² -2 x +4)/(x² +2 x +4) 3 for all real xThus the greatest value of given quadratic fraction is 3 and least value is 1/3.

Exercise

1. Solve: (i) x³ -3 x² -x +3 < 0      (ii) x4 -5 x² +4 0

2. Solve : 3. Find all real values of x which satisfy

(i) x³ (x -1)(x -2)² > 0(ii) x² (x -1)(x -2) 0

4. Solve : (x² +6 x -11)/(x + 3) < -1 5. Solve : (x² -3 x +240/(x² -3 x +3) < 4 6. Find the range of values of x for which (x² +x + 1)/(x +1) < 1/3, x being real. 7. Find the set of all x for which 2 x/(2 x² +5 x +2) > 1/(x+1)

8. Let y = Find all the real values of x for which y takes real values. 9. Find all integral values of x for which (5 x -1) < (x + 1)² < 7 x -3

10. Solve (i)      (ii) 11. If x is real, show that (x² -x +1)/(x² +x +1) takes values from 1/3 to 3. 12. If x be real, show that the value of (x -1)(x +3)]/[(x -2)(x +4)] cannot lie between 4/9

and 1. 13. If for real values of x, the greatest value of (x² -x +a)/(x² +x +a) is 3, then prove that

value of a cannot be less than 1. 14. Show the expression (2x² +4x +1)/(x² +4x +2) can have any real value if x is real. 15. Show that if x is real, the expression (x² -bc)/(2x -b -c) has no real values between b

and c. 16. Find the minimum and maximum values of (x² -1)/(x² +1). Is maximum value

obtained? 17. Let f(x) =( x² +6 x -8)/( +6 x -8x²). Find the interval of values of a for which f(x)

takes all real values for real values of x.

Answers

1. (i) x < -1 or 1 < x < 3         (ii) x -2 or -1 x 1 or x 22. x < -1 or x > 13. (i) x < 0 or x > 1, x 2        (ii) x = 0 or 1  x    24. x < -8 or -3 < x < 1           5. x < -1 or x > 46. -1 < x < -1/2                      7. -2 < x < -1 or -2/3 < x < 1/28. -1 x < 2 or x > 3            9. x = 310. (i) All real numbers            (ii) x -2 or x -116. -1, 1; No                         17. 2     14  

Representations of Complex Numbers

Geometric representation

The complex number z = x +iy can be uniquely represented by the point P(x, y) in the co-ordinate plane and conversely corresponding to the point P(x, y) in the plane there exists a unique complex number z = x +iy. The plane is called the complex plane and the representation of complex numbers as points in the plane is called Argand diagram.

Notice that length OP = x² +y² = |z|Also note that every real number x = x +0i is represented by point (x, 0) lying on x-axis, and every purely imaginary number iy is represented by point (0, y) lying on y-axis. Consequently, x-axis is called the real axis and y-axis is called the imaginary axis.

If z1 = x1 +iy1 and z2 = x2 +iy2 are two complex numbers, then the distance between two corresponding points P1(x1, y1) and P2(x2, y 2) is|P1P2| = [(x1 -x2)² +(y1 -y2)²] = |z1 -z2| = |z2 -z1|.

If P1(x1, y1) and P2(x 2, y2) are two points, then the point Q dividing [P1P2] in the ratio m

: n is given

Putting m = n = 1, the mid-point of z1 and z2 =

Trigonometric representation

z = x +iy = r cos +i r sin   = r (cos +i sin )This form of z is called trigonometric form or polar form. Thus if modulus of z is r and amp(z) = , then z = r (cos +i sin ) = r cis .Notice that if x = 0, then = /2; if y > 0 and = - /2 if y < 0.If x 0, then tan = r sin /r cos = y/x, so that     amp (z) = = tan-1(y/x)The unique value of q such that - < is called principal value of amplitude or argument.

For example, let z = 1 +i. Thenr = [(1)² + (1)²] = 2 and = tan-1 1 = /4.As another example, let r = 2 , = - /4Then z = r (cos +i sin )

  

 

Remark

Frequently we have to convert the complex number z = x +i y to its polar form z = r cis . Do the calculations as follows:    r = x² +y²To find :If x = 0 i.e. if z is purely imaginary, then             = /2 if y> 0, = - /2 if y < 0.If y = 0 i.e. if z is purely real, then            = 0 if x > 0, = if x < 0.

Otherwise, let be such that 0 < < /2Then = if x > 0, y > 0 (i.e. z is in first quadrant)        = - if x < 0, y > 0 (i.e. z is in second quadrant)        = + if x < 0, y < 0 (i.e. z is in third quadrant)        = - if x>0, y<0 (i.e. z is in fourth quadrant).Conversely, given z = r cis , convert it to standard form z = x +iy by using x = r cos , y = r sin .

Illustrative Examples

Example

If z is any complex number, show that -|z| Re(z) |z|. When do the equality signs hold?

Solution

Let z = x +iy = r cis = r (cos +i sin ) where r = |z| 0 and = amp(z).We know that -1 cos 1 for all => -r r cos r          (as r 0)=> -|z| Re(z) |z|.      (because Re(z) = r cos )Now -|z| = Re(z)  <=>     - [x²+y²] = x     <=>    y = 0 and x 0.Also Re(z) = |z|    <=>      x = [x²+y²]      <=>    y = 0 and x 0.Hence -|z| = Re(z) = |z|     <=> y = 0 and x = 0 => z = 0.

Example

Show that the area of the triangle on the Argand plane formed by the complex numbers z, iz and z +iz is |z|²/2

Solution

Note that the points O(0), P(z), R(z +iz) and Q(iz) form parallelogram OPRQ.

Also |OP| = |z| = |iz| = |OQ| and POQ = 90°Thus 0, z, z +iz, iz form a square of side |z|.Hence area of triangle with vertices z, iz and z +iz   = (1/2)|z|.|z| = (1/2)|z|²

           

Exercise

1. Represent the following complex numbers in polar form(i) 3 +i(ii) (iii) sin 90° +cos 90°(iv) tan -i(v) 1 +sin +i cos(vi) -1 -i

2. Represent the following complex numbers in standard form x +iy(i) cis 2 /3

(ii) 3. If z is a complex number, represent -z and -iz in complex plane. 4. Using distance formula, prove that the points 1, (-1 + 3i)/2 and (-1 - 3i)/2 form the

vertices of an equilateral triangle in complex plane. 5. Using distance formula or otherwise, prove that the points -2 +3i, -1 +2i and 2 +5i are

collinear.[Hint. Let the points A, B, C represent the complex numbers -2 +3i,-1 +2i and 2 +5i respectively. Show that AB +BC = AC.]

6. If z1, z2, z3, z4 are complex numbers, show that they are vertices of a parallelogram in the Argand diagram if and only if z1 +z3 = z2 +z4.[Hint. What is mid-point of AC? BD?]

Answers

1. (i) 2 cos /6   (ii) cis 0      (iii) cis 0

    (iv)        (v)     (vi) 2 cis 3 /42. (i)(1/2) + [( 3)/2] i                      (ii) 1 -i  

Locus / Regions in Argand Plane

Since any complex number z = x +iy corresponds to point (x,y) in complex plane (also called Argand plane), so many kinds of regions and geometric figures in this plane can be represented by complex equations or inequations. The equation of circle of radius r and center at origin is |z| = r.Note that for the circle |z -z0| = r,all points on the circumference are given by |z -z0| = r,all points within the circle are given by |z -z0| < r,while all points outside the circle are given by |z -z0| > r.

Illustrative Examples

Example

What is the equation of the circle in complex plane with radius 2 and center at 1 +i? Does the origin lie within this circle, on the circle or outside it? Does any real number lie on this circle?

         

Solution

Here center z0 = 1 +i, radius = 2.So the equation of the circle is |z -z0| = r,that is, |z -(1+i)| = 2Now, the distance of origin from center          (0 -1) ² +(0 -1)² = 2 < radius 2.Thus the origin lies within the circle.Now let the real number (a,0) lie on the circle.            So (a -1)² +(0 -1)² = 2=>  (a -1)² +1 = 4=>  (a -1)² = 3=>   a -1 = ± 3=>   a = 1 ± 3Now you try to find out if any purely imaginary number lies on this circle.

Example

(i) Interpret the loci arg z = /4 in complex plane.(ii) Interpret the loci arg (z -1) = /4 in complex plane.(iii) Represent |z +i| = |z -2| in Argand plane.(iv) How would you represent the line x -y = 0 in terms of complex number z?

Solution

(i) arg (z) = /4 represents a half ray as shown in the following diagram.

             Note that the point z = 0 is not included as arg z is not defined for z = 0.(ii) arg (z -1) = /4 represents a half ray as shown in following diagram.

             Note that the point z = 1 is not included.(iii) Let z = x +iy, then      |z +i| = |z -2|

=>  |x +iy +i| = |x +iy -2|=>  |x +(y +1)i| = |(x -2) +iy|=>  x² +(y +1)² = (x -2)² + y²=> 4x +2y = 3, which represents a line in Argand plane

           Note that intuitively, |z +i| = |z -2| represents all points equidistant from -i and 2 i.e. it represents the perpendicular bisector of join of -i and 2.(iv) The line x -y = 0 is shown in the following diagram.

        We note taht the points A(-1,i) and B(1,-i) are such that the line x -y = 0 is perpendicular bisector of AB.Hence required equation in terms of z is|z -(-1 +i)| = |z -(1 -i)|i.e. |z +1 -i| = |z -1 +i |.

Exercise

1. Write the equation or inequalities for the following:(i) a circle of radius 2 with center at origin.(ii) all points lying outside the circle of radius 2 and center at -1 -i.(iii) a circle with center at 1 +i and passing through origin.(iv) all points lying in first or fourth quadrant.(v) the y-axis.(vi) all points lying in first quadrant.

2. Which regions are given by following? Also indicate on complex plane.(i) 2 |z| 3(ii) |z -2 -3i| > 2(iii) Re(z) > 2(iv) Re(1/z) < 1/2(v) arg(z -1 -i) = /4(vi) |z -1| +|z +1| = 3(vii) |z -1|² +|z +1|² = 4(viii) |z +i| |z +2|(ix) |z -2i| = |z +2i|

3. Find the locus of a complex number z = x +iy satisfying the relation |z +i| = |z +2|. Illustrate the locus of z in the Argand plane.

4. Given z1 = 1 +2i. Determine the region in the complex plane represented by 1 < |z -z1| 3. Represent it with the help of an Argand diagram.

5. Find the locus of a complex number z = x +iy satisfying the relation arg(z -a) = /4, aR. Illustrate the locus of z in Argand diagram.

6. Find the locus of a complex number z = x +iy satisfying the relation . Illustrate the locus of z in the Argand diagram.

Answers

1. (i) |z| = 2                (ii) |z +1 +i|> 2   (iii) |z -1 -i| = 2   [ Hint. radius = distance between origin and 1 +i.]

   (iv) Re(z)> 0             (v) Re(z) = 0   (vi) Re(z)> 0 and Im(z)> 02. (i) it represents circular annulus lying between concentric circles of radii 2 and 3 centered at origin. All points on circumference of two circles are included. (You draw the diagram!)(ii) all points lying outside circle of radius 2 centered at 2 +3i.(iii) all points to the right of line x = 2.(iv) exterior of a circle of radius 1 with center (1,0).(v) portion of a line of inclination 45° passing through point 1 +i of the line x-y = 0.(vi) an ellipse with foci at (1,0) and (-1,0), and major axis of length 3.(vii) a circle with points (-1,0), (1,0) as end points of a diameter; in other words, a circle of radius 1 centered at origin.(viii) all points to the "left" of the line 4x-2y+3=0.(ix) The x-axis.3. The line 4x-2y+3=0, it is the perpendicular bisector of the join of the points -i and -2. (Please draw the diagram!)4. It represents circular annulus lying between concentric circles of radius 1 and 3 centered at (1,2). The region includes all points on the circumference of the outer circle but excludes all point on the circumference of the inner circle. (You draw the diagram!)5. It represents a portion of the line x-y-a=0. (Please draw the diagram!)6. It represents the circle x² +y² +12 y +4  = 0 with center at (0,-6) and radius = 4 2 units.  

Square roots of Complex numbers

Let z = a +i b, and let the square root of z be the complex number x +i y.Then (a +i b) = x +i ySquaring both sides, a +i b = (x +i y)² = (x² -y²) +i (2 x y).Equating real and imaginary parts, we get x² -y² = a, 2 x y = b.Solving these two equations for real x, y yields x + i y.There are two square roots of a complex number, of the form ± (x +i y).

Illustrative Examples

Example

Find square roots ofi. z = -3 +4 i. Verify your result. ii. z = i iii. z =

Solution

i. Let (-3 +4 i) = x +i y.Squaring both sides, we get -3 +4 i = (x +i y)² = (x² -y²) +2 x y i.Equating real and imaginary parts, we get x² -y² = -3, 2 x y = 4

=> y = 2; x² - = -3=> x4 - 4 = -3 x²   =>   x4 +3 x² -4 = 0=> (x² +4)(x² -1) = 0   =>   x² = -4 or x² = 1.Now x being real, x² = -4 is ruled out, so x² = 1  =>  x = ±1.When x = 1, y = 2/x = 2;when x = -1, y = 2/x = -2So we get (-3 + 4i) = ± (1 +2 i)Let us verify our result:[± (1 +2 i)]² = (1 +2 i)² = 1 +4 i² +4 i = 1 -4 +4 i = -3 +4 i

ii. Let i = x +i y    =>   i = (x² -y²) +2 i x y=>  x² -y² = 0; 2xy = 1

=>  y = 1; x² - = 0=>  x² = 1/(4x²)   =>    x4 = 1/4=> x² = ±1/2Now x being real, x² = -1/2 is ruled outx² = 1/2   =>   x = ± 1/ 2

When When x = -1/ 2, y = 1/(2x) = -1/ 2

Verification: . iii. Given z = , which is a real number.

So square roots of z are ±

Exercise

1. Find square roots of(i) -i(ii) 1(iii) -1(iv) e(v) -8

2. Find square roots of(i) -1 +2 2 i(ii) 5 -12 i(iii) -8 i(iv) 3 -4 i(v) 3 +4 i

3. If (a +b i) = ± (x +i y) prove that (a -b i) = ± (x -i y).

Answers

1. (i) ± (1-i)/ 2    (ii) ±1   (iii) ±i    (iv) ± e      (v) ±2 2 i2. (i) ±(1 + 2 i)     (ii) ±(3 -2 i)     (iii) ±2 (1 -i)    (iv) ±(2 -i)         (v) ±(2 +i)3. [Hint. (a +b i) = ±(x +i y)   =>  a +i b = (x² + y²) +2 i x y     =>    a -i b = x² +y² -2 i x y = (x -i y)²]

De-Moivre's Theorem (for rational index)

The theorem is stated in two steps as follows:(i) (cos +i sin )n = cos n +i sin n , if n is an integer (positive, zero or negative)(ii) cos n +i sin n is one of the values of (cos + i sin )n if n is a non integral rational number.

Remarks

1. If z is a complex number, and n is a positive integer, then zn is a unique complex number. However z1/n takes n distinct complex values. Thus a complex number has two square roots, three cube roots,..., n nth roots etc.

2. (cis )(cis ) = cis ( + ) and cis /cis = cis ( - ). 3. (cos +i sin )-n = cos (-n ) +i sin (-n ) = cos n -i sin n .

Thus, if z = cos +i sin then   1/z = cos -i sin . 4. (cos -i sin ) n = (cos(- ) +i sin(- ))n = cos (-n ) +i sin (-n ) = cos n -i sin n .

Roots of a complex number

Let z = r cos = r cos ( +2 m ), where m I

       z1/n = r1/n Note that r1/n is a real, positive number. Putting m = 0, 1, 2, ... n -1 we get n distinct roots of z,

and then the values of start repeating.

Cube roots of unity

Since 1 = 1 +0 i = cis 0 = cis (0 +2m ) = cis 2m , m I,11/3 = (cis 2m )1/3 = cis 2m /3 where m = 0, 1, 2= cis 0, cis 2 /3, cis 4 /3

      = 1, (-1 + 3 i)/2, (-1 - 3 i)/2Putting w = (-1 + 3 1)/2

  = (1 -3 -2 3 i)/4 = (-1 - 3 i)/2Hence three cube roots of unity are 1, w, w² where w = (-1 + 3 i)/2

Remarks

1. 1 +w +w² = 1 + (-1 + 3 i)/2 + (-1 - 3 i)/2 = 0

2. 1. w. w² = , thus w³ = 1. 3. It is easy to find cube roots of "a" in term of w, if a is a real number. Thus cube roots

of 27 are 3, 3w, 3w²; cube roots of -8 are -2, -2w, -2w² etc. 4. Note that 1, w, w² are vertices of an equilateral triangle in complex plane.

Illustrative Examples

Example

If z = (13 -5 i)/(4 -9 i), find z6.

Solution

Given z = (13 -5 i)/(4 -9 i) =[(13 -5 i)(4 +9 i)]/[(4 -9i)(4 +9i)]= (97 +97 i)/97= 1 + i/4 = 2 [cos( /4) + i sin ( /4)]Using Demoivre theorem,

z6 = = 8(cos 3 /2 +i sin 3 /2) = 8[0 +i(-1)] = -8i

Example

Convert the following number to standard form x +iy    (cos +i sin )7 (cos -i sin )³

Solution

(cos +i sin )7 (cos -i sin )³ = (cis )7 (cis (- ))³     = (cis )7 ((cis )-1)³ = (cis )7 (cis )-3

     = (cis )4 = cis 4 = cos 4 +i sin 4 .

Example

Given that one of the roots of x4 -2 x³ +3 x² -2 x +2 = 0 is 1 +i, find the other three roots.

Solution

Since all the coefficients of the given equation are real numbers, the roots, if complex, occur in conjugate pairs. Thus two roots are 1 +i, 1 -i.Hence x -(1 +i), x -(1 -i) are two factors of given equation,i.e. (x -(1 +i))(x -(1 -i)) = x² -2 x +2 is a factor of given equation.Now it is easy to see that x4 -2 x³ +3 x² -2 x +2 = (x² -2x +2)(x² +1).Also we know that x² +1 = (x -i)(x +i).Thus the four roots of the given equation are ±i, 1 ±i.

Example

Find the fourth roots of unity.

              

Solution

1 = 1 +0 i = cis 0   = cis (0 +2 m ) = cis 2m ,        where m I,11/4 = (cis 2 m )1/4 = cis 2m /4, where m = 0, 1, 2, 3       = cis 0, cis /2, cis , cis 3 /2       = 1, i, -1, -i.Hence the fourth roots of unity are ±1, ±i.

Example

Show that(i) (1 +w -w²)6 = 64.(ii) (1 -w +w²)7 +(1 +w -w²)7 = 128.(iii) (2 -w)(2 -w²)(2 -w10)(2 -w11) = 49.

Solution

(i) (1 +w -w²)6 = (-w² -w²)6          [using 1 +w +w² = 0]         = (-2w²)6 = 26 w12 = 64(w³)4 = 64 [using w³ = 1](ii) (1 -w +w²)7 +(1 +w -w²)7 = (-w -w)7 +(-w² -w²)7

    = -27. w7 -27 w14 = -27[(w³)². w +(w³)4. w²]    = -27. (w +w²) = -27. (-1) = 128.(iii) (2 -w)(2 -w²)(2 -w10)(2 -w11)    = [4 -2(w +w²) +w³] [4 -2w10(1 +w) +w21]    = [4 -2(-1) +1] [4 -2w (-w²) +1]                     (why?)    = (4 +2 +1)(4 +2. 1 +1) = 7. 7  = 49.

Exercise

1. Express each of the following in standard form x +iy(i) (cos +i sin ) 10(cos 2 +i sin 2 )4

(ii)   2. Solve the equation x4 -4 x² +8 x +35 = 0, given that one root is 2 + -3. 3. Find

(i) (4i)        (ii) -i         (iii) cube roots of -1

(iv)             (v) cube roots of 64.Also find the product of roots in case of (iv) and (v).

4. Prove that the complex cube roots of unity are reciprocal of each other. 5. If 1, w and w² are cube roots of unity, show that

(i) (1 +w)(1 +w²)(1 +w4)(1 +w8) = 1(ii) (1 +w)(1 +w²)(1 +w4)(1 +w8)... upto 2n factors = 1(iii) (1 -w)(1 -w²)(1 -w4)(1 -w8) = 9

6. Prove that = w or w²

7. Prove that is equal to(i) 2 if n is a multiple of 3(ii) -1 if n is not a multiple of 3.

Answers

1. (i) cos 2 +i sin 2    (ii) cos /12 -i sin /122. Roots are 2 ± 3 i, -2 ±i.3. (i) ± 2 (1 +i)       (ii) ±(1 -i)/ 2   (iii) -1, (1 ± 3 i)/2; that is -1, -w, -w²   (iv) ± (1 ±i)/ 2; product of roots is 1    (v) 4, 4 w, 4 w², where w = (-1+ 3 i)/2; product of roots is 64.

6. [Hint. Let = x; then [-1-x] = x]  

Angle

If rotation is anticlockwise, the angle is positive. If rotation is clockwise, the angle is negative. One full rotation indicates 360°.

An angle is said to be acute angle if 0° < 90°;right angle if = 90°; obtuse angle if 90° < < 180°;a straight angle if = 180°; anda reflex angle if 180° < < 360°.

There are three systems of measurement of an angle:

1. Sexagesimal systemIn this system an angle is measured in degrees, minutes and seconds. A complete rotation describes 360°.1 right angle = 90° (Since right angle is 1/4 th of full rotation)A degree is further subdivided as1degree = 60 minutes, written as 60'and 1 minute = 60 seconds, written as 60''.Thus 30·25° = 30° 15', 1·5' = 1' 30'' etc.We say that 30·25° is in degrees notation; 30° 15' is in degree-minute-second notation.

2. Centesimal systemIn this system an angle is measured in grades, minutes and seconds.Here 1 right angle = 100 grades, written as 100g.1 grade = 100 minutes, written as 100` and1 minute = 100 seconds, written as 100``.Thus 30·25g = 30g25`, 1·5` = 1`50`` etc.

Circular system

In this system an angle is measured in radians. The circular measure of an angle is the number of radians it contains.A radian is an angle subtended at the center of a circle by an arc whose length is equal to the radius. A radian is a constant angle.

Conversion Formula

       radians = 2 right angles = 180° = 200 g.

Length of an arc of a circle

If an arc of length s subtends an angle radians at the center of a circle of radius r,then s = r

Area of a sector of a circle

Area of sector = (1/2)r² = (1/2) s

Illustrative Examples

Example

Express in degrees, radians as well as in grades the fourth angle of a quadrilateral, which has three angles 46° 30' 10'', 75° 44' 45'', 123° 9' 35'' respectively. (Take = 355/113)

Solution

The sum of three given angles           = 46° 30' 10'' +75° 44' 45''+123° 9' 35''           = 245° 24' 30'' (since 90'' =1'30'' and 84' =1°24')The sum of all four angles of quadrilateral = 360°.Fourth angle = 360° -(245° 24' 30'') = 114° 35' 30''                                      (since 360° = 359° 59' 60'')To convert it into radians,    114° 35' 30'' = 114° +(35 + 30/60)'         = 114°(71/2)' = 114° +(71/120)°         = (13751/120)°         = (13751/120) x ( /180) radians        (As 180° = radians)          = (13751/120) x (1/180) x (355/113) = 2 radians nearlyTo convert the angle into centesimal system,(13751/120)° = (13751/120) x (100/90) grades (since 90° = 100 grades)   = 127·3241 grades = 127g 32` 41``.

Example

If G, D, denote respectively, the number of grades, degrees and radians in an angle, prove that

i. G/100 = D/90 = 2 / ii. G -D = 20 /

Solution

We know that 1 right angle = 100g = 90° = /2 radians.Let the angle be X right angles.Then G = 100 X, D = 90 X, = ( /2) X         ...(1)

i. From (1), X = G/100 = D/90 = 2 / Hence the result.

ii. From (1), G -D = 100 X -90 X = 10 X                         = 10 x 2 / = 20 /

Exercise

1. Draw diagrams for the following angles:(i) -135° (ii) 740°. In which quadrant do they lie?(iii) Find another positive angle whose initial and final positions are same as that of -135°, and indicate on the same diagram.

2. If lies in second quadrant, in which quadrant the following will lie?(i) /2     (ii) 2       (iii) - .

3. Express the following angles in radian measure as well as centesimal measure:(i) 45°     (ii) 40° 37' 30''.

4. The circular measure of an angle is 1·5. Express it in English as well as French system. Take = 3·14.

5. The wheel of a carriage is 91 cms in diameter and makes 5 revolutions per second. How fast is the carriage running?

6. A wheel makes 180 revolutions in a minute. Through how many radians does it turn in one second?

7. Large hand of a clock is 21 cm long. How much distance does its extremity move in 20 minutes?

8. Find the angle between the hands of a clock at 7.20 P.M. 9. Sum of two angles is 80 grades and difference is 18°. Find the angles in degrees. 10. The difference between two acute angles of a right-angled triangle is /3 in circular

measure. Find these angles in degrees. 11. The circular measures of two angles of a triangle are 1/2 and 1/3. Find the third angle

in English system. 12. The difference of two angles is 1° while their sum is 1 in circular measure. Find the

angles in degrees in terms of . 13. The angles of a triangle are in A.P. and the greatest is double the least. Find all the

angles in circular system. 14. Express the angle 236·345° in

(i>) degree-minute-second notation  (ii) radians.

Answers

1. (i) Third quadrant    (ii) First quadrant    (iii) 225°2. (i) First quadrant    (ii) Third or fourth quadrant  (iii) Third quadrant3. (i) /4 radians; 50 grades   (ii) 65 /228 radians; 45 g 13` 89``4. 85° 59' 14'' or 95g 54` 14``5. 51· 48 km / hour            6. 67. 44 cm                            8. 100°9. 45° and 27°                   10. 63°, 27°11. 132° 16' 22''12. (180 + )/2 , (180 + )/213. 2 /9, /3, 4 /9  radians14. (i) 236° 20' 42''  (ii) 4·125 radians  

Sign, Domain and range of T-ratios

Sign of t-ratios

Quadrant I II III IV

T-ratios which are +ve

Allsin cosec

tan cot

cos sec

This table can be memorised with the help of phrase "Add Sugar To Coffee"

Add Sugar To Coffee

All Sin Tan Cos

I II III IV

Domain of T-ratios

Function Domain

sin, cos all real numbers

tan, sec all real numbers other than (2 n +1) /2, n Z

cot, cosec all real numbers other than n , n Z

Limits of T-ratios (i.e. range)

The maximum and minimum values of sin and cos are +1 and -1 respectively.Thus, -1 cos 1 and -1 sin 1

Function Range

sin, cos [-1, 1]

tan, cot any real value

sec, cosec any real value except (-1, 1)

Illustrative Examples

Example

If tan = -2, find the values of the remaining trigonometric ratios of .

Solution

Given tan = -2 which is - ve, therefore, lies in second or fourth quadrant.Also sec² = 1 + tan² = 1 +(-2)² = 5=>   sec = ± 5Two cases arise:Case I. When lies in the second quadrant, sec is - ve.sec = - 5   =>   cos = -1/ 5sin = (sin /cos ).cos = tan . cos = (-2).(-1/ 5) = 2/ 5=>   cosec = ( 5)/2Also tan = -2   =>   cot = -1/2Case II. When lies in the fourth quadrant, sec is + ve.sec = 5    =>    cos = 1/ 5sin = (sin /cos ).cos = tan . cos = (-2).(1/ 5) = -2/ 5=>  cosec = -( 5)/2Also tan = -2    =>    cot = -1/2

Example

Prove that sin = x + 1/x is not possible for real x.

Solution

When x > 0, x +1/x = ( x - 1/ x)² +2 2When x < 0, let x = -y where y > 0. Then

Thus x +1/x 2 or x +1/x -2=>    sin 2   or sin -2Also, we know that -1 sin 1Hence sin = x +1/x is not possible for any real x.Aliter.sin = x +1/x =>  x 2 -sin . x +1 = 0It is a quadratic in x. As x is real, it has real roots=>    (-sin )² -4. 1. 1 0             (discriminant 0)

=>   sin² 4  =>    |sin | 2=>    sin -2 or sin 2Also we know that 1 sin -1Hence sin = x +1/x is not possible for any real x.

Example

Is the equation 2 sin² -cos +4 = 0 possible?

Solution

2 sin² -cos +4 = 0=>     2 (1 -cos² ) -cos +4 = 0=>   -2 cos² +cos -6 = 0=>   2 cos² +cos -6 = 0=>   (2 cos -3)(cos + 2) = 0=>   2 cos -3 = 0 or cos +2 = 0=>    cos = 3/2 or cos = -2, both of which are impossible as -1 cos 1.Hence the equation 2 sin² -cos +4 = 0 is not possible.

Exercise

1. Which of the six t-ratios are positive for the angles(i) 240°       (ii) -420°?

2. Find the other five t-ratios if(i) cos A = 1/2 and A lies in the second quadrant(ii) sin A = 3/5 and /2 < A < (iii) tan A = 3/2 and A does not lie in first quadrant(iv) cot A = 12/5 and < A < 3/2

3. In which quadrant does lie if(i) cos is positive and tan is negative(ii) both sin and cos are negative(iii) sin = 4/5 and cos = 3/5(iv) sin = 2/3 and cos = 1/3

4. For what real values of x is the equation 2 cos = x +1/x possible? 5. If sin sec = -1 and lies in the second quadrant, find sin and sec . 6. If sec A = x +1/4x, prove that sec A +tan A = 2 x or 1/2x. 7. If sin = 12/13 and lies in the second quadrant, show that

sec +tan = -5. 8. If sin : cos : : 3 : 1, find sin , cos .

Answers

1. (i) tan, cot     (ii) cos, sec2. (i) sin A = 3/2 , tan A = - 3, cot A = -1/ 3, sec A = -2, cosec A = 2/ 3    (ii) cos A =  4/5, tan A =   3/4, cot A =  4/3, sec A =  5/4, cosec A = 5/3    (iii) sin A =  3/5, cos A =  4/5, cot A = 4/3, sec A =  5/4 , cosec A =  5/3     (iv) sin A =    5/13 , cos A =  12/13, tan A = 5/12 , sec A =  13/12, cosec A = -13/53. (i) fourth    (ii) third       (iii) second    (iv) not possible as we must have sin² +cos² = 14. x = ±1 only5. 1/ 2, - 28. sin = 3/2, cos = 1/2 or sin = - 3/2, cos = 1/2  

T-ratios of Standard Angles

0°

sin 0 1/2 1/ 2 ( 3)/2 1

cos 1 ( 3)/2 1/ 2 1/2 0

tan 0 1/ 3 1 3 not defined

cot not defined 3 1 1/ 3 0

sec 1 2/ 3 2 2 not defined

cosec not defined 2 2 2/ 3 1

For memorising, we can use the following table:

0°

sin

cos

Illustrative Examples

Example

Solve for a and c in the given triangle. Also find the area of the ABC.

Solution

          sin A = height / hypotenuse = BC/AC=> sin 30° = a/12=> a = 12 sin 30° = 12.(1/2)= 6Similarly cos A = AB/AC => cos 30° = c/12=> c = 12 cos 30° = 12.( 3)/2 = 6 3Area of ABC = (1/2) x base x height = (1/2) x c x a = (1/2) x 6 3 x 6= 18 3 sq. units.

Example

If A, B, A +B, A -B are positive acute angles, find the values of A and B from the equations:     sin (A -B) = 1/2, cos (A +B) = 1/2

Solution

The given equations aresin (A -B) = 1/2 = sin 30° =>  A -B = 30°           ...(i)cos (A +B) = 1/2 = cos 60°   =>   A +B = 60°     ...(ii)Solving (i) and (ii) simultaneously, we getA = 45°, B = 15°

Exercise

1. Show that(i) sin 30° cos 0° +sin 45° cos 45° +sin 60° cos 30° = 7/4(ii) 4 sin ( /6) sin²( /3) +3 cos( /3) tan ( /4) +cosec²( /2) = 2 sec² ( /4)

2. Evaluate sec 30° tan 60° +sin 45° cosec 45° +cos 30° cot 60° 3. Taking A = 30°, verify that

(i) sin2 A +cos 2 A = 1  (ii) sin 3 A = 3 sin A -4 sin³ A(iii) sin 2 A = (2 tan A)/(1 -tan² A)(iv) cos 2 A = (1 -tan² A)/(1 +tan²A)

4. Taking A = 60°, B = 30°, verify that(i) sin (A +B) sin A +sin B(ii) cos (A +B) cos A +cos B

5. Find the value of (0° < < 90°) satisfying(i) cos / (cosec +1) + cos / (cosec -1) = 2(ii) (cos² -3cos +2)/ sin² = 1(iii) 2 sin² = 1/2(iv) 3 cos = 2 sin² (v) 2 cosec = 3 sec² (vi) tan + cot = 2(vii) sec² = 1 + tan

6. Assuming A, B, A +B, A -B to be positive acute angles, find A and B when(i) sin (A +B) = ( 3)/2, cos (A -B) = ( 3)/2(ii) tan (A +B) = 3, tan (A -B) = 1

Answers

2. 7/25. (i) 45°       (ii) 60°      (iii) 30°        (iv) 60°    (v) 30°      (vi) 45°     (vii) 45° (discarding = 0° as 0° < < 90°)6. (i) A = 45°, B = 15° (ii) A = 52·5°, B = 7·5°  

T-ratios of Complementary Angles

Two angles are called complementary iff the sum of their measures is 90° (or /2 radians)

     sin (90° - ) = cos , cos (90° - ) = sin ,tan (90° - ) = cot , cot (90° - ) = tan ,sec (90° - ) = cosec , cosec (90° - ) = sec .

Illustrative Examples

Example

Without using trigonometric tables, evaluate(i) sin23°/cos67°         (ii) tan65°/cot25°(iii) sin 18° -cos 72°

Solution

(i) sin 23°/cos 67° = sin 23°/cos(90°-23°)      = sin 23°/sin23°      (because cos (90° - ) = sin )      = 1(ii) tan 65°/cot25° = tan (90° -25°)/cot25°      = cot25°/cot25°      (because tan (90° - ) = cot )     = 1(iii) sin 18° -cos 72° = sin 18° -cos (90° -18°)      = sin 18° -sin 18° = 0

Exercise

Without using trigonometric tables, evaluate the following (1-2):1. (i) cos18°/sin72°         (ii) cosec 31°/sec 59°

(iii) cosec17°30'/sec 72°30' 2. (i) sin 62° -cos 28°      (ii) cosec 35° -sec 55°

(iii) sin 35° sin 55° -cos 35° cos 55° 3. Express each of the following in terms of t-ratios of angles between 0° and 45°.

(i) tan 81° +cos 72°     (ii) cot 49° +cosec 87°

Answers

1. (i) 1          (ii) 1       (iii) 12. (i) 0          (ii) 0       (iii) 03. (i) cot 9° +sin 18°    (ii) tan 41° +sec 3°  

T-ratios of Allied Angles

Two angles are said to be allied if their sum or difference is a multiple of 90°. For example, 30° and 60° are allied, 150° and -30° are allied.

Aid to Memory for formulae of t-ratios of allied angles

Any function of an angle (n ± ) treating as acute is numerically equal to same function of , with sign depending upon the quadrant in which the revolving line terminates. The proper sign can be ascertained by "All-Sin-Tan-Cos" formula. For example sin (180°+ ) = -sin ; -ve sign was chosen because 180° + lies in third quadrant and sin is -ve in third quadrant.

Any function of an angle , treating as acute, is numerically equal to co-function of , with sign depending upon the quadrant in which the revolving line terminates. Note that sin and cos are co-functions of each other; tan and cot are co-functions of each other; sec and cosec are co-functions of each other.

For memorising, we can use the following table:

  - 90°- 90° + 180°- 180° + 270°- 270° + 2n - 2n +

sin -sin cos cos sin -sin -cos -cos -sin sin

cos cos sin -sin -cos -cos -sin sin cos cos

tan -tan cot -cot -tan tan cot -cot -tan tan

Illustrative Examples

Example

Find the values of(i) cos 495°      (ii) sin 1230°   (iii) tan (-1590°).

Solution

i. cos 495° = cos (360° +135°) = cos (135°)               = cos (90° +45°) = -sin 45° = -1/ 2

ii. sin 1230° = sin (3 x 360 +150)° = sin 150°               = sin (180° -30°) =  sin 30° = 1/2

iii. tan (-1590°) = -tan 1590° = -tan (4 x 360 +150)°       = -tan 150° = -tan (180° -30°) = -(-tan 30°)       = tan 30° =   1/ 3

Example

i. If tan A = - 3, find all possible values of A between 0° and 360°. ii. Find all values of x lying between 0 and 360 such that sin 2x° = 0·6428.

Solution

i. We know that for first quadrant, tan 60° = 3.Now tan A = - 3, so A lies in second or fourth quadrant.Since required values of A are 180°-60° or 360°-60° i.e. 120° or 300°,we get A = 120° or 300°.

ii. From trigonometric tables, we find that 0·6428 = sin 40°.As sin 2x° = 0·6428, a positive value, 2x° lies in first or second quadrant,we get 2x° = 40° or 180°-40° i.e. 40° or 140°=> x = 20 or 70.

iii. Since 0 < x° < 360°, so 0 < 2x° < 720°, therefore,360° +40° and 360° +140° should also be considered.So we may have 2 x = 400 or 500 =>  x = 200 or 250.Hence required values of x are 20, 70, 200 or 250.

Example

i. Which is greater: sin 40° or cos 40°? ii. If = -400°, determine the sign of (sin +cos ).

Solution

i. In first quadrant, 1 > 2   =>  sin 1 > sin 2, as value of sin steadily increases from 0 to 1 as   increases from 0° to 90°.Now cos 40° = cos (90° -50°) = sin 50°.Therefore sin 50° > sin 40°  =>  cos 40° > sin 40°.

ii. sin (-400°) = -sin (400°) = -sin (360° +40°) = -sin (40°)cos (-400°) = cos (400°) = cos (360° +40°) = cos 40°= cos (90° -50°) = sin 50°Hence sin (-400°) +cos (-400°) = -sin 40° +sin 50°= sin 50° -sin 40° > 0.              (since  sin 50° > sin 40°)

Example

If ABCD is a cyclic quadrilateral, then show that                cos A +cos B +cos C +cos D = 0.

Solution

Since ABCD is a cyclical quadrilateral,A +C = , B +D = C = -A, D = -BHence cos A +cos B +cos C +cos D                = cos A +cos B +cos ( -B)                = cos A +cos B -cos A -cos B = 0.

Exercise

1. Find all the t-ratios of (i) 120° (ii) 150° (iii) 180° 2. Evaluate

(i) sin 930°(ii) cos (-870°)(iii) tan (-2025°)(iv) cot (-315°)(v) sin 19 /4(vi) cos 2 /3(vii) tan (- /3)

3. Express the following as functions of angles less than 45°:(i) sin (-1785°)   (ii) cosec (-7498°).

4. (i) Which is bigger: sin 55° or cos 55°?(ii) If = 100°, determine the sign of (sin +cos ).

5. Evaluate(i) 2sin 135° cos 210° tan 240° cot 300° sec 330°(ii) sin 690° cos 930° +tan (-765°) cosec (1170°)

6. If 8 = , show that cos 7 +cos = 0. 7. If cos = -( 3) /2, find all possible values of between -180° and 180°. 8. If cos = sin 200°, find all possible values of between -180° and 60°. 9. Given that cos = -0·5150, sin is positive and that lies between 500° and 900°, find

the values of .

10. Find x from the following equation:  cosec (90° + ) + x cos cot (90° + ) = sin (90° + ).

11. Find all values of satisfying 0 < < and tan² +cot² = 2. 12. If A, B, C, D are angles of a cyclical quadrilateral, prove that

(i) cot A +cot B +cot C +cot D = 0(ii) sin A +sin B +sin C +sin D = 0      [Hint. A +C = B +D = .]

Answers

1. (i) sin 120° = ( 3)/2, cos 120° = 1/2, tan 120° = - 3,       cot 120° = -1/ 3, sec 120° = -2,       cosec 120°  = 2/ 3.  (ii) sin 150° = 1/2, cos 150°  = -( 3)/2,      tan 150° = -1/ 3,  cot 150° = - 3,       sec 150° = -2/ 3,  cosec150° = 2.(iii) sin 180° = 0, cos 180° = -1, tan 180° = 0,     sec 180° = -1;     cot 180° and cosec 180° are undefined.2. (i) 1/2  (ii) -( 3)/2   (iii) -1         (iv) 1/ 2(v) 1/2 (vi) 1/2        (vii) - 3.3. (i) sin 15°   (ii) sec 28°4. (i) sin 55°   (ii) +ve5. (i) 1            (ii) ( 3)/4 +17. 120°, -120°8. -110°, 110°, 250°9. 841°10. x = tan 11. /4, 3 /4  

Periodicity of Circular Functions

A function f is said to be periodic if there exists a constant real quantity p such thatf (x +p) = f (x) for all x Df

There may exist more than one value of p satisfying the above relation. The least positive value of p satisfying above relation is called the period of f.We know that sin (x + p) = sin x for all real x, where p = ± 2 , ± 4 , ± 6 , ...In general, sin (x +2 n ) = sin x for all real x and n ZThus the period of sin x (or sin ) is 2 (or 360°).The period of cos, sec, cosec is also 2 (or 360°).However for tangent and cotangent functions, we havetan (x + n ) = tan x,cot (x + n ) = cot x.So the period of tan and cot is (or 180°).In general, period of [a sin (bx +c)] or [a cos (bx +c)] is 2 /| b |

Exercise

1. Draw the graphs of following functions. Also mention their range, amplitude and period of cycle.(i) sin 3 x(ii) 3 sin x(iii) 0·3 sin 3 x

2. Draw a graph of sin and cosec in the same diagram. 3. Draw the graphs of the following:

(i) cos (x - /2)(ii) cos (x - /4)(iii) cos (x + /4)(iv) 3 + 2 cos (2x - /6)(v) cos x -sin x(vi) sin²x(vii) | sin x |

4. Graphically solve the equation 3 cos x +2 = 0, where 0 < x < . 5. Draw the graph of y = cos 2 x +cos x for values of x from 0 to . On the same

diagram, draw the graph of y = x. Hence estimate the positive root of the equation x = cos 2 x +cos x.

Answers

1. (i) range is -1 to 1, amplitude is 1, period is 2 /3 (i.e. 120°)   (ii) range is - 3 to 3, amplitude is 3, period is 2 (i.e. 360°)   (iii) range is -0·3 to 0·3, amplitude is 0·3, period is 2 /3 (i.e. 120°)4. x = 2·30 (radians)5. x = 0·77 (radians)  

Addition and Subtraction Formulae

(i) sin (A +B) = sin A cos B +cos A sin B(ii) cos (A +B) = cos A cos B -sin A sin B(iii) tan (A +B) = (tan A + tan B)/(1 - tan A tan B)(iv) sin (A -B) = sin A cos B - cos A sin B(v) cos (A -B) = cos A cos B + sin A sin B(vi) tan (A -B) = (tan A - tan B)/(1 + tan A tan B)

Corollaries

(i) tan( /4 +A) = tan(45° + A) = (1 + tan A)/(1 -tan A)(ii) tan( /4 -A) = tan(45° -A) = (1 - tan A)/(1 + tan A)(iii) cot (A +B) = (cot A cot B - 1)/(cot B + cot A)(iv) cot (A -B) = (cot A cot B + 1)/(cot B -cot A)(v) tan (A +B +C) = (tan A +tan B +tan C -tan A tan B tan C)/(1 - tan A tan B - tan B tan C - tan C tan A)(vi) sin (A +B) sin (A -B) = sin²A -sin²B(vii) cos (A +B) cos (A -B) = cos²A -sin²B

Illustrative Examples

Example

Using t-ratios of 45° and 60°, evaluate(i) sin 105°   (ii) tan (13 /12).

Solution

i. sin 105° = sin (60° +45°) = sin 60° cos 45° + cos 60° sin 45°=(( 3)/2)(1/ 2) + (1/2)(1/ 2) = ( 3 + 1)/2 2

ii.

   == (-1 + 3)(1 - (-1) 3) = ( 3 - 1)/( 3 + 1)

Example

i. Prove that cos - sin = 2 cos( + /4) ii. Find the maximum and minimum values of 7 cos + 24 sin .

Solution

i. cos - sin = 2 [(cos ) (1/ 2) - (sin )(1/ 2)]=    2(cos cos /4 -sin sin /4)= 2 cos ( + /4)

ii. Since 7² +24² = 49 +576 = 625 = 25²,     7 cos + 24 sin = 25 [(cos )(7/25) + (sin )(24/25)]Now, we can find an angle such that cos = 7/25 and sin = 24/257 cos + 24 sin = 25 (cos cos + sin sin ) = 25 cos ( - )Now -1 cos ( - ) 1=> -25 25 cos ( - ) 25=> -25 (7 cos + 24 sin ) 25

Hence the given expression varies between -25 and 25.Therefore, the maximum and minimum values of 7 cos + 24 sin are 25 and -25 respectively.

In general, a cos + b sin = (a² +b²). cos ( - ), where is given bycos = a/ (a² +b²), sin = b/( (a² +b²)Thus a cos +b sin varies between - (a² +b²) and (a² +b²).

Exercise

1. Prove that sin 75° = [ 6 + 2]/4 2. Find tan 15° and hence show that tan 15° +cot 15° = 4 3. Evaluate (i) cos 195°   (ii) sin(3 /4) 4. Simplify by reducing to a single term:

(i) sin 38° cos 22° +cos 38° sin 22°(ii) cos 70° cos 10° +sin 70° sin 10°(iii) sin (x -y) cos x -cos (x -y) sin x(iv) (tan 69° +tan 66°)/(1 -tan 69° tan 66°)

5. Evaluate(i) cos 105° +sin 105°  (ii) cos 15° -sin 15°(iii) cot 105° -tan 105°

6. (i) A positive acute angle is divided into two parts whose tangents are 1/2 and 1/3. Show that the angle is /4.(ii) Prove that tan 22° +tan 23° +tan 22° tan 23° = 1.(iii) If A +B = 45°, show that (1 +tan A)(1 +tan B) = 2 and hence find the value of tan 22½°.(iv) If A +B = 225°, prove that tan A +tan B = 1 -tan A tan B.

7. Prove that(i) tan 70° = tan 20° + 2 tan 50°(ii) tan 13A = tan 4A + tan 9A + tan 4A tan 9A tan 13A(iii) (1 + tan A)(1 + tan B) = 2 tan A, if A -B = 45°(iv) tan (x - y) + tan (y - z) + tan (z - x) = tan (x -y) tan (y - z) tan (z - x)(v) tan 56° = (cos 11° +sin 11°)/(cos 11° -sin 11°)

8. Prove thatsin (x + y) sin (x - y) + sin (y + z) sin (y - z) + sin (z + x) sin (z - x) = 0.

9. Prove that sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C-tan A tan B tan C).

10. Prove thatsin A sin (B - C) + sin B sin (C - A) + sin C sin (A -B) = 0.

11. In any quadrilateral ABCD, show that  cos A cos B -cos C cos D = sin A sin B -sin C sin D.

Answers

2. ( 3 -1)/( 3 +1)3. (i) -( 3 + 1)/2 2    (ii) 1/ 24. (i) ( 3)/2       (ii) 1/2    (iii) - sin y     (iv) -15. (i) 1/ 2      (ii) 1/ 2      (iii) 2 36. (iii) 2 -1  

Converting Product into Sum/Difference and vice versa

A, B formulae

i. 2 sin A cos B = sin (A +B) +sin (A -B) ii. 2 cos A sin B = sin (A +B) -sin (A -B) iii. 2 cos A cos B = cos (A +B) +cos (A -B) iv. 2 sin A sin B = cos (A -B) -cos (A +B)

C, D formulae

i. sin C +sin D = 2 sin (C +D)/2 cos (C -D)/2 ii. sin C -sin D = 2 cos (C +D)/2 cos (C -D)/2 iii. cos C +cos D = 2 cos (C +D)/2 cos (C -D)/2 iv. cos C -cos D = - 2 sin (C +D)/2 sin (C -D)/2 = 2 sin (C+D)/2 sin (D - C)/2

Illustrative Examples

Example

Show that cos 10° +cos 110° +cos 130° = 0

Solution

cos 10° +cos 110° +cos 130° = (cos 110° +cos 10°) +cos 130°= 2 cos (110° +10°)/2 cos (110° -10°)/2 +cos (180° -50°)= 2 cos 60° cos 50° -cos 50° = 2.(1/2) cos 50° -cos 50° = 0

Example

Prove that cos 20° cos 40° cos 80° = 1/8

Solution

cos 20° cos 40° cos 80° = (1/2) cos 40° (2 cos 80° cos 20°)= (1/2) cos 40° (cos (80° +20°) + cos (80° -20°))= (1/2) cos 40° (cos 100° + cos 60°) = 1/2 cos 40° (cos 100° +1/2)= (1/4) cos 40° +(1/4) (2 cos 100° cos 40°)= (1/4) cos 40° + (1/4) (cos 140° +cos 60°)= (1/4) cos 40° - (1/4)cos 40° +(1/4)(1/2)                    (Since cos 140° = cos (180° -40°) = -cos 40°)= 1/8

Exercise

1. Convert the following products into sums or differences(i) 2 sin 3 cos 2 (ii) 2 cos 3 sin 2 (iii) 2 sin 4 sin 2 (iv) 2 cos 7 cos 3

2. Evaluate (i) 2 cos 45° cos 15° (ii) 2 sin 75° sin 45° 3. Express each of the following as the product of sines and cosines:

(i) sin 10 + sin 6 (ii) sin 10 - sin 6 (iii) cos 10 + cos 6 (iv) cos 10 - cos 6 (v) cos 25° - cos 37°(vi) sin 36° + cos 36°(vii) sin 80° - cos 70°

4. Prove that(i) cos 52° + cos 68° + cos 172° = 0(ii) cos 20° + cos 100° + cos 140° = 0

5. Prove that(i) cos A +cos (120° -A) +cos (120° +A) = 0(ii) cos /8 +cos 3 /8 +cos 5 /8 +cos 7 /8 = 0(iii) cos 2 cos /2 -cos 3 cos 9 /2 = sin 5 sin 5 /2

6. Prove that(i) sin 10° sin 50° sin 70° = 1/8(ii) sin 10° sin 30° sin 50° sin 70° = 1/16(iii) sin 20° sin 40° sin 60° sin 80° = 3/16

7. Prove that(i) cos 20° cos 40° cos 60° cos 80° = 1/16(ii) cos 10° cos 30° cos 50° cos 70° = 3/16

8. Prove that(i) tan 20° tan 40° tan 80° = tan 60°(ii) sin 12° sin 48° sin 54° = 1/8

9. Prove that(i) sin + sin + sin -sin ( + + ) = 4 sin ( + )/2 sin( + )/2 sin ( + )/2(ii) sin + sin +sin + cos ( + + )= 4 cos ( + )/2 cos ( + )/2 cos ( + )/2

10. Prove that(i) sin (B -C) cos (A -D) +sin (C -A) cos (B -D) + sin (A -B) cos (C -D) = 0(ii) sin (B +C -A) +sin (C +A -B) +sin (A +B -C) - sin (A +B +C) = 4 sin A sin B sin C

11. If sin x +sin y = a and cos x +cos y = b, find the values of(i) tan (x +y)/2(ii) tan (x -y)/2

Answers

1.(i) sin 5 +sin              (ii) sin 5 -sin     (iii) cos 2 -cos 6        (iv) cos 10 +cos 4 2. (i) ( 3 +1)/2               (ii) ( 3 +1)/23. (i) 2 sin 8 cos 2         (ii) 2 cos 8 sin 2     (iii) 2 cos 8 cos 2       (iv) -2 sin 8 sin 2     (v) 2 sin 31° sin 6°         (vi) 2 cos 9°  (vii) cos 50°

11. (i) a/b                        (ii) ±  

T-ratios of Multiple and Sub-multiple angles

T-ratios of 2A in terms of those of A

(i) sin 2A = 2 sin A cos A(ii) cos 2 A = cos² A -sin² A = 1 -2 sin² A = 2 cos² A -1

T-ratios of 2 A in terms of tan A

(i) sin 2 A = 2 tan A / (1 + tan²A)(ii) cos 2 A = (1 - tan² A)/(1 + tan²A)(iii) tan 2 A = 2 tan A / (1 - tan²A)

T-ratios of A in terms of cos 2 A

(i) cos² A = (1 +cos2A)/2 i.e. cos A = ±

(ii) sin² A = (1 -cos2A)/2 i.e. sin A = ±

(iii) tan² A = (1 -cos2A)/(1 +cos2A) i.e. tan A = ±

T-ratios of sub-multiple angles

Replacing A by A/2 in the above formulae, we get(i) sin A = 2 sin A/2 cos A/2(ii) cos A = cos² A/2 - sin²A/2 = 1 -2 sin²A/2 2 cos²A/2 - 1(iii) sin A = 2 tan A/2 /(1 + tan² A/2)(iv) cos A = (1 - tan² A/2)/(1 + tan² A/2)(v) tan A = 2 tan A/2 / (1 - tan² A/2)

(vi) cos A/2 = ±

(vii) sin A/2 = ±

(viii) tan A/2 = ±

T-ratios 3A in terms of those of A

(i) sin 3 A = 3 sin A -4 sin³ A(ii) cos 3 A = 4 cos³ A -3 cos A(iii) tan 3 A = (3tan A -tan³A)/(1 -3 tan² A)

T-ratios of some special angles

sin 18° = ( 5 -10/4, cos 18° = ( [10 + 2 5]) /4

cos 36° = ( 5 +1)/4sin 36° = ( [10 -2 5]) /4sin 72° = sin (90° -18°) = cos 18° = ( [10 + 2 5]) /4cos 72° = cos (90° -18°) = sin 18° = ( 5 -1)/ 4sin 54° = sin (90° -36°) = cos 36° = ( 5 +1)/4cos 54° = cos (90° -36°) = sin 36° = ( [10 -2 5]) /4

tan = 2 -1

Illustrative Examples

Example

Prove that i. cos 6° cos 42° cos 66° cos 78° = 1/16 ii. sin /15 sin 2 /15 sin 3 /15 sin 4 /15 = 5/16

Solution

i. L.H.S. = cos 6° cos 42° cos 66° cos 78°           = (1/4)(2 cos 66° cos 6°)(2 cos 78° cos 42°)           = (1/4) [cos (66° +6°) +cos (66° -6°)] [cos (78° +42°) +cos (78° -42°)]           = (1/4)(cos 72° +cos 60°)(cos 120° +cos 36°)           = (1/4)(sin 18° +cos 60°)(cos 120° +cos 36°)

           =

           = = R.H.S. ii. L.H.S. = sin /5 sin 2 /5 sin 3 /5 sin 4 /5

           = sin /5 sin 2 /5 sin           = sin /5 sin 2 /5 sin 2 /5 sin /5

           = = (sin 36° sin 72°)² = (sin 36° cos 18°)²

           =             = (100 - 20)/16 = 80/16 = 5 = R.H.S.

Example

Prove that = 2 cos

Solution

L.H.S. =            = [2 + 2 cos 2 ] = [2+2(2cos² -1)]           = [4cos² ] = 2 cos = R.H.S.

Exercise

1. Evaluate without using tables or calculator:

(i) 2 cos sin (ii) 2 cos² 15° -1(iii) 8 cos³20° -6 cos 20°(iv) 3 sin 40° -4 sin³ 40°

2. Prove that(i) sin 6° sin 42° sin 66° sin 78° = 1/16(ii) cos 36° cos 42° cos 60° cos 78° = 1/16(iii) tan 6° tan 42° tan 66° tan 78° = 1

3. Prove that(i) cos /7 cos 2 /7 cos 4 /7 = - 1/8(ii) (1+cos /8) (1+cos 3 /8) (1+cos 5 /8)(1+cos 7 /8) = 1/8

4. Prove that tan A +cot A = 2 cosec 2 A and deduce that tan 75° + cot 75° = 4. 5. If 2 cos = x + 1/x, prove that cos 3 = (1/2)(x³ + 1/x³) 6. Prove that

(i) cos 4 = 1 -8 sin² cos² = 1 -8 cos² +8 cos4 (ii) sin 4 = 4 sin cos³ -4 cos sin³ (iii) sin 5 = 5 sin -20 sin³ +16 sin5

7. Given that cos A/2 = 12/13, calculate without the use of tables, the values of sin A, cos A and tan A.

8. Given that tan x = 12/5, cos y = 3/5 and the angles x and y are in the same quadrant, calculate without the use of tables the values of(i)sin (x +y)    (ii) cos y/2

9. If 0 x < 2 , find sin x/2, cos x/2 and tan x/2 if(i) tan x = -4/3, x lies in second quadrant(ii) cos x = 1/3, x does not lie in second quadrant(iii) sin x = 1/4, x does not lie in first quadrant.

10. Using tan (x -y) = (tan x - tan y)/(1 + tanx tan y) , evaluate tan 13 /12[ Hint: Take x = 5 /4 and y = /6]

Answers

1. (i) 1/ 2     (ii) 3/2    (iii) 1     (iv) 3/27. 120/169, 119/169, 120/1698. (i) 56/65     (ii) -1/ 5

9. (i) 2/ 5, 1/ 5, - 2    (ii) , -1/ 3, - 2

(iii) 10. ( 3 - 1)/( 3 + 1)  

Conditional Identities

You have studied identities like (x +y)² = x² +y² +2xy, which hold for all values (of x and y). Conditional identity is an identity which holds if variables satisfy a given condition.

Method of solving conditional identities

1. If the identity involves sines and cosines of angles, then sums/differences should be converted into products using A-B formulae and then simplification should be done, by using C-D formulae or other relevant formulae.

2. If the identity involves squares of sines or cosines, first the squares should be changed into cosines of double angles by using the formulae   cos²A= (1 +cos2A)/2 and sin²A = (1 -cos2A)/2

3. If tangents or cotangents are involved, express the sum of two angles in terms of third angle, (from given relation) and then take tangents of both sides and expand, and simplify.

Illustrative Examples

Example

If A, B, C are angles of a triangle, prove that     sin 2A +sin 2C = 4 sin A sin B sin C

Solution

Since A, B, C are angles of a triangle, we have A +B +C = = 180°.L.H.S. = (sin 2A +sin 2B) +sin 2C

= 2 sin [(2A +2B)/2] cos [(2A -2B)/2] + sin 2C= 2 sin (A +B) cos (A -B) + 2 sin C cos C= 2 sin ( -C) cos (A -B) + 2 sin C cos ( -(A +B))= 2 sin C cos (A -B) -2 sin C cos (A +B)= 2 sin C [cos (A -B) -cos (A +B)]= 2 sin C (2 sin A sin B) = 4 sin A sin B sin C = R.H.S.

Example

If A +B +C = , prove that tan A +tan B + tan C = tan A tan B tan C

Solution

A +B +C = => A +B = -C=> tan (A +B) = tan ( - C)=> (tan A +tan B)/(1 -tan A tan B) = -tan C=> tan A +tan B = -tan C +tan A tan B tan C=> tan A +tan B +tan C = tan A tan B tan C.

Exercise

1. If A +B +C = , prove that(i) sin 2A +sin 2B -sin 2C = 4 cos A cos B sin C(ii) sin 2A -sin 2B +sin 2C = 4 cos A sin B cos C(iii) cos 2A +cos 2B +cos 2C = -1 -4 cos A cos B cos C(iv) cos 2A -cos 2B +cos 2C = -1 -4 sin A cos B sin C.

2. If A, B, C are angles of a triangle, prove that(i) cos²A + cos²B +cos²C = 1 -2 cos A cos B cos C(ii) sin²A -sin²B + sin²C = 2 sin A cos B sin C

3. If A +B +C = , prove that(i) tan 2A +tan 2B +tan 2C = tan 2A tan 2B tan 2C(ii) tan A/2 tan B/2 + tan B/2 tan C/2 +tan C/2 tan A/2 = 1

4. If A +B +C = , prove thatsin A/2 +sin B/2 +sin C/2 = 1 + 4 sin ( -A)/4 sin ( -B)/4 sin ( -C)/4

5. If A +B +C = 90°, prove that(i) sin 2A +sin 2B +sin 2C = 4 cos A cos B cos C(ii) cos²A + cos²B +cos²C = 2 +2 sin A sin B sin C(iii) tan A tan B +tan B tan C +tan C tan A = 1.

 

The law of Sines or Sine Rule

The sides of a triangle are proportional to the sines of the angles opposite to them i.e.   a/ sin A = b/sin B = c/sin C

Napier Analogies (Laws of tangents)

In any ABC,i. tan (B -C)/2 = [(b -c)/(b+c)]cot A/2 ii. tan (C -A)/2 = [(c - a)/(c+a)]cot B/2 iii. tan (A -B)/2 = [(a -b)/(a+b)]cot C/2

Law of cosines or cosine formula

In any ABC,i. cos A = (b² +c² -a²)/2bc ii. cos B = (a² +c² -b²)/2ca iii. cos C = (a² +b² -c²)/2ab

Projection formulae

In any triangle ABC,i. a = b cos C +c cos B ii. b = a cos C +c cos A iii. c = a cos B +b cos A

Illustrative Examples

Example

Using sine formula, prove the cosine formula.

Solution

Let ABC be any triangle. Using sine rule,    a/sin A = b/sin B = c/sin C = k (say)=> a = k sin A, b = k sin B, c = k sin CNow a² +b² - c² = k² sin² A + k² sin² B -k² sin² C= k² sin² A + k² (sin² B -sin² C)= k² sin² A + k² sin (B +C) sin (B -C)= k² sin² A + k² sin A sin (B -C)         [ A +B +C =   =>  B +C = -A   =>  sin (B +C) = sin ( -A) = sin A]= k² sin A [sin A +sin (B -C)]= k² sin A [sin (B +C) +sin (B -C)] = k² sin A (2 sin B cos C)= 2 (k sin A)(k sin B) cos C = 2 a b cos C=>    cos C = (a² +b² -c².)/2abSimilarly the other two cosine formulae can be proved.

Example

In a ABC right angled at B, prove that

i.   ii. tan (C -A)/2 = (c-a)/(c+a)

Solution

i. By Napier analogy, tan (B -C)/2 = [(b -c)/(b+c)] cot A/2           ....(1)If B = 90°, we get A + C = 90°=>   B - C = 90° -(90° - A) = A=> tan (B-C)/2 = tan A/2Hence from (1), we get tan A/2 = [(b-c)/(b+c)].[1/tan A/2]tan²A/2= (b-c)/(b+c)

So tan A/2 = (Taking +ve value as A/2 < 90°) ii. By Napier analogy, tan (C -A)/2 = [(c -a)/(c+a)] cot B/2

       = [(c -a)/(c+a)]cot 90°/2 = [(c -a)/(c+a)]

Example

If a cos A = b cos B, then prove that either the triangle is isosceles or right-angled.

Solution

Let a/sin A = b/sin B = c/sin C = k (say)=>   a = k sin A, b = k sin B, c = k sin CNow a cos A = b cos B   =>   k sin A cos A = k sin B cos B=> 2 sin A cos A = 2 sin B cos B   =>  sin 2 A = sin 2 B=> sin 2 A -sin 2 B = 0   => 2 cos (A +B) sin (A -B) = 0=> cos (A +B) = 0 or sin (A -B) = 0=> either A +B = 90° or A -B = 0° (0° < A +B < 180°)=> C = 180° -(A +B) = 180° -90° = 90° or A = B=> triangle is rt d or triangle is isosceles

Exercise

In any triangle ABC, prove the following (1 -14):1. cos (B -C)/2 = 2 sin A/2, where b +c = 2 a 2. (c -b cos A)/(b -c cos A) = cos B/cos C

3. (a +b) cos C +(b +c) cos A +(c +a) cos B = a +b +c 4. (c -b)² cos² A/2 +(c +b)² sin² A/2 = a² 5. a²sin (B -C)/[sin B +sin C ] +b²sin (C -A)/[sin C +sin A] + c² sin (A - B)/[sin A +sin B] =

0 6. (a² -b²)/(cos A +cos B) + b² -c²)/(cos B +cos C) + (c² -a²)/(cos C +cos A) = 0 7. b² sin 2 C +c² sin 2 B = 2 b c sin A 8. [(b² -c²)/a] cos A +[(c² -b²)/b] cos B + [(a² -b²)/c] cos C = 0 9. (cos A)/a + (cos B)/b + (cos C)/c = (a² +b² +c²)/(2abc) 10. b² cos 2 A -a² cos 2 B = b² -a² 11. a cos (A +B +C) -b cos (B +A) -c cos (A +C) = 0 12. (i) sin B +sin C > sin A.

(ii) (sin A +sin B)(sin B +sin C)(sin C +sin A) > sin A sin B sin C[Hint. Use a +b > c etc.]

13. (c² -a² + b²) tan A = (a² - b² +c²) tan B = (b² - c² +a²) tan C. 14. sin³ A cos (B -C) +sin³ B cos (C -A) +sin³ C cos (A -B) = 3 sin A sin B sin C. 15. If in a triangle ABC, sin 2 A +sin 2 B = sin 2 C, prove that either A = 90° or B = 90°. 16. If in a triangle ABC, sin² A +sin² B = sin² C, prove that ABC is right angled. 17. The sides of a triangle are of lengths 5 x +4 y, 4 x +5 y and 7 x +7 y units, where x

and y are positive. Show that the triangle is obtuse. 18. If the angles of a triangle are in the ratio 1: 2 :3, prove that the corresponding sides of

the triangle are in ratio 1 : 3 2. 19. In a ABC, a = 1, b = and C = /6. Find the other two angles and the third side. 20. If the angles A, B, C of a triangle ABC are in the ratio 3:4:4 prove that a +c 2 = 2 b. 21. In a ABC, the sides are 3, 5, 7 units. Using cosine formula, find the angles in

degree measure (upto one decimal). 22. If a = 29· 45 cm, b = 30·12 cm and B = 66°, find the remaining side and angles of

the triangle. 23. If a = 2· 0 ft, b = 3·0 ft, A = 23·1° and B is obtuse, show that c = 1·1 ft. 24. Show that there is no triangle satisfying the conditions a = 2·0 ft,b = 6·0 ft and A =

23·1°.[Hint. Try computing sin B using the law of sines.]

25. Let b = 1, a = 2 and B = 30°. Show that A = 45° or 135°.(i) Taking A = 45°, determine the remaining parts of ABC.(ii) Taking A = 135°, determine the remaining parts of ABC.

Answers

19. A = 30°, B = 120°, c = 121. 21·8°, 38·2°, 120°22. A = 63·3°, C = 50·7°, c = 25·5 cm25. (i) C = 105°, c = 1·93 (approx.)        (ii) C = 15°, c = 0·52  

Half angle formulae or Semi-sum formulae

Let s be the half perimeter of the ABC i.e. 2 s = a +b +c, then

(i)    (ii)

(iii) tan A/2 =

Area of a Triangle

In a triangle ABC, the area is given by      =(1/2)bc sin A = (1/2)ca sin B = (1/2)ab sin C

Heron formula

Approximately 2000 years ago, Heron of Alexandria derived a formula for area of a triangle. Modern derivative of this formula is      = [s (s -a)(s -b)(s -c)]

Illustrative Examples

Example

In a ABC, a = 3, b = 5, c = 6. Calculate(i) sin A     (ii) cos A    (iii) tan A    (iv) sin A/2   (v) cos A/2(vi) tan A/2                (vii) area of triangle i.e.

Solution

Here 2s = a +b +c = 3 +5 +6 = 14,So s = 7, s -a = 4, s - b = 2, s -c = 1

i. sin A = 2 /bc= [2 s (s -a)(s -b)(s -c)]/bc = (2 7.4.2.1)/(5.6) = (2 14)/15 ii. cos A =[ b² +c² -a²] / (2bc) = [5² +6² -3²] / (2.5.6) = [25 +36 -9]/60 = 52/60 = 13/15 iii. tan A = sin A/cos A = (2 14)/13

iv.   = 1/ 15

v. = 1/ 15

vi.   vii. = s (s -a)(s -b)(s -c) = 7.4.2.1 = 2 14

Exercise

In a ABC, prove that (1 -5) :1. = [a² sin B sin C]/ 2 sinA = [b² sin C sin A]/2 sinB = [c² sin A sin B]/2 sinC 2. = s (s -a) tan A/2 = s(s -b) tan B/2 = s (s -c) tan C/2 3. (i) cot A/2 +cot B/2 +cot C/2 = s²/

(ii) tan A/2 tan B/2 tan C/2 = /s² 4. tan A/2 tan B/2 = (a +b -c)/(a +b +c) = (s -c)/s 5. a² sin 2 B +b² sin 2 A = 4 6. If a = 18, b = 24, c = 30, calculate the following :

(i)             (ii) sin A        (iii) cos A         (iv) tan A(v) sin A/2  (vi) cos A/2    (vii) tan A/2

Answers

6. (i) 216    (ii) 0·6    (iii) 0·8    (iv) 0·75     (v) 0·316     (vi) 0·947   (vii) 0·333  

Circles connected with a given Triangle

The circle which passes through three vertices of a triangle is called the circumscribing circle or circumcircle.The center of this circle is called circumcenter, usually denoted as O, and its radius is called circumradius, usually denoted as R. It can be shown that circumcenter is the point of intersection of right bisectors of the sides of a triangle.The (unique) circle which is drawn within a triangle so as the three sides touch this circle is called the inscribed circle or incircle. The center of this circle is called incenter, usually denoted by I, and its radius is called inradius, usually denoted by r. It can be shown that incenter is the point of concurrence of the bisectors of the three (internal) angles of the triangle.The circle which lies outside the triangle and touches the side BC and also the sides AB and AC produced is called escribed circle or excircle opposite to angle A. Its center is called excenter and its radius is called exradius. It can be shown that excenter is the point of concurrence of internal bisector of angle A and external bisectors of angles B and C. Similarly there are two more excircles, one opposite to angle B and one opposite to angle C. The three excenters are usually denoted as I1, I2, I3 and the three ex-radii are usually denoted as r1, r2, r3.

        

       

The Circumradius

In any ABC,i. R = a/(2 sin A) = b/(2 sin B) = c/(2 sin C), and ii. R = abc/4

The Inradius

In any ABC,i. r = /s ii. r = (s -a) tan A/2 = (s -b) tan B/2 = (s -c) tan C/2

iii.  

The Exradius

In any ABC,i. r1 = /(s-a) = s tan A/2 = (s -c) cot B/2 = (s -b) cot C/2

ii.  

Illustrative Examples

Example

In any ABC, prove that    1/r1 +1/r2 +1/r3 = 1/r

Solution

 = (3s - 2s)/ = s/   = 1/r

Example

In a ABC, a = 18, b = 24, c = 30 cms. Find:i.ii. R iii. r iv. r1, r2 , r3

v. Area of circumcircle vi. Area of incircle vii. Area of escribed circle opposite angle B

Solution

2s = a +b +c = 18 +24 +30 = 72 cm =>    s = 36 cmi. = [s (s -a)(s -b)(s -c)] = [36.18.12.6] = 216 cm² ii. R = abc/4 = (18.24.30)/(4.216) = 15 cm iii. r = /s = 216/36 = 6 cm iv. r1 = /(s-a) = 216/18 = 12 cm,

r2 = /(s-b) = 216/12 = 18 cm,r3 = /(s-c) = 216/6 = 36 cm

v. Area of circumcircle = R² = (15)² = 225 cm² vi. Area of incircle = r² = (6)² = 36 cm² vii. Area of escribed circle opposite to angle B

           = r2² = (18)² = 344 cm²

Exercise

In a ABC, prove the following (1 -9):1. s = 4 R cos A/2 cos B/2 cos C/2 2. = 2R² sin A sin B sin C 3. (a +b) sec [(A-B)/2] = 4R cos C/2 4. R(a² +b² +c²) = abc(cot A +cot B +cot C) 5. (s/a -1)(s/b -1)(s/c -1) = r/R 6. (b +c) tan A/2 +(c +a) tan B/2 + (a +b) tan C/2 = 4(R +r) 7. 8Rr(cos²A/2 +cos²B/2 +cos²C/2) = 2(ab +bc +ca)-(a² +b² +c²) 8. (i) r r1 r2 r3 = ²

(ii) (r1 -r) (r2 -r) (r3 -r) = 4 R² r(iii) rr1/(r2r3) = tan² A/2

9. (r1 +r2)(r2 +r3)(r3 + r1) = 4 R s² 10. If r1 = r2 +r3 +r, prove that A = 90° 11. If a² +b² +c² = 8 R², prove that the triangle is right angled. 12. If the diameter of an excircle is equal to the perimeter of the triangle, show that the

triangle is right angled. 13. If in a ABC, a = 4 cm, b = 5 cm, c = 6 cm, calculate

(i)       (ii) R      (iii) r       (iv) r1, r2, r3

(v) lengths of altitudes p1, p2, p3

(vi) area of circumcircle (correct upto 2 decimals. Use = 22/7)(vii) area of incircle(viii) area of escribed circle opposite to angle C(ix) length of bisector of angle A(x) sin A             (xi) sin A/2

14. If in a ABC, a : b : c = 5 : 6 : 7, find the ratio of the radius of the circumcircle to that of incircle.

15. If the area of a triangle is 24 sq. cm. and its ex-radii are 4 cm, 1 cm, 12 cm, find the lengths of its three sides.

16. If the angles of a triangle are in ratio 1 : 2 : 3 and its circumradius is 10 cm, show that the lengths of its sides are 10 cm, 20 cm, 10 3 cm.

Answers

13. (i) 9·921 sq. cm     (ii) 3· 024 cm      (iii) 1·323 cm    (iv) 2·83, 3·97, 6·61 cm      (v) 5·67, 7·94, 13·23 cm    (vi) 28·74 cm²       (vii) 5·50 cm²      (viii) 137·32 cm²    (ix) 5·102 cm        (x) 0·66               (xi) 0·35314. 35 : 16               15. 6, 8, 10 cm  

Techniques of solving a Triangle

The three angles and the three sides of a triangle are called the six elements of the triangle. If three elements of a triangle are given (at least one of them being a side), the other three can be found out by using sine rule, law of cosines and / or semi-sum formulae. This process of finding the unknown elements is called solving a triangle.

Solution of right angled triangle

If we know the two sides and one angle (i.e. 90°), we can use Pythagoras theorem to find the third side and then the sine rule to find other two angles.If one side and another angle are known, we can find the third angle from A +B +C =180°, and then use sine rule to find the other two sides.

Solution of oblique triangles (i.e. not right angled)

We categorise such cases in four categories:(i) When three sides are given. (SSS)(ii) When two sides and the included angle are given. (SAS)(iii) When one side and two angles are given. (SAA)(iv) When two sides and an angle opposite to one of them are given. (SSA)(1) When three sides are given (SSS)As a, b, c are known, first we find the semi-perimeter s = (a +b +c)/2Then we calculate two angles by using any of the following formulae:

   

     

     sin A = 2 /bc, cos A = (b²+c² -a²)/2bcThen the third angle can be found by using A +B +C = 180°(2) When two sides and the included angle are given (SAS)Let sides b, c and angle A be given (b > c).Then A +B +C = 180°  =>  B +C = 180° -AAfter this, we use Napier Analogy: tan (B -C)/2 = [(b -c)/(b+c)]cot A/2From this (B -C)/2 and thus, B -C can be found.Now, since B +C and B -C are known, we can find B and C by adding and subtracting. The side a can be found by using sine rule.(3) When one side and two angles are given (SAA)We can find the third angle using A +B +C = 180°. Then we calculate the other two sides using sine rule.(4) When two sides and an angle opposite to one of them is given (SSA)Let the two side b, c be given, and also angle B (opposite to given side b) be given.To find angle c, we use sine rule:          b/sin B = c/sin C     => sin C = c sin B/bThree cases arise:(i) If c sin B > b, then sin C > 1, which is not possible. Hence there is no solution.(ii) If c sin B = b, we get sin C = 1  => C = 90°. Then the triangle can be solved.(iii) If c sin B < b, then sin C < 1, and two values of C are possible, say C1 (0 < C1 < 90°) and C2 = 180° -C1 (90° < C2 < 180°). If B +C2 < 180°, two solutions of given triangle are possible, and this is called ambiguous case. If B +C2 > 180°, we have to reject this value of C (i.e. C2) and solve the triangle by using value C1 only.

Illustrative Examples

Example

Solve the right angled triangle where(i) C = 90°, A = 30° and c = 10(ii) A = 90°, C = 56° and a = 50.

Solution

i.   B = 180° - A - C

= 180° -30° -90° = 60°From the figure, we getsin 30° = BC/AB = a/10a = 10 sin 30° = 10.(1/2) = 5b = AC = [AB² -BC²] = [10² -5²] = 75 = 5 3 units

ii.      B = 180° - A - C = 180° -90° -56° = 34°

Now a/sin A = b/sin B=>         50/sin 90° = b/sin 34°b = 50 sin 34°/sin 90° = 27·96Again, a/sin A = b/sin C=>  c = a sin C/sin A = 50.sin 56°/sin 90° = 41·45

Example

Solve the ABC, if a = 24·2, b = 37·5, c = 28·9.

Solution

2 s = a +b +c = 24·2 +37·5 +28·9 = 90· 6    =>   s = 45·3s -a = 21·1, s -b = 7·8, s -c = 16· 4

Now log tan A/2 = (1/2) [log 7·8 +log 16·4 -log 45·3 -log 21·1]= (1/2) [0·8921 +1·2148 -1· 6561 -1·3243]= (1/2) [2·1069 -2·9804] = [-0·8735] = -0· 4367 = ·5633=>    A/2 = 20° 6'  =>  A = 40° 12'

Similarly =>    log tan B/2 = [log 16·4 +log 21·1 -log 45·3 -log 7·8]= (1/2) [1·2148 +1·3243 -1·6561 -0·8921]= (1/2) [2·5391 -2·5482] = [-0·0091] = -0·0045 = 1·9955B/2 = 44° 42' =>   B = 89° 24'Now C = 180° -(A +B) = 180° -(40° 12' +89° 24') = 180° -129° 36' = 50° 24'Hence A = 40° 12', B = 89° 24', C = 50° 24'.

Example

In a triangle ABC, C = /6, = 3 and a = 1. Find the other two angles and the side.

Solution

C = /6 = 30°   =>   A +B = 180° -C = 180° -30° = 150°Now tan (B -A)/2 = [(b -a)/(b+a)] cot C/2 = [( 3 -1)/( 3 +1)]cot 15°= [( 3 -1)/( 3 + 1)].[( 3 +1)/( 3 - 1)] = 1 = tan 45°=>    (B -A)/2 = 45°  =>    B -A = 90°From B +A = 150°, B -A = 90°, we get by adding

(B +A) +(B -A) = 150° +90°  =>   2 B = 240°  => B = 120°A = 150° -B = 150° -120° = 30°Now a/sin A = c/sin C    =>      1/sin 30° = c/sin 30°      =>  c = 1Hence A = 30°, B = 120°, c = 1

Example

In a triangle, A = 30°, B = 70° and a = 10. Solve the triangle.

Solution

C = 180° -(A +B) = 180° -(30 +70°) = 80°Now a/sin A = b/sin B = c/sin C=> 10/sin 30° = b/sin 70° = c/sin 80°b = 10 sin 70°/sin 30° = 2 x 10 x 0·9397 = 18·794and c = 10 sin 80°/sin 30° = 2 x 10 x 0·9848 = 19· 696

Example

Solve the triangle ABC when b = 7, c = 10, B = 51°.

Solution

Using sine formula, sin C/c sin B/b    =>  sin C = c sin B/blog sin C = log c -log b +log sin B= log 10 -log 7 +log sin 51°= 1 -0·8451 +1·8905= 0· 0454, which is +ve=>   sin C > 1, which is impossible.Thus there is no solution.

Exercise

Solve the following triangles (1 -8):1. B = 90°, a = 3·5, b = 5· 6 2. C = 90°, b = 3·3, c = 4· 4 3. C = 90°, a = 50· 4, b = 26·2 4. C = 90°, A = 61°, c = 64 5. B = 90°, C = 49°, b = 100 6. a = 2, b = 6, c = 3 -1. (Use cosine formulae) 7. a = 25, b = 26, c = 27. (Use cosine formulae) 8. a = 7, b = 8, c = 9 9. The sides of a triangle are 2, 3 and 4. Find the greatest angle f log 2 = 0·30103, log 3

= 0· 4771213, L tan 52° 14' = 10·1108395 and L tan 52° 15' = 10·1111004. 10. The sides of a triangle are 5, 8, 11. Find the greatest angle if log 7 = 0·8450980, L

sin 56° 47'= 9·9225205, L sin 56° 48' = 9·9226032. 11. If a = 3+1, b = 3 -1, c = 60°, solve the triangle. 12. If A = 132°, b = 50, c = 80, solve the triangle. 13. If a = 21, b = 11 and C = 34° 42'30'', find A and B, it being given that log 2 = 0·30103

and L tan 72° 38'45'' = 10·50515. 14. The two sides of a triangle are in the ratio 13 : 7, and the angle included by them is

60°. Find the remaining angles, it being given that log 3 = 0·4771213, L tan 27°27' = 9·7155508, and the difference for 1' is 3087.

15. Solve the triangle ABC when c = 72, A = 56°, B = 65°. 16. If A = 50° 10', C = 72° and b = 302, solve the triangle. 17. In ABC, A = 60° 15', B = 54° 30' and c = 100 cm. Solve the triangle. 18. Solve the triangle in which b = 50, c = 63 and B = 54°. 19. Solve the triangle in which b = 3 +1, c = 2 2, B = 75°. 20. Solve the triangle in which b = 100, c = 100 2 and B = 30°.

Answers

1. A = 38° 40', C = 51° 20', c = 4·372. B = 49°, A = 41°, a = 2·93. A = 62° 32', B = 27° 28'  , c = 56·804. B = 29°, a = 55·97, b = 31·03

5. A = 41°, a = 65·6, c = 75·56. A = 45°, B = 120°, C = 15°7. A = 56° 15' 12'', B = 59° 51' 12'', C = 63° 53' 36''8. A = 48° 10', B = 58° 24', C = 73° 26'9. 104° 28'40''10. 113°34'42''11. A = 105°, B = 15°, c = 612. B = 18°10', C = 29°51', a = 119·613. A = 117° 38'45''  and  B = 27° 38'45''14. 87° 27' 25·5'', 32° 32' 34·5''15. a = 69·6, b = 76·12, C = 59°16. b = 273·96, c = 339·81, B = 57° 50'17. a = 95·59, b = 89·64, C = 65°15'18. No solution exists19. a = 3 -1, A = 15°, C = 90°20. a = 50, A = 105°, C = 45° or a = 50( 6  - 2), A = 15°, C = 135°  

Practical use of Trigonometry

Trigonometric knowledge can be used to estimate heights of objects or distances betweens points, for example, height of a mountain or breadth of a river.

Angles of elevation and depression

If a horizontal line is drawn from the eye of the observer (O), and an object P is above this line OX, then POX is called angle of elevation. If an object Q is below the horizontal line OX, then QOX is called angle of depression. Note that from P, the point O is situated at an angle of depression = X'PO = POX =

                    

Bearings of a point

The bearing of a line in horizontal plane is the angle made by it from North-South line. The angle (measured from north in clockwise direction) is generally given in three figures (e.g. 045° instead of 45°). So this kind of measurement is called three figure bearing.

                            

           However, more frequently, quadrant bearings are used. Here, firstly we see whether the line makes acute angle on North or South side. Then we measure acute angle from North or South towards East or West as the case may be, and use that measure (in degrees) and

letter E or W accordingly.

                               

Some useful figures

While solving problems on heights and distances, the following figures should be remembered.     2 1· 414, 3 1·732, 5 2·236

sin 0° = 0, sin 15° = ( 3 -1)/[2 2 , sin 36° = ( 5 - 1)/[2 2]sin 18° = ( 5 -1)/4, sin 30° = 1/2, sin 60° = ( 3)/2 ,sin 75° = ( 3 -1)/2 2sin 90° = 1

Illustrative Examples

Example

A tree casts shadow 3 times its height. Find the angle of elevation of the sun.

Solution

Let BC = h be the height of the tree, so that length of shadow, AB = 3 h. If is the angle of the elevation of the sun, we see from the diagram that tan = BC/AB = h/[h 3] = 1/ 3 = tan 30°

            = 30°               

Example

From the top of a tower 60 m high, the angles of depression of the top and bottom of a building are observed to be 30° and 60° respectively. Find the height of the building. Also find the distance between the tower and the building.

Solution

Let AB be the building h meters high, and PQ be the tower 60 m high situated at distance BQ = x meters away.Then AC = BQ = x, CQ = AB = h,PC = 60 -hFrom BPQ, cot 60° = BQ/PQ = x/60=> x = 60 cot 60°From APC, cot 30° = AC/PC = x/(60 -h)=>    60 -h = x/ cot 30° = 60 cot 60° tan 30° = 60 (1/ 3)(1/ 3) = 20

=>   h = 60 -20 = 40        Thus the building is 40 meters high and is situated 34·64 meters away from the tower.

Example

Two boats leave a place at the same time. One travels 56 km in the direction N 50° E, while the other travels 48 km in the direction S 80° E. What is the distance between the boats?

                                                    

Solution

Let points A and B represent the position of two boats. We have to find AB, and we know that OA = 56 km, OB = 48 km,

AOB = 180° -50° -80° = 50°Using cosine formula in OAB, we getcos 50° =[ (56)²+(48)² - (AB)²]/[ 2(56)(48)]AB² = (56)² +(48)² -2 (56)(48) cos 50°= 3136 +2304 -5376 x 0·6424= 5440 -3455·69 = 1984·31=> AB = 198.31 = 44·54 meters

Example

A radio transmitter antenna of height h stands at the top of a tall building. At a point on the ground, the angle of elevation of the bottom of the antenna is and that of the top of the antenna is . Prove that the height of the building is h tan /(tan -tan ).

                                                   

Solution

Let AB be the building of height x, CA be the antenna of height h, and y be the distance of the observer from the foot of the building.In OBC, tan = (x +h)/yIn OAB, tan = x/yEliminating y between these two equations,(x +h)/x = tan / tan   =>    x tan +h tan = x tan=>    h tan = x tan -x tan=>   x  = h tan /(tan -tan )

Exercise

1. A kite with a string 150 ft. makes an angle of 45° with the ground. Assuming that the string is straight, how high is the kite?

2. A tree 10 meters high casts a 17·3 meter shadow. Find the angle of elevation of the sun.

3. From the top of a hill, the angles of depression of two consecutive kilometre stones due east are 30 and 45 degrees. How high is the hill?

4. A 12 meter ladder is inclined to the vertical at angle 15°. How far is it from the base? 5. An observer in a lighthouse is 66 feet above the surface of the water. The observer

sees a ship and finds the angle of depression to be 0·7°. Estimate the distance of the ship from the base of the lighthouse. Round the answer to the nearest 5 feet.

6. From the point on a ground level, you measure the angle of elevation to the top of a mountain to be 45°. Then you walk 200 m further away from the mountain and find

that the angle of elevation is now 30°. Find the height of the mountain. Round the answer to the nearest meter.

7. A surveyor stands 30 yards from the base of a building. On top of the building is a vertical radio antenna. Let denote the angle of elevation when the surveyor sights to the top of the building. Let denote the angle of elevation when the surveyor sights to the top of the antenna. Express the length of the antenna in terms of the angles and .

8. A helicopter hovers 800 feet directly above a small island. From the helicopter, the pilot takes a sighting to a point directly ashore on the mainland, at the waters edge. If the angle of depression is 30°, how far off the coast is the island?

9. A ladder 18 feet long leans against a building. The ladder forms an angle of 60° with the ground.(i) How high up the side of the building does the ladder reach?(ii) Find the horizontal distance from the foot of the ladder to the base of the building.

10. Two satellite tracking stations, located at points A and B in a desert, are 200 miles apart. At a prearranged time, both stations measure the angle of elevation of a satellite as it crosses the vertical plane containing A and B. If the angles of elevation from A and from B are and respectively, express the altitude h of the satellite in terms of and .

11. From the base of a 30 m high building, the angle of elevation of a tower is 60°, and from the top of the building, it is 30°.Find the height of the tower.

12. A man on the top a building 30 meters high observes a man coming directly towards it at a uniform speed. If it takes 15 minutes for the angle of depression to change from 30° to 45°, how much time will it take after this for the man to reach the base of the building? Round your answer to nearest minute.

13. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°.If after 10 seconds, the elevation is observed to be 30°, find the uniform speed per hour of the plane.

14. A man standing south of a lamppost observes his shadow on the horizontal plane to be 24 ft long. On walking eastwards 300 ft, he finds his shadow as 30 ft. If his height is 6 ft, find the height of the lamp above the plane.

15. Two poles of equal height are standing opposite to each other on either side of a road, which is 30 ft wide. From a point between them on the road, the angles of elevation of the tops are 30° and 60°. Find the height of each pole, rounded to nearest feet.

16. A vertical tower is surmounted by a vertical flagstaff of height 10 ft. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are and respectively. Prove that the height of the tower is 10 tan /(tan -tan ) feet.

17. An aeroplane when 6000 m high passes vertically above another plane at an instant when their angles of elevation at the same observing point are 60° and 45° respectively. How many meters higher is the one than the other?

18. At the foot of a mountain, the elevation of its peak is 45°. After ascending 100 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain (to the nearest meter).

19. A vertical pole is struck by a speeding car and breaks into two, the top striking the ground at an angle of 30° and at a distance of 10 feet from the foot of the pole. Find the total height of the pole.

20. The breadth of a street between two houses is 9 meters and the angle of depression of the top of one, as observed from the top of the other, which is 12 meters high, is 30°. Find the height of the other house.

21. A town B is 13 km south and 18 km west of town A. Find the bearing and distance of B from A.

Answers

1. 75 2 ft                           2. Approx 30°                   3. 2·37 km4. 3·10 m                             5. 5400 ft                          6. 273·22 m7. 30 (tan -tan )                  8. 800 3m9. (i) 9 3 ft  (ii) 9 ft              10. 200 /(cot -cot ) miles11. 45 m                               12. 41 minutes                   13. 240 3 km/hr14. 106 ft                              15. 13 feet                         17. 2536 m18. 137 m                             19. 17·32 ft                        20. 6·8 m21. 22·2 km, S 54°10' W  

Distance Formula

Distance between two points P(x,y) and Q(x,y) is|PQ| = [(x2 -x1)² +(y2 -y1)²].Hence the distance of the point P (x, y) from the origin (0, 0)    = [(x -0)² +(y -0)²] = [x² +y²]

Remarks

To prove that a quadrilateral is a(i) rhombus, show that all the sides are equal.(ii) square, show that all the sides are equal and the diagonals are also equal.(iii) parallelogram, show that the opposite sides are equal.(iv) rectangle, show that the opposite sides are equal and the diagonals are also equal.

Illustrative Examples

Example

Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right angled triangle.

Solution

Let the points be A (7, 10), B (-2, 5) and C (3, -4), then|AB| = [(- 2 -7)² +(5 -10)²] = [81 +25] = 106,|BC| = [(3 +2)² +(11 -3)²] = [25 +81] = 106 and|CA| = [(7- 3)² +(10 -(-4))²] = [16 +196] = 212=> AB² = 106, BC² = 106 and CA² = 212,Hence AB² +BC² = 106 +106 = 212 = CA²=> ABC is right angled and it is right angled at B.Also |AB| = 106 = |BC| => ABC is isosceles.

Example

Show that the points (-1, -1), (2, 3) and (8, 11) are collinear.

Solution

Let the points be A (-1, -1), B (2, 3) and C (8, 11), then|AB| = [(2 -(-1))² +(3 -(-1))²] = [32 +42] = [9 +16] = 25 =5,|BC| = [(8 -2)² +(11-3)²] = [36 +64] = 100 = 10 and|CA| = [(8 -(-1))² +(11 -(-1))²] = [92 +122] = [81 +144] = 225 =15.Hence |AB| +|BC| = 5 +10 = 15 = |CA|=> the given points are collinear.

Example

The vertices of a triangle are A (1, 1), B (4, 5) and C (6, 13). Find cos A.

Solution

If a, b, c are the sides of ABC, thena = |BC| = [(6 -4)² +(13 -5)²] = [4 +64] = 68 = 2. 17,b = |CA| = [(6 -1)² +(13 -1)²] = [25+144] = 169 = 13, andc = |AB| = [(4 -1)² +(5 -1)²] = [9 +16] = 25 = 5By cosine formula, we havecos A = (b² +c² -a²)/ 2bc = (169 +25 -68)/(2.13.5) = 126/130 = 63/65

Exercise

1. A is a point on y-axis whose ordinate is 5 and B is the point (-3, 1). Compute the length of AB.

2. The distance between A (1, 3) and B (x, 7) is 5. Find the values of x. 3. What point (or points) on the y-axis are at a distance of 10 units from the point (8, 8)? 4. Find point (or points) which are at a distance of 10 from the point (4, 3) given that

the ordinate of the point (or points) is twice the abscissa. 5. Find the abscissa of points whose ordinate is 4 and which are at a distance of 5 units

from (5, 0). 6. What point on x-axis is equidistant from the points (7, 6) and (-3, 4)? 7. Find the value of x such that |PQ| = |QR| where P, Q, R are (6, -1), (1, 3) and (x, 8)

respectively. Are the points P, Q, R collinear? 8. Show that the points (4, 2), (7, 5) and (9, 7) are collinear. 9. Show that the points (2, 3), (-4, -6) and (1, 3/2) do not form a triangle.

[Hint. Show that the given points are collinear.] 10. Show that the points A (2, 2), B (-2, 4) and C (2, 6) are the vertices of a triangle.

Prove that ABC is an isosceles triangle. 11. Show that the points:

(i) (-2, 2), (8, -2) and (-4, -3) are the vertices of a right-angled triangle.(ii) (0, 0), (5, 5) and (-5, 5) are the vertices of a right-angled isosceles triangle.(iii) (1, 1), (-1, -1),(- 3 , 3)are the vertices of an equilateral triangle.(iv) (2 a, 4 a), (2 a, 6 a), (2 a + 3 a, 5 a) are the vertices of an equilateral triangle.

12. Show that the points:(i) (2, 1), (5, 4), (4, 7), (1, 4) are the angular points of a parallelogram.(ii) (7, 3), (3, 0), (0, -4), (4, -1) are the vertices of a rhombus.(iii) (2, -2), (8, 4), (5, 7), (-1, 1) are the vertices of a rectangle.(iv) (3, 2), (0, 5), (-3, 2), (0, -1) are the vertices of a square.

13. The points A (0, 3), B (-2, a) and C (-1, 4) are the vertices of a right-angled triangle at A, find the value of a.

14. Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of equal sides is 3 units.

Answers

1. 5                     2. 4 or -2            3. (0, 2), (0, 14)4. (1, 2), (3, 6)    5. 2 or 8             6. (3, 0)7. 5 or -3; No     13. 114. (2 +( 11)/2, 5/2) or (2 -( 11)/2, 5/2)  

Section Formula

When the Point divides the line segment Internally

Let P (x1, y1) and Q (x2, y2) be two given points in the co-ordinate plane, and R (x, y) be the point which divides the segment [PQ] internally in the ratio m1 : m2 i.e.    PR/RQ = m1 / m2, where m1 0, m2 0, m1 + m2 0Then the coordinates of R are (m1 x2 +m2 x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)Note. [PQ] stands for the portion of the line PQ which is included between the points P and Q including the points P and Q. [PQ] is called segment directed from P to Q. It may be observed that [QP] is the segment directed from Q to P. If a point R divides [PQ] in the ratio m1 : m2 then it divides [QP] in the ratio m2 : m1.

When the Point divides the line segment Externally

Let P (x1, y1) and Q (x2, y2) be two given points in the co-ordinate plane, and R (x, y) be the point which divides the segment [PQ] externally in the ratio m1 : m2 i.e.     PR/RQ = m1 / m2, where m1 0, m2 0, m1 - m2 0Then the co-ordinates of R are m1 x2 -m2 x1)/(m1 -m2), (m1y2 -m2y1)/(m1 -m2)

Mid-point formula

The co-ordinates of the mid-point of [PQ] are ((x1 +x2)/2, (y1 +y2)/2)

Illustrative Examples

Example

Find the co-ordinates of the point which divides the line segment joining the points P (2, -3) and Q (-4, 5) in the ratio 2 : 3(i) internally(ii) externally.

                                            

Solution

i. Let (x, y) be the co-ordinates of the point R which divides the line segment joining the points P (2, -3) and Q (-4, 5) in the ratio 2 : 3 internally, thenx = [2.(-4) +3.2]/(2+3) = - 2/5 andy = [2.5 +3.(-3)]/(2+3) = 1/5Hence the co-ordinates of R are (-2/5, 1/5)

ii. Let (x, y) be the co-ordinates of the point R which divides the line segment joining the points P (2, - 3) and Q (-4, 5) in the ratio 2 : 3 externally i.e.internally in the ratio 2 : -3.

                             x = [2.(-4) + (-3).2]/[2 +(-3)] = -14/1 = 14and y = [2.5 + (-3)(-3)]/[2 +(-3)] = 19/(-1) -19Hence the co-ordinates of R are (14, -19).

Example

In what ratio is the line segment joining the points (4, 5) and (1, 2) divided by the y-axis? Also find the co-ordinates of the point of division.

Solution

Let the line segment joining the points A (4, 5) and B (1, 2) be divided by the y-axis in the ratio k : 1 at P.By section formula, co-ordinates of P are ((k +4)/(k+1), (2k +5)/(k+1)).But P lies on y-axis, therefore, x-coordinate of P = 0=>    (k +4)/(k+1) = 0   =>     k +4 = 0      =>    k = -4The required ratio is -4 : 1 or 4 : 1 externally.Also the co-ordinates of the point of division are    (0, (2.(-4) +5)/(-4+1)) i.e (0, 1)

Exercise

1. Find the co-ordinates of the point which divides the join of the points (2, 3) and (5, -3) in the ratio 1 : 2(i) internally(ii) externally.

2. Find the co-ordinates of the point which divides the join of the points (2, 1) and (3, 5) in the ratio 2 : 3(i) internally(ii) externally.

3. Find the co-ordinates of the point that divides the segment [PQ] in the given ratio:(i) P (5, -2), Q (9, 6) and ratio 3 : 1 internally.(ii) P (-7, 2), Q (-1, -1) and ratio 4 : 1 externally.

4. Find the co-ordinates of the points of trisection of the line segment joining the points (3, - 1) and (-6, 5).

5. Find point (or points) on the line through A (- 5, -4) and B (2, 3) that is twice as far from A as from B.

6. Find the point which is one-third of the way from P (3, 1) to Q (-2, 5). 7. Find the point which is two third of the way from P(0, 1) to Q(1, 0). 8. Find the co-ordinates of the point which is three fifth of the way from (4, 5) to (-1, 0). 9. If P (1, 1) and Q (2, -3) are two points and R is a point on PQ produced such that PR

= 3 PQ, find the co-ordinates of R. 10. In what ratio does the point P (2, -5) divide the line segment joining the points A (- 3,

5) and B (4, -9)?

11. In what ratio is the line joining the points (2, - 3) and (5, 6) divided by the x-axis? Also find the co-ordinates of the point of division.

12. In what ratio is the line joining the points (4, 5) and (1, 2) divided by the x-axis? Also find the co-ordinates of the point of division.

13. In what ratio is the line joining the points (3, 4) and (- 2, 1) divided by the y-axis? Also find the co-ordinates of the point of division.

14. Point C (-4, 1) divides the line segment joining the points A (2, - 2) and B in the ratio 3 : 5. Find the point B.

15. The point R (-1, 2) divides the line segment joining P (2, 5) and Q in the ratio 3 : 4 externally, find the point Q.

16. Find the ratio in which the point P whose ordinate is 3 divides the join of (-4, 3) and (6, 3), and hence find the co-ordinates of P.

17. By using section formula, prove that the points (0, 3), (6, 0) and (4, 1) are collinear. 18. Points P, Q, R are collinear. The co-ordinates of P, Q are (3, 4), (7, 7) respectively

and length PR = 10 unit, find the co-ordinates of R. 19. The mid-point of the line segment joining (2 a, 4) and (-2, 3 b) is (1, 2 a +1). Find the

values of a and b. 20. The center of a circle is (-1, 6) and one end of a diameter is (5, 9), find the co-

ordinates of the other end. 21. Show that the line segments joining the points (1, - 2), (1, 2) and (3, 0), (-1, 0) bisect

each other. 22. Show that the points A(-2, -1), B (1, 0), C (4, 3) and D (1, 2) from a parallelogram. Is

it a rectangle? 23. The vertices of a quadrilateral are (1, 4), (- 2, 1), (0, -1) and (3, 2). Show that the

diagonals bisect each other. What does quadrilateral become? 24. Three consecutive vertices of a parallelogram are (4, - 11), (5, 3) and (2, 15). Find

the fourth vertex.

Answers

1. (i) (3, 1)      (ii) (-1, 9)           2. (i) (12/5, 13/5)    (ii) (0, - 7)3. (i) (4, 8)    (ii) (1, - 2)           4. (0, 1) and (-3, 3)5. (-1/3, 2/3) and (9, 10)          6. (4/3, 7/3)7. (2/3, 1/3)                               8. (1, 2)9. (4, -11)                                 10. 5 : 2 internally11. 1 : 2 internally; (3, 0)            12. 5 : 2 externally; (-1 , 0)13. 3 : 2 internally                      14. (- 14, 6)15. (3, 6)                                    16. 3 : 2 internally; (2, 3)18. (11 , 10)                               19. a = 2, b = 220. (-7 , 3)                                 22. No23. Parallelogram                        24. (1, 1)  

Centroid and Incenter

The point which divides a median of a triangle in the ratio 2 : 1 is called the centroid of the triangle. Thus, if AD is a median of the triangle ABC and G is its centroid, then      AG/GD = 2/1

         By section formula, the co-ordinates of G are

       The symmetry of the co-ordinates of G shows that it also lies on the medians through B and C. Hence the medians of a triangle are concurrent.Incenter of a triangleThe point of the intersection of any two internal bisectors of the angles of a triangle is called the incenter of the triangle. It is usually denoted by I.

         If the internal bisector of A of a ABC meets the side BC in D, then        BD/DC = AB/ACBy section formula, the co-ordinates of I are

The symmetry of the co-ordinates of I shows that it also lies on the internal bisector of C. Hence the internal bisectors of the angles of a triangle are concurrent.

Illustrative Examples

Example

Find the co-ordinates of the incenter of the triangle whose vertices are(-2,4), (5,5) and (4,-2).

Solution

Let A (-2,4), B (5,5) and C (4,-2) be the vertices of the given triangle ABC, thena = | BC| = [(4 -5)² +(-2 -5)²] = [1 +49] = 50 = 5 2,b = |CA| = [(4 -2)² +(-2 -4)²] = [36 +36] = 72 = 6 2 andc = |AB| = [(5 -2)² +(5 -4)²]   = [49 +1] = 50 = 5 2.The co-ordinates of the incenter of ABC are

          

     

      

Exercise

1. Find the centroid of the triangle whose vertices are (-1,4), (2,7) and (-4,-3). 2. Find the point of intersection of the medians of the triangle whose vertices are (3,-5),

(-7,4) and (10,-2). [Hint. Find centroid.] 3. Find the third vertex of a triangle if two of its vertices are (3,-5) and (-7,4), and the

medians meet at (2,-1). 4. Find the centroid of the triangle ABC whose vertices are A(9,2), B(1,10) and C(-7,-6).

Find the co-ordinates of the middle points of its sides and hence find the centroid of the triangle formed by joining these middle points. Do the two triangles have same centroid?

5. If (-1,5), (2,3) and (-7,9) are the middle points of the sides of a triangle, find the co-ordinates of the centroid of the triangle.

6. If A(1, 5), B (-2,1) and C(4,1) are the vertices of ABC, and internal bisector of A meets side BC at D, find |AD|. Also find the incenter of ABC.

7. Find the co-ordinates of the center of the circle inscribed in a triangle whose angular points are (-36,7), (20,7) and (0,-8).

Answers

1. (-1,8/3)          2. (2,-1)                  3. (10,-2)4. (1,2); mid-points of BC, CA, AB are (-3,2), (1,-2), (5,6); (1,2); Yes5. (-2,17/3)       6. 4 units; (1,5/2)    7. (-1,0)  

Slope of a Straight Line

Slope (or gradient) of a straight line

If ( 90°) is the inclination of a straight line, then tan is called its slope (or gradient). The slope of a line is usually denoted by m.Remark. Since tan is not defined when = 90°, therefore, the slope of a vertical line is not defined.

Slope of the line joining two points

The slope m of a non-vertical line passing through the points P(x1 , y1) and Q(x1, y1) is given by     slope = m = (y2 -y1)/(x2 -x1)

Remarks

Two (non-vertical) lines are parallel iff their slopes are equal. Two (non-vertical) lines are perpendicular iff the product of their slopes = -1. Slope of a perpendicular line is the negative reciprocal of the slope of the given line.

Illustrative Examples

Example

Without using Pythagoras theorem, show that the points A (1, 2), B (4, 5) and C (6, 3) are the vertices of a right-angled triangle.

Solution

In ABC, we havem1 = slope of AB = (5 -2)/(4-1) = 3/3 = 1 andm2 = slope of BC = (3 -5)/(6-4) = 2/2 = -1m1 m2 = 1. (-1) = -1    => AB BCHence, the given points are the vertices of a right-angled triangle.

Example

Using slopes, show that the points A (6, -1), B (5, 0) and C (2, 3) are collinear.

Solution

Slope of AB = [0 -(-1)]/(5-6) = 1/(-1) = -1and slope of AC = [3 -(-1)]/(2-6) = 4/(-4) = -1=>  slope of AB = slope of AC=>  AB and AC are parallel.But AB and AC have point A is common, therefore, the given points A, B and C are collinear.

Exercise

1. Find the slope of a line whose inclination is:(i) 30°         (ii) 2 /3     (iii) /3

2. Find the inclination of a line whose gradient is:(i) 1/ 3     (ii) -1       (iii) - 3

3. Find the gradient of the line containing the points(i) (-2, 3) and (5, -7)             (ii) (3, -7) and (0, 2)

4. A line passes through the points (4, -6) and (-2, -5). Does it make an acute angle with the positive direction of x-axis?

5. Find the equation of the locus of all points P such that the slope of the line joining origin and P is - 2.

6. Show that the line joining (2, -3) and (-5, 1) is(i) parallel to the line joining (7, -1) and (0, 3)(ii) perpendicular to the line joining (4, 5) and (0, - 2)

7. State, whether the two lines in each of the following problems are parallel, perpendicular or neither:

(i) through (2, -5) and (-2, 5); through (6, 3) and (1, 1)(ii) through (5, 6) and (2, 3); through (9, -2) and (6, - 5)(iii) through (8, 2) and (-5, 3); through (16, 6) and (3, 15)

8. Find y if the slope of the line joining (-8, 11), (2, y) is - 4/3. 9. Find the value of x so that the line through (x, 9) and (2, 7) is parallel to the line

through (2, - 2) and (6, 4). 10. Without using Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are

the vertices of a right angled triangle.

Answers

1. (i) 1/ 3            (ii) - 3        (iii) not defined2. (i) 30°                (ii) 135°        (iii) 120°3. (i)- 10/7              (ii) -34. No                      5. 2x +y = 07. (i) Perpendicular  (ii) parallel        (iii) neither8. -7/3                    9. 10./3  

General Form of a Circle

The equation x² +y² +2 gx +2 fy +c = 0 represents a circle iff g² +f² -c > 0.Its center is (-g, -f) and radius = [g² +f² -c].Concentric circles. Circles having same center are called concentric circles.Equal circles Circles having equal radius are called equal circles.

Illustrative Examples

Example

One end of a diameter of the circle x² +y² -6 x +5 y -7 = 0 is (-1, 3). Find the co-ordinates of the other end.

Solution

The given equation is   x² + y² -6 x +5 y -7 = 0It is easy to see that this represents a circle with center C (3, - 5/2)Let A(-1,3) and B( , ) be the ends of the diameter.Since C is mid-point of [AB], we get   ( - 1)/2 = 3 and ( +3) / 2 = - 5/2    = 7 and = -8Hence the other end of the diameter is (7, -8).

Example

Show that the points (7, 5), (6, -2), (-1, -1) and (0, 6) are concyclic. Also find the radius and the center of the circle on which they lie.

Solution

Let us find the equation of the circle passing through the points(7, 5), (6, - 2) and (-1, -1).Let the equation of this circle be   x² + y² + 2gx + 2fy + c = 0            ...(i)As the points (7, 5), (6, -2) and (-1, -1) lie on it, we get49 +25 +14 g +10 f +c  => 14 g +10 f +c +74 = 0        ...(ii)36 +4 +12 g -4 f +c = 0     =>      12 g -4 f + c +40 = 0     ...(iii)1 -2 g -2 f +c = 0   =>  2 g +2 f -c -2 = 0          ...(iv)Adding (ii) and (iv), we get16 g +12 f +72 = 0   =>   4 g +3 f +18 = 0 ...(v)Adding (iii) and (iv), we get14 g -2 f +38 = 0      =>     7 g - f +19 = 0 ...(vi)Solving (v) and (vi) simultaneously, we get g = -3, f = -2.     ...(vi)From (ii), we get c = -14 (-3) -10 (-2) -74 = -12.Substituting these values of g, f and c in (i), we get

x² +y² -6 x -4 y -12 = 0 ...(vii)The fourth point (0, 6) will lie on (vii) if 0 +36 -0 -24 -12 = 0 i.e. if 0 = 0, which is true.Hence the given points are concyclic.Also, (vii) is the equation of the circle on which these points lie.Its center is (3, 2) and radius = [9 +4 -(-12)] = 5.

Exercise

1. Which of the following equations represent a circle? If so, determine its center and radius:(i) x² + y² +4 x -4 y -1 = 0(ii) 2 x² +2 y² = 3 x -5 y +7(iii) x² +y² +4 x +2 y +14 = 0(iv) 2 x² +2 y² = 5 x +7 y + 3(v) (x +3)² +(y -2)² = 0(vi) x² + y² - a x - b y = 0

2. Find the value of p so that x² + y² +8 x; +10 y +p = 0 is the equation of a circle of radius 7 units.

3. The radius of the circle x² +y² -2 x +3 y +k = 0 is 2. Find the value of k. Find also the equation of the diameter of the circle which passes through the point.

4. (i) Find the equation of the circle the end points of whose one diameter are the centers of the circles x² +y² +6 x -14 y +5 = 0 and x² + y² -4 x +10 y +7 = 0.(ii) One end of a diameter of the circle x² + y² -3 x +5 y -4 = 0 is (2, 1), find the co-ordinates of the other end.

5. Find the equation of the circle concentric with the circle x² +y² -8 x +6 y -5 = 0 and passing through the point (-2, -7).

6. Find the equation of the circle which passes through the center of the circle x² +y² -4 x -8 y -41 = 0 and is concentric with the circle x² +y² -2 y +1 = 0.

7. Find the equation of the circle concentric with the circle 2 x² +2 y² +8 x +10 y -35 = 0 and with area 16 square units.

8. Find the equation of the circle which is concentric with the circle x² + y² -4 x +6 y -3 = 0 and of double its (i) circumference (ii) area.

9. Prove that the centres of three circles x² + y² -4 x -6 y -14 = 0, x² + y² +2 x +4 y -5 = 0 and x² + y² -10 x -16 y +7 = 0 are collinear.

10. Prove that the radii of the circles x² + y² = 1, x² + y² -2 x -6 y -6 = 0 and x² + y² -4 x -12 y -9 = 0 are in A.P.

11. Find the equation of the circle which has its center on the line y = 2 and which passes through the points (2, 0) and (4, 0).

12. Find the equation of the circle which passes through the points (1, - 2), (4, -3) and has its center on the line 3 x +4 y +10 = 0.

13. Find the equation of the circle passing through the three points(i) (0, 0), (0, 1) and (2, 3)(ii) (0, 2), (3, 0) and (3, 2);Also find its center and radius.

Answers

1. (i) circle; (- 2 , 2), 3     (ii) circle; (3/4, 5/4), (3 10)/4     (iii) empty set   (iv) circle; (5/4, 7/4), (v) point circle; (-3, 2), zero     (vi) circle;(a/2, b/2)2. -8                  3. -3; 2 x -2 y = 54. (i) x² + y² +x -2 y -41 = 0      (ii) (1, -6)5. x² +y² -8 x +6 y -27 = 06. x² + y² -2 y -12 = 07. 4 x² +4 y² +16 x +20 y -23 = 08. (i) x² + y² -4 x +6 y -51 = 0    (ii) x² + y² -4 x +6y -19 = 011. x² +y² -6 x -4 y +8 = 012. x² +y² -4 x +8 y +15 = 013. (i) x² +y² -5 x - y = 0 -y = 0  ; (5/2, 1/2), 26 /2      (ii) x² +y² -3 x -2 y = 0  ;  (3/2, 1), 13 /2  

Equation of a Straight Line

Important Definitions, Results and Formulae

1. The angle which a line makes with the positive direction of x-axis measured in the   anti-clockwise direction is called the inclination of the line.

2. If ( 90°) is the inclination of a line, then tan is called its slope (or gradient). 3. Slope of a line.

(i) If the inclination of a line is , its slope = m = tan .(ii) Slope of a line through (x1, y1) and (x2, y2) is given by     m = (y2 -y1)/(x2 -x1)

4. Equation of a straight line.(i) Equation of a line parallel to x-axis is y = b.(ii) Equation of a line parallel to y-axis is x = a.(iii) Equation of a line with slope m and y-intercept c is y = m x +c.(iv) Equation of a line through (x1, y1) and with slope m is y -y1 = m (x -x1).

5. Conditions of parallelism and perpendicularity.Two lines with slopes m1 and m2 are(i) parallel if and only if m1 = m2

(ii) perpendicular if and only if m1m2 = -1 6. Reflection of P( , ) in the line y = x is P'( , ).

Exercise

1. Find the inclination of a line whose gradient is(i) 1   (ii) 3   (iii) 1/ 3

2. Find the equation of a straight line parallel to x-axis and passing through the point (2, -7).

3. Find the equation of a straight line whose(i) gradient = 3 , y-intercept = -4/3(ii) inclination = 30°, y-intercept = -3

4. Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12. 5. The equation to the line PQ is 3y -3x +7 = 0.

(i) Write down the slope of the line PQ.(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.

6. The given figure represents the lines y = x +1 and y = 3 x -1. Write down the angles which the lines make with the positive direction of x-axis. Hence determine .[Hint. Ext. = sum of two opposite int. s; 60° = +45°]

                7. Given that (a, 2a) lies on the line& y/2 = 3x -6, find the value of a. 8. The graph of the equation y = mx +c passes through the points (1, 4) and (-2, -5).

Determine the values of m and c. 9. Find the equation of a straight line passing through (-1, 2) and having y-intercept 4

units. 10. Find the equation of a st. line whose inclination is 60° and passes through the point

(0, -3). 11. Given that the line y/2 = x -p and the line ax +5=3y are parallel, find the value of a. 12. Find the value of m, if the lines represented by 2mx -3y = 1 and y = 1 -2x are

perpendicular to each other. 13. If the lines 3x +y = 4, x -ay +7 = 0 and bx +2y +5 = 0 form three consecutive sides of

a rectangle, find the values of a and b. 14. Find the equation of a straight line perpendicular to the line 2x +5y +7 = 0 and with y-

intercept - 3 units. 15. Find the equation of a straight line parallel to the line 2x +3y = 5 and having the same

y-intercept as x +y +4 = 0. 16. Find the equation of the line which is parallel to 3x -2y -4 = 0 and passes through the

point (0, 3). 17. Write down the equation of the line perpendicular to 3x +8y = 12 and passing through

the point (-1, -2). 18. The co-ordinates of two points E and F are (0, 4) and (3, 7) respectively. Find

(i) the gradient of EF(ii) the equation of EF(iii) the co-ordinates of the point where the line EF intersects the x-axis.

19. Find the equation of the line passing through the points (4, 0) and (0, 3). Find the value of k, if the line passes through (k, 1·5).

20. If A (-1, 2), B (2, 1) and C (0, 4) are the vertices of a ABC, find the equation of the median through A.

21. Find the equation of a line passing through the point (-2, 3) and having x-intercept 4 units.[Hint. Since x-intercept is 4, the line passes through (4, 0)]

22. Find the equation of the st. line containing the point (3, 2) and making positive equal intercepts on the axes.

23. The intercepts made by a st. line on the axes are -3 and 2 units. Find:(i) the gradient of the line.(ii) the equation of the line.(iii) the area of the triangle enclosed between the line and the co-ordinate axes.

24. A line through the point P (2, 3) meets the co-ordinate axes at points A and B. If PA = 2 PB, find the co-ordinates of A and B. Also find the equation of the line AB.

         25. Calculate the co-ordinates of the point of intersection of the lines represented by x +y

= 6 and 3x -y = 2. 26. The line joining the points P (4, k) and Q (-3, -4) meets the x-axis at A (1, 0) and y-

axis at B. Find(i) the value of k.   (ii) the ratio of PB : BQ.

27. Find the equations of the diagonals of a rectangle whose sides are x = -1, x = 2, y = -2 and y = 6.

28. Find the co-ordinates of the image of (3, 1) under reflection in x-axis followed by reflection in the line x =1.

29. If P' (-4, -3) is the image of the point P under reflection in the origin, find(i) the co-ordinates of P.(ii) the co-ordinates of the image of P under reflection in the line y = -2.

30. Find the co-ordinates of the image of the point P (4, 3) under reflection in the x-axis followed by reflection in the line x = -2.

Answers

1. (i) 45° (ii) 60° (iii) 30°                2. y +7 = 03. (i) 3 3x -3y -4 = 0  (ii) x - 3y -3 3 = 0.    4. -2/3; 45. (i) 1 (ii) 45°              6. 45°, 60°; 15°           7. 38. m = 3, c = 1             9. y = 2x +4                 10. 3x -y -3 = 011. 6                             12. 3/4                         13. a = 3, b = 614. 5x -2y -6 = 0          15. 2x +3y +12 = 0      16. 3x -2y +6 = 017. 8x -3y +2 = 0         18. (i) 1 (ii) x -y +4 = 0 (iii) (-4, 0)19. 3x +4y -12 = 0; 2   20. x -4y +9 = 0           21. x +2y -4 = 022. x +y -5 = 023. (i) 2/3      (ii) 2x -3y +6 = 0 (iii) 3 sq. units24. A (6, 0), B(0,9/2); 3x +4y -18 = 025. (2, 4).                       26. (i) 3 (ii) 4 : 327. 8x -3y +2 = 0, 8x +3y -10 = 0                    28. (-1, -1)29. (i) (4, 3) (ii) (4, -7)                                       30. (-8, -3)  

Equation of a Straight Line in different forms

Slope-Intercept form

The equation of a straight line having slope m and making an intercept c on y-axis is            y = m x +c

Point-slope form

The equation of a straight line passing through the fixed point (x1, y1) and having slope m is       y -y1 = m (x -x1)Corollary. The equation of a line passing through origin and having slope m is y = m x.

Two-point form

The equation of the line passing through two fixed points A (x1, y1) and B (x2,y2) isy - y1 = [(y2 - y1)/(x2 - x1)](x -x1)

Intercept form

The equation of the line cutting off intercepts a and b on the axes is      x/a + y/b = 1

Normal (or perpendicular) form

The equation of a straight line in terms of the length of perpendicular from the origin upon it and the angle which this perpendicular makes with the positive direction of x-axis is given by     x cos + y sin = p

General form

Every straight line can be represented by an equation of the first degree in x and y, and conversely every first degree equation in x, y represents a straight line.The equation A x +B y +C = 0 (where at least one of A and B is non-zero) is called the general form.

Illustrative Examples

Example

Find the equation of a straight line whose inclination is 5 /6 and which cuts off an intercept of 4 units on negative direction of y-axis.

Solution

Let m be the slope of the line, thenm = tan 5 /6 = tan 150° = tan (180° -30°) = -tan 30° = 1/ 3Also c = y-intercept = -4So the equation of the line is y =(-1/ 3) x +(-4)               (As y = m x +c)i.e.      x + 3y + 4 3 = 0

Example

Show that the points (a, 0), (0, b) and (3 a, -2 b) are collinear. Also find the equation of the line containing them.

Solution

The equation of the line through (a, 0) and (0, b) isy -0 =[(b-0)/(0-a)](x -a)                  y - y1 = [(y2 - y1)/(x2 - x1)](x -x1)i.e. -ay = bx -ab       i.e. bx +ay -ab = 0The third point (3 a, -2 b) lies on it if b. 3 a + a. (-2 b) - ab = 0 i.e. if 0 = 0, which is true.Hence the given points are collinear and the equation of the line containing them is     bx +ay -ab = 0.

Example

In what ratio is the line joining the points (2, 3) and (4, -5) divided by the line joining the points (6, 8) and (-3,-2)?

Solution

The equation of the straight line joining the points (6, 8) and (-3, -2) isy -8 = [(-2 -8)/(-3 -6)](x -6)                                  (Two-point form)=> y -8 = (10/9) (x -6)   =>  9 y -72 = 10 x -60=> 10 x -9 y +12 = 0         ...(i)Let the line joining the points (6, 8) and (-3, -2) i.e. the line (i) divide the line segment joining

the points (2, 3) and (4, -5) at the point P in the ratio k : 1, then the co-ordinates of point P are((k.4 +1.2)/(k+1), (k (-5) +1 .3)/(k+1)) i.e. ((4k +20)/(k+1), (-5k +3)/(k+1))Since P lies on (i), we get10[(4k +20)/(k+1)] -9[(-5k +3)/(k+1)] + 12 = 0=> 40 k +20 +45 k -27 +12 k +12 = 0=> 97 k +5 = 0 => k = -5/97Hence the required ratio is -5/97 i.e. 5 : 97 externally.

Example

Prove that the locus of the point which is equidistant from the points (-3, 7) and (2, -5) is a straight line.

Solution

Let A (-3, 7) and B (2, -5) be the given points and P (x, y) be any point on the locus, then | AP | = | BP | (given)=>  (x +3)² +(y -7)² = (x -2)² +(y +5)²=> x² +6 x +9 + y² -14 y +49 = x² -4 x +4 + y² +10 y +25=> 10 x -24 y +29 = 0, which is a first degree equation in x and y, and so it represents a straight line.Hence the locus is a straight line. In fact, this Is the perpendicular bisector of the line segment joining the two given points.

Exercise 1

1. Find the equation of a straight line parallel to x-axis at a distance(i) 3 units above it(ii) 3 units below it.

2. Find the equation of a straight line parallel to y-axis at a distance(i) 2 units to the right(ii) 2 units to the left of it.

3. Find the equation of a horizontal line passing through (5, -2). 4. Find the equation of a vertical line passing through (-7, 3). 5. Find the equation of a line which is equidistant from the lines y +5 = 0 and y -2 = 0. 6. Find the equation of a line whose

(i) gradient = -1, y-intercept = 3(ii) slope = - 2/7, y-intercept = -3(iii) inclination = 3 /4, y-intercept = -5

7. Find the equations of the straight lines cutting off an intercept of 3 units from the negative direction of y-axis and equally inclined to the axes.[Hint. The lines which are equally inclined to the axes have slope = ± 1]

8. Find the equation of the line passing through the point (2, -5) and making an intercept of -3 on the y-axis.

9. If the straight line y = mx +c passes through the points (2, 4) and (-3, 6) find the values of m and c.

10. Find the equation of a straight whose y-intercept is -5 and which is(i) parallel to the line joining the points (3, 7) and (-2, 0)(ii) perpendicular to the line joining the points (-1, 6) and (-2, -3).

11. Find the equation of a straight line passing through origin and making an angle of 120° with the positive direction of x-axis.

12. Find the equations of the bisectors of the angles between the co-ordinate axes. 13. Find the equation of the straight line passing through the point (5, 7) and inclined at

45° to x-axis. If it passes through the point P whose ordinate is -7, what is the abscissa of P?

14. Find the equation of the line through the point (-5, 1) and parallel to the line joining the points (7, -1) and (0, 3).

15. Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining (1, 4), (2, 3).

Answers

1. (i) y -3 = 0             (ii) y +3 = 02. (i) x -2 = 0             (ii) x +2 = 03. y +2 = 0                  4. x +7 = 05. 2y +3 = 06. (i) x +y - 3 = 0        (ii) 2 x +7 y +21 = 0   (iii) x +y +5 = 0

7. x +y +3 = 0, x -y -3 = 08. x +y +3 = 0              9. m = -2/5, c = 24/510. (i) 7 x -5 y -25 = 0 (ii) x +9 y +45 = 011. 3 x +y = 0           12. x - y = 0, x +y = 013. x - y +2 = 0; -9      14. 4 x +7 y +13 = 015. x -y +3 = 0

Exercise 2

1. State the geometrical meaning of the constants involved in(i) x/a + y/b = 1(ii) x cos + y sin = p(iii) (x -x1)/cos = (y -y1)/ sin = r

2. Find the equation of the line which cuts off intercepts 3 and 4 from the axes. 3. Prove that the straight line whose intercepts on the axes are 2 and -3 respectively

passes through the point (4, 3). 4. Find the equation of a straight line which passes through the point (1, -3) and makes

an intercept on y-axis twice as long as on x-axis. 5. Find the equation of the straight line which passes through the point (3, -4) and

makes(i) equal intercepts on the axes(ii) intercepts equal in magnitude but opposite in sign on the axes.

6. Find the equation of the straight line which passes through the point (3, -2) and cuts off positive intercepts on the x-axis and y-axis which are in the ratio 4 : 3.

7. A straight line passes through the point (2, 3) and the portion of the line intercepted between the axes is bisected at this point, find its equation.

8. A straight line is such that the portion of it intercepted between the axes is bisected at the point (x1 , y1). Find its equation.

9. A straight line passes through the point (1, 1) and the portion of the line intercepted between the axes is divided at this point in the ratio 3 : 4, find its equation.

10. Find the equation of a line which passes through the point (-5, 2) and whose segment between the axes is divided by this point in the ratio 2 : 3.

11. If the straight line x/a + y/b = 1passes through the points (12, -15) and (8, -9), find the values of a and b.

12. Find the equation of a line which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on y-axis by 5.

13. Find the equation of a line which passes through (-3, 10) and sum of its intercepts on the axes is 8.

14. A straight line passes through the points (a, 0) and (0, b), the length of the line segment contained between the axes is 13 and the product of the intercepts on the axes in 60. Calculate the values of a and b and find the equation of the straight line.

15. The area of a triangle formed by a line and the co-ordinate axes is 6 sq. units and the length of the segment intercepted between the axes is 5 units. Find the equation (s) of line.

Answers

1. (i) a and b are the intercepts made by the line on x-axis and y-axis respectively.    (ii) p is the length of the perpendicular drawn from origin on the line and the angle which this perpendicular makes with the x-axis.   (iii) (x1, y1) is the point through which the line passes, is the inclination of the line and r is the directed distance of any point (x, y) on the line from the point (x1, y1).2. 4 x +3 y = 124. 2 x +y +1 = 05. (i) x +y +1 = 0 (ii) x -y = 76. 3 x +4 y = 17. 3 x +2 y -12 = 08. x/x1 + y/y1 = 29. 4 x +3 y = 710. 3 x -5 y +25 = 011. a = 2, b = 312. 6 x +11 y -66 = 0, x +2 y -10 = 013. 2 x - y +16 = 0, 5 x +3 y -15 = 014. a = 12, b = 5, 5 x +12 y = 60; a = 5, b = 12, 12 x +5 y = 60;     a = -12, b = -5, 5 x +12 y +60 = 0; a = -5, b = -12, 12 x +5 y +60 = 015. 3 x +4 y = 12, 4 x +3 y = 12; 3 x +4 y +12 = 0, 4 x +3 y +12 = 0  

Angle between two Lines

The angle between two non-vertical and non-perpendicular lines

Let l1 and l2 be the two non-vertical and non-perpendicular lines with slopes m1 and m2 respectively. Let 1 and 2 be their inclinations, then m1 = tan 1 and m2 = tan 2. There are two angles and - between the lines l1 and l2, given bytan = ± (m1-m2)(1+m1m2)

                              

Illustrative Examples

Example

Find the angle between the lines joining the points(-1,2), (3,-5) and (-2,3), (5,0).

Solution

Here, m1 = slope of the line joining (-1,2) and (3,-5)   = (-5-2)/(3+1) = -7/4 and m2 = slope of the line joining (-2,3) and (5,0)   = (0-3)/(5+2) = -3/7Let be the acute angle between the given lines, then

tan = .

          =Hence the acute angle between the lines is given by   tan = 37/49

Exercise

1. Find the angle between the following pairs of lines:(i) 3 x -7 y +5 = 0 and 7 x +3 y -11 = 0(ii) 3 x +y -7 = 0 and x +2 y +9 = 0(iii) y = (2 - 3) x +9 and y = (2 + 3) x +1(iv) 2 x -y +3 = 0 and x +y -2 = 0[Hint. (iv) It will be found that acute angle is given by tan = 3which gives as 71° 34', by using tables of natural tangents]

2. Find the angle between the lines joining the points (0,0), (2,3) and (2,-2), (3,5). 3. If A(-2,1), B(2,3) and C(-2,-4) are three points, find the angle between the lines AB

and BC. 4. Find the angles between the lines x +1 = 0 and 3 x +y -3 = 0. 5. Find the angle between the lines which make intercepts on the axes a,-b and b,-a

respectively. 6. Find the measures of the angles of the triangle whose sides lie along the lines x +y -5

= 0,  x -y +1 = 0 and y -1 = 0. 7. Find the equations of the two straight lines passing through the point (4,5) which

make an acute angle of 45° with the line 2 x-y +7 = 0. 8. Find the equations of the two straight lines passing through the point (1,-1) and

inclined at an angle of 45° to the line 2 x -5 y +7 = 0. 9. A vertex of an equilateral triangle is (2,3) and the equation of the opposite side is x

+y +2 = 0. Find the equations of the other two sides.

10. One diagonal of a square lies along the line 8 x -15 y = 0 and one vertex of the square is at (1,2). Find the equations of the sides of the square passing through this vertex.

11. If (1,2) and (3,8) are a pair of opposite vertices of a square, find the equations of the sides and the diagonals of the square.

Answers

1.(i) 90° (ii) 45° (iii) 60° (iv) 71° 34'2.   25° 34'   3. 33° 42'   4. 30°

5. The acute angle is given by tan =6. 45°, 45°, 90°7. 3 x +y -17 = 0, x -3 y +11 = 08. 7 x -3 y -10 = 0, 3 x +7 y +4 = 09. (2 + 3) x -y -1 -2 = 0, (2 - 3) x -y -1 +2 = 010. 23 x -7y -9 = 0, 7x +23 y -53 = 0.11. Sides are 2 x +y -4 = 0, x -2 y +3 = 0,     2 x +y -14 = 0, x -2 y +13 = 0 and      diagonals are 3 x -y -1 = 0, x +3 y -17 = 0  

Position of two Points relative to a Line

The points P (x1, y1) and Q (x2 , y2) lie on the same side or on opposite sides of the line a x +b y +c = 0 according as the expressions a x1 + b y1 +c and a x2 + by2 +c have same sign or opposite signs.

Illustrative Examples

Example

The sides of a triangle are given by the equations 3 x +4 y = 10, 4 x -3 y = 5 and 7 x +y +10 = 0. Show that the origin lies with in the triangle.

                                       .

Solution

The given lines are3 x +4 y -10 = 0            ...(i)4 x -3 y -5 = 0               ...(ii)and 7 x + y +10 = 0       ...(iii)Let ABC be the triangle formed by these lines.Solving these equations simultaneously, taking two at a time, the vertices of the triangle areA (-2, 4), B (2, 1) and C (-1, -3)On substituting (-2, 4) in L.H.S. of (ii), we get -8 -12 -5 = -25and substituting (0, 0) in L.H.S. of (ii), we get 0 -0 -5 = -5Since both have same sign, therefore, origin and A lie on the same side of BC.Similarly origin and B lie on the same side of CA; and origin and C lie on the same side of AB. (Please check it)From these results, it follows that the origin lies with in the triangle formed by the given lines.

Exercise

1. Are the points (2, 3) and (-1, 5) on the same side or on opposite sides of the line y = 2 x +5.

2. Show that the points (3, 5) and (-3, -2)lie on the same side of the line 3 x = 7 y + 8. 3. Which of the points (1, 1), (-1, 2), (2, 3) and (-3, 0) lie on the same side of the line 4 x

+3 y = 5 on which the origin lies? 4. Find by calculation whether the points (13, 8), (26, -4) lie in the same, adjacent or

opposite angles formed by the straight line 5 x + 6 y -112 = 0 and 10 x +11 y - 217 = 0.

5. Prove that the point P (x1, y1) and the origin lie on the same side or on opposite sides of the line ax +by +c = 0 according as ax1 + by1 + c and c have same sign or opposite signs.

Answers

1. Opposite sides3. (-1, 2) and (-3, 0)4. Opposite  

Distance of a Point from a Line

The perpendicular distance d of a point P (x 1, y 1) from the line ax +by +c = 0 is given by           d =| ax1 +by1 +c|/[ (a² +b²)]

                        

Rule to find the distance between parallel lines

i. Choose a point on one of the given parallel lines. ii. Find the perpendicular distance from this point to the other line.

Illustrative Examples

Example

Find the equation of a straight line, with a positive gradient, which passes through the point (-5, 0) and is at a perpendicular distance of 3 units from the origin.

Solution

Let m (> 0) be the gradient of the line, then any line through (-5, 0) and with gradient m is      y -0 = m(x +5) i.e. mx -y +5m = 0 ...(i)It will be the required line if its perpendicular distance from origin (0, 0) is 3 units=> |m.0 -0 +5 m| /[ (m² +(-1)²)] = 3      => | 5 m | = 3 [m² +1]=> 25 m² = 9 (m² +1)      => 16 m² = 9     => m = 3/4        (m > 0)Substituting this value of m in (i), the equation of the required line is      (3/4) x -y + 5.3/4 = 0 or 3 x -4 y +15 = 0

Example

Find the distance between the lines 3 x -4 y +7 = 0 and 6 x -8 y = 18.

Solution

The given lines are       3 x -4 y +7 = 0       ...(i)and 6 x -8 y -21 = 0      ...(ii)We note that the slope of (ii) = - 6/(-8) = 3/4 = the slope of (i)=> the given lines are parallel.To find distance between these lines, we choose a point on (i).

On putting x = 0 in (i), we get -4y +7 = 0 => y = 7/4Thus (0, 7/4) is a point on (i).Required distance between given parallel lines   = perpendicular distance from (0, 7/4) to the line (ii)

    = = |-35 | = units

Exercise

1. Find the distance of the point P from the line AB in the following cases:(i) P (2, -3), line AB is 2 x -3 y -25 = 0(ii) P (4, 1), line AB is 3 x -4 y -9 = 0(iii) P (0, 0), line AB is h (x +h) +k (y +k) = 0

2. Find the distance of the point (0, - 1) from the line joining the points (1, 3) and (-2, 6). 3. Calculate the length of the perpendicular from (7, 0) to the straight line 5 x +12 y -9 =

0 and show that it is twice the length of perpendicular from (2, 1). 4. Find the value (s) of k, given that the distance of the point (4, 1) from the line 3 x -4y

+k = 0 is 4 units. 5. The points A (0, 0), B (1, 7), C (5, 1) are the vertices of a triangle. Find the length of

perpendicular from A to BC and hence the area of ABC. 6. Find the lengths of altitudes of the triangle whose sides are given by x -4 y = 5, 4 x

+3 y = 5 and x +y = 1. 7. Find the length of perpendicular from the point (4, -7) to the line joining the origin and

the point of intersection of the lines 2 x -3 y +14 = 0 and 5 x +4 y -7 = 0. 8. A vertex of a square is at the origin and its one side lies along the line x -4 y -10 = 0.

Find the area of the square.

Answers

1. (i) 12/ 13 units         (ii) 1/5 units           (iii) [h² +k²] units2. 5/ 2 units                   3. 2 units               4. 12, -285. 17/ 13 units, 17 sq. units6. 1 unit, 1/7 units, 1/[5 2] units7. 1 unit                              8. 4 sq. units  

Families of Lines

One parameter family of lines

It may seem that the equation of a straight line ax +by +c = 0 contains three arbitrary constants. In fact, it is not so. On dividing it by a (or b, which ever is non-zero), we get    x + y(b/a) + (c/a),which can be written as x + By + C = 0 where B =b/a and C = c/a.It follows that the equation of a straight line contains two arbitrary constants, and the number of these arbitrary constants cannot be decreased further. Thus, the equation of every straight line contains two arbitrary constants, consequently, two conditions are needed to determine the equation of a straight line uniquely.One condition yields a linear relation among two arbitrary constants and hence each arbitrary constant determines the other. Therefore, the lines which satisfy one condition contain a single arbitrary constant. Such a system of lines is called one parameter family of lines and the unknown arbitrary constant is called the parameter.

Examples of one parameter families

i. The equation y = m x +2, for different real values of m, represents a family of lines with y-intercept 2 units. A few members of this family are shown in figure below.

                          ii. The equation 2 x +3 y + k = 0, for different real values of k, represents a family of

lines with slope -2/3.A few members of this family are shown in figure below.

                     iii. The equation y -y1 = m (x -x1), for different real values of m, represents a family of

lines which pass through the fixed point (x1, y1) except the vertical line x = x1. iv. The equation x = a, for different real values of a, represents the family of lines

parallel to y-axis (including the y-axis itself). v. The equation a x +b y +k = 0, for different real values of k, represents a family of

lines parallel to the line ax +by +c = 0. vi. The equation b x -a y +k = 0, for different real values of k, represents a family of lines

perpendicular to the line ax +by +c = 0. vii. If l1 = a1 x +b1 y +c1 = 0 and l2 = a2 x +b2 y +c2 = 0 then l1 +k l2 = 0, for different real

values of k, represents a family of lines passing through the point of intersection of the lines l1 and l2.

Illustrative Examples

Example

Find the equation of the family of lines with x-intercept -4.

Solution

Since the x-intercept of the family is given to be -4, therefore, each member of the family passes through the point (-4, 0).By using point-slope form, the equation of such a family of lines is y -0 = m (x -(-4)) i.e. y = m (x +4), where m is a parameter.Note. The above equation of the family does not give the vertical line through the point (-4, 0). However, the equation of this line is x = -4 i.e. x +4 = 0.

Example

Find the equation of the straight line which is parallel to 3 x -7 y = 11 and makes x-intercept 3 units.

Solution

The given line is 3x -7 y -11 = 0 ...(i)The equation of the family of lines parallel to (i) is   3 x -7 y+k = 0 ...(ii) where k is a parameter.To find x-intercept, put y = 0, we get 3 x +k = 0  =>  x =k/3For the required member of the family which makes x-intercept 3 units,   k/3 = 3 =>   k = -9Substituting this value of k in (ii), the equation of the required line is    3 x -7 y -9 = 0

Exercise

1. Write the equations of the family of lines:(i) with y-intercept -3(ii) with slope 2(iii) parallel to 2 x +3 +3 y -5 = 0(iv) perpendicular to 3 x +7 +7 y = 8

2. Find the equation of the line through the intersection of the lines 4 x -3 y +7 = 0 and 2 x +3 y +5 = 0 and the point (-4, 5).

3. Find the equation of the straight line parallel to 2 x -5 y +3 = 0 and having x-intercept -4.

4. Find the equations of two straight lines which are parallel to the line x +7 y +2 = 0 and at a unit distance from the point (2, -1).

5. Find the equations of straight lines which are perpendicular to the line 3 x +4 y -7 = 0 and are at a distance of 3 units from (2, 3).

6. Find the equations of the two straight lines drawn through the point (0, a) on which the perpendiculars dropped from the point (2 a, 2) are each of length a.

7. Find the equation of the line which lies mid-way between the lines 2 x +3 y +7 = 0 and 2 x +3 y +5 = 0.

8. Find the equations of straight lines parallel to the lines 3 x - y -3 = 0 and 3 x - y +5 = 0 and whose distances from these lines are in the ratio 3 : 5. Point out the line which lies between the given lines.

9. Find the equation of a straight line parallel to 2 x +3 y = 10 and which is such that the sum of its intercepts on the axes is 15.

10. A line is drawn perpendicular to 5 x = y +7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 5 sq. units.

Answers

1. (i) y = m x -3, m parameter                 (ii) y = 2 x +c, c parameter   (iii) 2x + 3 + 3y + k = 0, k parameter    (iv) 7 x -3 y + k = 0, k parameter2. 8 x +3 +3 y +17 = 0                             3. 2 x-5 y +8 = 04. x +7 y +5 +5 ( 2 +1) = 0, x +7 y -5 ( 2 -1) = 05. 4 x -3 y +16 = 0, 4x + 16 = 0, 4 x -3 y -14 = 06. y -a = 0, 4 x -3 y +3 +3 a = 07. 2 x +3 +3 y +6 = 08. 3 x - y = 0, 3 x - y -15 = 0 ; 3 x - y = 09. 2 x +3 +3 y -18 = 010. x +5 y -5 2 = 0, x +5 y +5 2 = 0  

Orthogonal Circles

Angle of intersection of two circles

An angle between the two tangents to the two circles at a point of intersection is called an angle of intersection between two circles.

                         

Orthogonal circles

Two circles are said to cut orthogonally iff angle of intersection of these circles at a point of intersection is a right angle i.e. iff the tangents to these circles at a common point are perpendicular to each other.If r, r' are radii of circles S and S' respectively and d is the distance between their centers and is an angle of intersection of these circles, then               cos =(r² +r'² -d²)/(2 r r')If the circles S and S' cut orthogonally, then = 90°,cos = cos 90° = 0    => (r² +r'² -d²)/(2r r') = 0    => r² +r'² = d²Two circles S = x² +y² +2gx +2 f y +c = 0 andS' = x² +y² +2 g' x +2 f'y +c' = 0 cut orthogonally if 2 (g g' +f f') = c +c'.

Illustrative Examples

Example

Find the equation of the circle which intersects the circles x² +y² +2 x -2 y +1 = 0 and x² +y² +4 x -4 y +3 = 0 orthogonally and whose center lies on the line 3 x -y -2 = 0.

Solution

Let the equation of the required circle be     x² +y² +2 g x +2 f y + c = 0 ...(i)Since this circle cuts the circles x² +y² +2 x -2 y +1 = 0 and x² + y² +4 x -4 y +3 = 0 orthogonally, we get       2 (g. 1 + f. (-1)) = c +1=>   2 g -2 f - c -1 = 0              ...(ii)and 2 (g.2 + f.(-2)) = c +3=>   4 g -4 f -c -3 = 0               ...(iii)As center of (i) i.e. (-g, -f) lies on 3 x -y -2 = 0, we get     -3 g + f -2 = 0=>    3 g - f +2 = 0                   ...(iv)Subtracting (ii) from (iii), we get       2 g -2 f -2 = 0=>  g -f -1 = 0                            ...(v)Solving (iv) and (v) simultaneously, we get g = -3/2 and f = -5/2From (ii), we get c = 2 g -2 f -1 = -3 +5 -1 = 1Substituting these values of g, f and c in (i), we getx² +y² -3 x -5 y +1 = 0, which is the equation of the required circle.

Exercise

1. Determine the angle of intersection of the two circles (x -3)² +(y -1)² = 8 and (x -2)² +(y +2)² = 2.

2. Find the angle at which the circles x² + y² = 16 and x² +y² -2 x -4 y = 0 intersect each other.

3. Show that the circles x² + y² -4 x  -6 y +4 = 0 and x² +y² -10 x -14 y +58 = 0 cut orthogonally.

4. Show that the circles x² +y² -2 a x +2 b y + c = 0 and x² + y² +2 bx +2 a y -c = 0, a² + b² > | c |, cut orthogonally.

5. For what value of  do the circles x² + y² +5 x +3 y +7 = 0 and x² +y² -8 x +6 y + = 0 cut orthogonally?

Answers

1. /22. The acute angle is given by cos = 2/ 55. -18  

Equation of a Circle in different forms

A circle is the locus of a point which moves in a plane so that it remains at a constant distance from a fixed point. The fixed point is called the center and the constant distance is called radius. Radius is always positive.

Standard (or simplest) form

The equation of a circle with O(0,0) as center and r (>0) as radius is    x² +y² = r²

Central form

The equation of a circle with C(h,k) as center and r (>0) as radius is given by    (x -h)² +(y -k)² = r²

Diameter form

Let A(x1,y1) and B(x2,y2) be the extremities of a diameter of the circle.Then the equation of the circle is(x -x1)(x -x2) + (y -y1)(y -y2) = 0

       

Illustrative Examples

Example

Find the equation of a circle whose center is (3,-2) and which passes through the intersection of the lines 5x +7y = 3 and 2x -3y = 7.

Solution

Given lines are5x +7y -3 = 0    ...(i), and2x -3y -7 = 0    ...(ii)Solving (i) and (ii) simultaneously, we get x = 2, y = -1.So the point of intersection, say P, of the given lines is (2,-1).Since the center of the circle is C(3,-2) and it passes through the point P(2,-1), radius = |CP| = [(2 -3)² +(-1 +2)²] = [1 +1] = 2Hence the equation of the circle is    (x -3)² +(y +2)² = ( 2)²          (central form)or    x² +y² -6x +4y +11 = 0.

Example

Find the equation of a circle which touchesi. the y-axis at origin and whose radius is 3 units ii. both the co-ordinate axes and the line x = 3.

Solution

i. There are two circles satisfying given conditions. As the circles touch y-axis at the origin, their centers lie on x-axis. Since radius is 3 units, centers of the circles are (3, 0) and (-3, 0) and hence the equations of the circles are  (x ±3)² +(y -0)² = 3²     or    x² +y² ±6x = 0

ii.         There are two circles satisfying the given conditions. From above figure, clearly, the centers of these circles are and (3/2,3/2) and (3/2, -3/2) radius of each circle is 3/2. So the required equation is     x² +y² -3x ±3y + (9/4) = 0or  4x² +4y² -12x ±12y +9 = 0

Exercise

1. Find the equation of a circle whose(i) center is at the origin and the radius is 5 units.(ii) center is (-1, 2) and radius is 5 units.

2. Determine the equation of a circle whose center is (8, -6) and which passes through the point (5, -2).

3. Prove that the points (7, -9) and (11, 3) lie on a circle with center at the origin. Also find its equation.

4. Find the equation of the circle whose(i) center is (a, b) and passes through origin.(ii) center is (2, -3) and passes through the intersection of the lines 3x -2y -1 = 0 and 4x +y -27 = 0.

5. Find the equation of a circle whose center is the point of intersection of the lines 2x +y = 4 and x -y = 5 and passes through the origin.

6. Find the equation of a circle whose two diameters lie along the lines 2x -3y +12 = 0 and x +4y + 12 = 0 and x +4y -5 = 0 and has area 154 square units.

7. Find the equation of the circle whose center lies on the negative direction of y-axis at a distance 3 units from origin and whose radius is 4 units.

8. Find the equations of the circles of radius 5 whose centers lie on the x-axis and pass through the point (2,3).

9. Find the equation of the circle(i) whose center is (0, -4) and which touches the x-axis.(ii) whose center is (3, 4) and touches the y-axis.

10. Find the equations of circles which touch both the axes and(i) has radius 3 units(ii) touch the line x = 2a.

11. Find the equations of circles which pass through two points on x-axis at distances of 4 units from the origin and whose radius is 5 units.

12. Find the equations of circles(i) which touch the x-axis on the positive direction at a distance 5 units from the origin and has radius 6 units.(ii) passing through the origin, radius 17 and ordinate of the center is -15.(iii) which touch both the axes and pass through the point (2, 1).

13. Find the equations of circles which touch the y-axis at a distance of 4 units from the origin and cut off an intercept of 6 units along the positive direction of x-axis.

14. Find the equations to the circles touching axis of y at the point (0, 3) and making an intercept of 8 units on x-axis.

15. Find the equations to the circles which touch the x-axis at a distance of 4 units from the origin and cut off an intercept of 6 from the y-axis.

Answers

1. (i) x² +y² = 25  (ii) x² +y² -2x +4y = 02. x² +y² -16x + 12y +75 = 03. x² +y² -130 = 04. (i) x² +y² -2ax -2by = 0   (ii) x² +y² -4x +6y -96 = 05. x² +y² -6x +4y = 0.6. x² +y² +6x -4y -36 = 07. x² +y² +6y -7 = 0.8. x² +y² -12x +11 = 0,     x² +y² +4x -21 = 09. (i) x² +y² +8y = 0   (ii) x² +y² -6x -8y +16 = 010. (i) x² +y² ±6x ±6y +9 = 0    (ii) x² +y² -2ax ±2ay +a² = 011. x² +y² ±6y -16 = 012. (i) x² +y² -10x ±12y +25 = 0     (ii) x² +y² ±6x +30y = 0     (iii) x² +y² -2x -2y +1 = 0,  x² +y² -10x -10y +25 = 013. x² +y² -10x ±8y +16 = 014. x² +y² ±10x -6y +9 = 015. x² +y² ±8x ±10y +16 = 0  

Parametric form of Circle

                                   x = r cos , y = r sin ,0 < 2 represent the circle x² +y² = r², where is called parameter and the point P (r cos , r sin ) is called the point " " on the circle x² +y² = r².

Parametric form of the circle (x -h)² +(y -k)² = r²

Every point P on the circle can be represented as      x = h + r cos , y = k + r sin , 0 < 2 Thus, x = h + r cos , y = k + r sin , 0 < 2 , represent the circle (x -h)² +(y -k)² = r².is called parameter and the point (h +r cos , k +r sin ) is called the point " " on this circle.

Illustrative Examples

Example

Find the parametric equations of the circle x² +y² = 5

Solution

The given circle is x² + y² = 5We know that the parametric equations of the circle x² +y² = r² are     x = r cos , y = r sin , 0 < 2 The given circle is comparable with x² +y² = r², here r = 5Therefore, the parametric equations of the given circle x² +y² = 5 are    x = 5cos , y = 5 sin , 0 < 2

Example

Find the cartesian equations of the curves x = p +c cos , y = q +c sin , where is parameter. Do these equations represent a circle? If so, find center and radius.

Solution

Given x = p +c cos , y = q + c sin =>    x -p = c cos , y -q = c sin To eliminate the parameter , on squaring and adding these equations, we get     (x -p)² + (y -q)² = c² (cos² +sin² )=> (x -p)² +(y -q)² = c²,which represents a circle with center (p, q) and radius = | c |.

Exercise

1. Find the parametric equations of the following circles :(i) x² +y² = 13(ii) (x -2)² +(y +3)² = 36(iii) x² + y² +4 x - 6 y -12 = 0(iv) 2 x² +2 y² = 5 x +7 y +3(v) x² + y² -2 a x - 2 a y = 0(vi) x² + y² + p x +q y = 0

2. Find the cartesian equations of the following curves:(i) x = 2 cos , y = 2 sin (ii) x = 1 +5 cos , y = 2 +5 sin (iii) x = -3 + 7cos , y = 4 + 7 sin ,where is parameter. Do these equations represent circles? If so, find center and radius.

Answers

1. (i) x = 13 cos , y = 13 sin , 0 < 2     (ii) x = 2 +6 cos , y = -3 +6 sin , 0 < 2     (iii) x = -2 +5 cos , y = 3 +5 sin , 0 < 2     (iv) x = cos , y = sin , 0 < 2      (v) x = a +| a | cos , y = a +| a | sin , 0 < 2      (vi) x = -p/2 +(1/2) (p² +q²) cos , y = -p/2 + (1/2) (p² +q²) sin , 0 < 2 2. (i) x² +y² = 4; circle, (0, 0), 2    (ii) (x -1)² +(y -2)² = 25; circle, (1, 2), 5    (iii) (x +3)² +(y -4)² = 7; circle, (-3 , 4), 7  

Point and Circle

Let S be a circle with center C and radius r (> 0) and P be any point in the plane of the circle S, then

i. P is called exterior to S iff | C P | > r, ii. P is called interior to S iff |CP| < r and iii. P is said to lie on S iff |CP| = r.

If P is exterior to S then we say that P lies outside S, and if P is interior to S then we say that P lies inside SNotation. Let S = x² + y² +2 g x +2 f y + c = 0, g² + f² - c > 0, be a circle and P (x1, y1) be a point in the plane of S, then S1 = x1² +y1² +2 g x1 +2 f y1 + c.Let S be a circle and P (x1, y1) be a point in the plane of S, then

i. P is exterior to S iff S1 > 0 ii. P is interior to S iff S1 < 0 iii. P lies on S iff S1 = 0

Corollary.Let S be a circle and P (x1, y1), Q (x2, y2) be two points in the plane of S then they lie

i. on the same side of S iff S1 and S2 have same sign ii. on the opposite sides of S iff S1 and S2 have opposite signs.

Exercise

1. Among the points given below, point out which of these are interior, exterior or lie on the circle S with center (1, 2) and radius 3?(i) (2, 3)           (ii) (-1, -3)         (iii) (4, 2)

2. Among the points given below, point out which of these are exterior, interior or lie on the circle 3 x 2 +3 y 2 -2 x -11 = 0?

(i) (-1, 0)              (ii) (1, -2)                (iii) 3. Do the following pairs of points lie on the same side or on opposite sides of the circle

with center (- 1, 2) and radius?(i) (2, -3) and (1, 2)      (ii) (0, 4) and (1, 3)

Answers

1. (i) Interior       (ii) exterior      (iii) lies on2. (i) Interior       (ii) exterior      (iii) lies on3. (i) Opposite sides                   (ii) both lie on the circle  

Line and Circle

The condition that the line y = mx +c may intersect the circle x² + y² = a² is given by               a²(1 + m²) c²

Remark

The line y = m x + c will intersect the circle x² + y² = a² in two distinct points iff a² (1 +m²) > c², and the line will intersect the circle in one and only one point i.e. the line will be a tangent to the circle iff a²(1 +m²) = c², and the line will not intersect the circle iff a²(1 + m²) < c².

Corollary 1. Condition of the tangency

The line y = mx + c will touch the circle x² +y² = a²   iff a²(1 + m²) = c² i.e. iff c = ± a [1 +m²]

Corollary 2. Equations of tangents in slope form

Substituting the values of c = ± a [1 +m²] in equation y = mx + c, we get    y = mx ± a [1 +m²]Thus, there are two parallel tangents to the circle x² +y² = a² having m as their slope.

Length of intercept made by a circle on a line

Let a line l meet a circle S with center C and radius r in two distinct points. If d is the distance of C from l then the length of intercept = p [r² -d²]

Length of tangent

Let S be a circle and P be an exterior point to S, and PT1, PT2 be two tangents to S through P, then the distance |PT1| or |PT2| is called the length of tangent from P to the circle S.

The length of tangent =

                  

Illustrative Examples

Example

Find the locus of the point of intersection of perpendicular tangents to the circle x² +y² = a²

                     

Solution

he given circle is x² + y² = a²       ...(i)The equation of any tangent to the circle (i) in the slope form is       y = mx +a [1 +m²]                             .... (ii)Let (ii) pass through the point P ( , ), then     = m + a [1 +m²]      - m = a [1 +m²]=> ( - m )² = a²(1 + m²)=>  ² + m 2 ² - 2 m - a² - a² m² = 0=>  ( ² - a²) m² - 2 m + ( ² - a²) = 0,which is a quadratic in m having two roots, say m1, m2; and these represent slopes of two tangents passing through P ( , ).Since the tangents are at right angles, m1 m2 = -1=>   ² - a² = -1 =>   ² - a² = - ² + a² ² -a²=>   ² + ² = 2 a²The locus of P ( , ) is x² +y² = 2 a²Thus, the locus of point of intersection of perpendicular tangents to the circle x² +y² = a² is x² +y² = 2 a², which is a circle concentric with the given circle.This is known as director circle of the circle x² +y² = a².

Exercise

1. Determine the number of points of intersection of the circle x² +y² + 6x -4y +8 = 0 with each of the following lines:(i) 2 x + y -1 = 0(ii) x +1 = 0(iii) 4x +3y -12 = 0

2. Determine the points of intersection (if any) of the circle x² +y² +5 x = 0 with each of the following lines:(i) x = 0(ii) 3x - y +1 = 0(iii) 3x -4 y = 7

3. Find the points in which the line y = 2 x +1 cuts the circle x² + y² = 2. Also find the length of the chord intercepted.

4. (i) Find the points of intersection of the circle 3 x² +3 y² -29 x -19 y -56 = 0 and the line y = x +2. Also find the length of the chord intercepted.(ii) If y = 2 x is a chord of the circle x² + y² -10 x = 0, find the equation of the circle with this chord as diameter. Hence find the length of the chord intercepted.

5. Find the lengths of intercepts made by the circle x² + y² -4 x -6 y - 5 = 0 on the co-ordinate axes.

6. Find the length of the chord intercepted by the circle x² +y² -8 x -6 y = 0 on the line x -7 y -8 = 0.

7. Find the length of the chord intercepted by the circle x² +y² = 9 on the line x +2 y = 5. Determine also the equation of the circle described on this chord as diameter.[Hint. The center of the circle described on the chord x +2 y = 5 as diameter is the point of intersection of this line and the line through (0, 0) and perpendicular to this line.]

8. (i) Prove that the lines x = 7 and y = 8 touch the circlex² + y² -4 x -6 y -12 = 0. Also find points of contact.(ii) Find the co-ordinates of the center and the radius of the circle x² + y² -4 x +2 y -4 = 0. Hence, or otherwise, prove that x +1 = 0 is a tangent to the circle. Calculate the co-ordinates of the point of contact. If this point of contact is A, find the co-ordinates of the other end of the diameter through A.

9. Prove that the line y = x +a 2 touches the circle x² +y² = a². Also find the point of contact.

10. Prove that the line 4 x +y -5 = 0 is a tangent to the circle x² + y² +2 x -y -3 = 0, also find the point of contact.

11. Find the condition that the line l x +m y + n = 0 may touch the circle x² +y² = a². 12. Find the condition that the line l x + m y +n = 0 may touch the circle x² +y² +2 g x +2 f

y + c = 0. 13. If the circle 2 x² +2 y² = 5 x touches the line 3 touches the line 3 x + 4 y = k, find the

values of k. 14. (i) Find the equation of the circle with center (3, 4) and which touches the line 5x

+12y -1 = 0.(ii) Find the equation of the circle whose center is (4, 5) and touches the x-axis. Find the co-ordinates of the points at which the circle cuts y-axis.

15. Find the equation to the circle concentric with x² +y² -4 x -6 y -3 = 0 and which touches the y-axis.

16. Find the equation to the circle which is concentric with x² +y² -6 x +7 = 0 and touches the line x +y +3 = 0.

17. Find the length of the chord made by the x-axis with the circle whose center is (0, 3 a) and which touches the straight line 3 x +4 y = 37.

18. Show that 3 x -4 y +11 = 0 is a tangent to the circle x² + y² -8y +15 = 0 and find the equation of the other tangent which is parallel to the line 3 x = 4 y.

19. Find the equations of the tangents to the circle x² +y² = 25 which are parallel to the line y = 2 x +4.

20. Find the equations of the tangents to the circle x² +y² -2 x -4 y = 4 which are perpendicular to the line 3 x - 4 y -1 = 0.

Answers

1. (i) one point      (ii) two distinct points       (iii) none2. (i) (0, 0)          (ii) (-1, -2), (-1/10, 7/10) (iii) none

3.(-1, -1), 4. (i) (1, 3), (5, 7) ; 4 2      (ii) x² + y² -2 x -4 y = 0 ; 2 55. Intercept on x-axis = 6, intercept on y-axis = 2 146. 5 2                                 7. 4; x² +y² -2 x -4 y +1 = 08. (i) (7, 3), (2, 8)

   (ii) (2, -1), 3; point of contact (-1, -1), other end of diameter (5, -1)9. (-a/ 2, a/ 2)                   10. (1, 1)11. n = ± a [l² +m²]12. (l g + m f -n)² = (l² + m²)(g² + f² -c)13. 10, 5/214. (i) 169 (x² + y² -6 x -8 y) +381 = 0     (ii) x² + y² -8 x -10 y +16 = 0; (0, 2), (0, 8)15. x² + y² -4 x -6 y +9 = 016. x² + y² -6 x -9 = 017. 8 | a |                             18. 3 x -4 y +21= 019. 2 x - y ± 5 5 = 020. 4 x + 3 y +5 = 0, 4 x +3 y -25 = 0  

Tangent and Normal to a Circle at a Point

The equation of the tangent at a point on a circle

The equation of the tangent to the circle x² +y² +2 g x +2 f y +c = 0 at the point P (x1 , y1) is   xx1 +yy1 +g (x +x1) +f(y +y1) +c = 0

The equation of the normal at a point on the circle

The equation of the normal to the circle x² +y² +2 g x +2 f y +c = 0 at the point P (x1, y1) is  (y1 +f) x -(x1 +g) y +(g y1 -f x1) = 0Normal at a point on the circle passes through the center of the circle.

Illustrative Examples

Example

Find the equations of tangent and normal to the circle x² +y² -5 x +2 y +3 = 0 at the point (2, -3).

Solution

The given circle is x² +y² -5 x +2 +2 y +3 = 0 ... (i)The equation of the tangent to the circle (i) at the point P (2, - 3) isx. 2 + y. (-3) -5.(1/2).(x +2) +2.(1/2).(y/3) +3 = 0=>   4 x -6 y  -5 x -10 +2 y -6 +6 = 0=>   -x -4 y -10 = 0   =>   x +4 y +10 = 0The slope of the tangent at P = - 1/4=>    the slope of the normal at P = 4The equation of the normal to the circle (i) at P (2, -3) isy +3 = 4 (x - 2) i.e. 4x - y -11 = 0

Exercise

1. (i) Find the equation of the tangent to the circle x² + y²= a² at the point P (x1, y1) on it.(ii) Find the equation of the normal to the circle x² + y² = a² at the point P (x1, y1) on it.

2. Find the equations of the tangent and the normal to the following circles at the given points.(i) x² +y² = 169 at (12, - 5)(ii) 4 x² +4 y² = 25 at (3/2, -2)(iii) x² + y² -4 x +2 y +3 = 0 at (1, -2)(iv) 3 x² +3 y² -4 x -9 y = 0 at the origin.

3. Find the equations of the tangent and the normal to the following circles:(i) x² +y² = 10 at the points whose abscissa is 1.(ii) x² +y² -8 x -2 y +12 = 0 at the points whose ordinate is -1.

4. Show that the tangents drawn at the points (12, - 5) and (5, 12) to the circle x² + y² = 169 are perpendicular to each other.

Answers

1. (i) x x1 +y y1 = a²      (ii) y1 x -x1 y = 02. (i)12 x -5 y -169 = 0; 5x +12 y = 0

    (ii) 6x +8 y +25 = 0; 4x -3y = 0    (iii) x +y +1 = 0; x -y -3 = 0    (iv) 4 x + 9 y = 0; 9 x -4 y = 03. (i) x +3 y -10 = 0, x -3 y -10 = 0, 3 x -y = 0,    (ii) x -2 y -7 = 0, x +2 y -1 = 0, 2 x +y -9 = 0, 2 x - y -7 = 0  

Relative position of two Circles

Let S, S' be two (non-concentric) circles with centers A, B and radii r1 , r2 and d be the distance between their centers, then(i) One circle lies completely inside the other iff d < | r1 -r2 |(ii) The two circles touch internally iff d = | r1 -r2 |(iii) The two circles intersect in two points iff d > |r1 -r2| and d < r1 +r2

(iv) The two circles touch externally iff d = r1 +r2

(v) One circle lies completely outside the other iff d > r1 +r2

Note

1. If two circles intersect, then we can solve the equations of the circles simultaneously to find the points of intersection. In particular, the equation of the common chord is given by S -S' = 0.2. If the two circles touch (internally or externally), then the equation of their common tangent is given by S -S' = 0.

Illustrative Examples

Example

Show that the circles x² +y² -2 x = 0 and x² +y² +6 x -6 y +2 = 0 touch each other. Do these circles touch externally or internally? Find the point of contact and the common tangent.

                                              

Solution

The equations of the given circles areS = x² + y² -2 x = 0                     ...(i)and S = x² + y²+6x -6y +2 = 0    ...(ii)Their centers are A (1, 0) and B (-3, 3), and their radii arer1 = [1² +0² -0] = 1 and r2 = [9 +9 -2] = 4 respectivelyThe distance between their centers= d = [(- 3 -1)² +(3 -0)²] = 5 = 1 +4=>  d = r1 + r2

=>  the given circles (i) and (ii) touch externally and the point of contact P divides [AB] internally in the ratio r1 : r2 i.e. in the ratio 1 : 4The co-ordinates of the point of contact are      (1.(-3) +4.1)/(1+4), (1.3 +4.0)/(1+4) i.e. (1/5, 3/5)The equation of the common tangent is S -S' = 0=>  -8 x +6 y -2 = 0   =>    4 x -3 y +1 = 0

Exercise

1. Prove that the circle x² + y² -6 x -2 y +9 = 0 lies entirely inside the circle x² + y² = 18. 2. Prove that the circles x² + y² -4 x +6 y +8 = 0 and x² + y² -10 x -6 y +14 = 0 touch

each other externally. Find their point of contact and also the common tangent. 3. Show that the circles x² + y² +2 x -6 y +9 = 0 and x² + y² +8 x -6 y +9 = 0 touch

internally. Find their point of contact and also the common tangent. 4. Prove that the circles x² + y² -6x -2 y +1 = 0 and x² + y² +2 x -8 y +13 = 0 touch one

another and find the equation of the tangent at their point of contact.

5. Show that the circles x² + y² = 2 and x² + y² -6x -6 y +10 = 0 touch each other. Do these circles touch externally or internally? Also find their point of contact.

6. Find the equation of the circle whose radius is 3 and which touches the circle x² + y² -4 x -6 y -12 = 0 internally at the point (-1, -1).

Answers

2. (3, -1, x +2 y -1 = 03. (0, 3), x = 04. 4 x -3 y +6 = 05. Externally; (1, 1)6. 5 (x² + y²) -8x -14 y +32 = 0  

Families of Circles

A collection of circles is called a family or a system of circles.Let S and S' be two intersecting (or touching) circles, then S +k S' = 0, k -1, represents a family of circles through their points (or point) of intersection.

Remarks

1. If k = -1, then the equation S +kS' = 0 reduces to S -S' = 0 which represents the common chord in case of intersecting circles or common tangent in case of touching circles.

2. The equation S +kS = 0 represents all members of the family except the member S'. If we need the member S', then take the equation of the family as S' +kS = 0.

Illustrative Examples

Example

Find the equation of the family of circles passing through the points A(0,4) and B(0,-4).

Solution

Let the equation of the desired family of circles bex² +y² +2gx +2fy +c = 0     ...(i)As all these circles pass through the points A(0,4) and B(0,-4), we get0 +16 +0 +8f +c = 0 => 8f +c +16 = 0               ...(ii),0 +16 +0 -8f +c = 0 => -8f +c +16 = 0   ...(iii)On solving (ii) and (iii), we get f = 0, c = -16.Substituting these values in (i), we getx² +y² +2gx -16 =0 ...(iv)Note that for every real value of g,g² +f² -c = g² +16 > 0, therefore, (iv) represents a circle.Hence the equation x² +y² +2gx -16 = 0 for different real values of g represents the desired family of circles. It is one-parameter family of circles where g is the parameter.

Example

Find the equation of the circle which passes through the points of inter-section of x² +y² -4 = 0 and x² +y² -2x -4y +4 = 0 and touches the line x +2 y = 0.

Solution

The equations of the given intersecting circles areS = x² +y² -4 = 0, andS' = x² +y² -2 x -4y +4 = 0The equation of the common chord of these circles isS -S' = 0   => l = 2 x +4 +4 y -8 = 0The equation of the family of circles passing through the intersection of the given circles is

x² +y² -4 +k (2x +4 +4y -8) = 0 ...(i)Its center is (-k,-2k) andradius = [k² +4k² +4 +8k] = [5k² +8k +4].For the particular member of the family which touch the line x +2y = 0, we have|-k +2 (-2k)|/ [1² +2²] = [5k² +8k +4]=>   5|k|/ 5 = [5k² +8k +4]=> 5 k² = 5 k²+8 k +4=> 8 k +4 = 0 => k = -1/2Substituting this value of k in (i), the equation of the required circle isx² +y² -4 - (1/2)(2x +4y -8) = 0 i.e. x² + y² -x -2y = 0.

Exercise

1. Find the equation of the family of circles passing through the origin. 2. Find the equation of the family of circles passing through the origin and the point

(0,1). 3. Find the equation of the family of circles passing through the points A(5,0) and B(-

5,0). 4. Find the equation of the family of circles with radius 3 and whose centers lie on the x-

axis. 5. Find the equation of the family of concentric circles with center as (-4,2). Also find a

member of the family which touches the line x -y = 3. 6. Show that the equation of the family of circles which touch both the co-ordinate axes

can be put into the form x² +y² ±2rx ±2ry +r² = 0, where r is radius.[Hint. Let ( , ) be center of the family. Since it touches both the axesi.e. y = 0 and x = 0, so | | = | |   r  => = ±r, = ±r.]

7. Find the equation of the circle which passes through the origin and the points of intersection of the circles x² +y² +2x +2y -2 = 0 and x² +y² +4x -8y +4 = 0.

8. Find the equation of the circle through the points of intersection of the circles x² +y² +2x +3y -7 = 0 and x² +y² +3x -2y -1 = 0 and through the point (1,2).

9. Find the equation of the circle which passes through the point (1,-1) and through the points of intersection of the circles x² +y² +2x -2y -23 = 0 and 3x² +3y² +12x -4y -9 = 0.

10. Find the equation of the circle passing through the point (2,3) and through the points of intersection of the circle x² +y² +3x -4y +5 = 0 and the line x -y +2 = 0.

11. Find the equation of the circle through the intersection of the circles x² +y² -8x -2y +7 = 0 and x² +y² -4x +10y +8 = 0 and having its center on the x-axis.

Answers

1. x² +y² +2 gx +2 +2fy = 0, where g, f are any real numbers2. x² +y² +2gx -y = 0, where g is any real number3. x² +y² +2fy -25 = 0, where f is any real number4. (x -h)² +y² = 9, where h is any real number5. x² +y² +8x -4y +20 -r² = 0, where r is radius;   2(x² +y²) +16x -8y -41 = 07. 3(x² +y²) +8x -4y = 08. x² +y² +4x -7y +5 = 09. 32(x² +y²) +115x -47y -226 = 010. x² +y² -9x +8y -19 = 011. 6(x² +y²) -44x +43 = 0  

Conics

The curves known as conics were named after their historical discovery as the intersection of a plane with a right circular cone.

       Applonius (before 200 B.C.) realized that a conic (or conic section) is a curve of intersection of a plane with a right circular cone of two nappes, and the three curves so obtained are parabola, hyperbola and ellipse.

     Let l be a fixed line and F be a fixed point not on l, and e > 0 be a fixed real number. Let |MP| be the perpendicular distance from a point P (in the plane of the line l and point F) to the line l, then the locus of all points P such that |FP| = e |MP| is called a conic.

      The fixed point F is called a focus of the conic and the fixed line l is called the directrix associated with F. The fixed real number e (> 0) is called eccentricity of the conic.In particular, a conic with eccentricity e is called(i) a parabola iff e = 1 (ii) an ellipse iff e < 1 (iii) a hyperbola iff e > 1.

Four standard forms of the parabola

     

Main facts about the parabola

Equationsy²= 4ax (a>0)Right hand

y² = -4axa>0Left hand

x² = 4aya>0Upwards

x² = -4aya>0Downwards

Axis y=0 y = 0 x = 0 x = 0

Directrix x +a = 0 x -a = 0 y +a = 0 y -a = 0

Focus (a, 0) (-a, 0) (0,a) (0, -a)

Vertex (0,0) (0,0) (0,0) (0,0)

Length of Latus-rectum

4a 4a 4a 4a

Equation of Latus-rectum

x -a = 0 x +a = 0 y -a = 0 y +a = 0

Focal distance of the point(x,y)

x +a a -x y +a a -y

Another definition of ellipse

An ellipse is the locus of a point in a plane, sum of whose distances from two given points F and F' (in the plane) is a constant and greater than |FF'|.Remark. The focal property of ellipse gives us a practical method of drawing an ellipse. Fasten the ends of a string of length 2a > |FF'| at two distinct points F and F'. Keep the string taught by means of a pencil placed against the string and slide it along the string, the curve thus traced is an ellipse with F and F' as its foci.

     

Main facts about the ellipse

Equationx²/a² + y²/b² = 1a > b > 0

x²/b² + y²/a² = 1a > b> 0

Equation of major axis y = 0 x = 0

Length of major axis 2a 2a

Equation of minor axis x = 0 y = 0

length of minor axis 2b 2b

Vertices (a,0),(-a,0) (0, a),(0, -a)

Foci (ae, 0), (-ae,0) (0, ae), (0, -ae)

Directrices x - a/e = 0, x + a/e = 0 y - a/e = 0, y + a/e = 0

Length of Latus -rectuum 2b²/a 2a²/b

Equation of a latera-recta Center

x-ae = 0, x +ae = 0(0,0)

y -ae = 0,y +ae = 0

Focal distance of any point (x,y)

a -ex, a +ex a -ey, a +ey

   

Another definition of hyperbola

A hyperbola is the locus of a point in a plane, the difference of whose distances from two given points F and F' is 2a (constant), and 0 < a < (1/2)|FF'.

     

Main facts about the hyperbola

Equationx²/a² - y²/b²= 1a > 0,b > 0

y²/a² - x²/b² = 1a > 0,b > 0

Length of transverse axis y = 0 x = 0

Equation of transverse axis 2a 2a

Length of conjugate axis 2b 2b

Equation of conjugate axis x = 0 y = 0

Vertices (a, 0), (-a, 0) (0, a), (0, -a)

Foci (ae, 0), (-ae, 0) (0, ae), (0, -ae)

Directrices x - a/e = 0, x + a/e=0 y - a/e= 0, y + a/e =0

Length of lactus-rectum 2b²/a 2a²/b

Equation of latera-rectacenter

x -ae = 0, x + ae = 0(0,0)

y -ae = 0,y + ae = 0(0,0)

Focal Distance point(x,y) |ex -a|, |ex +a| |ey -a|,|ey +a|

Illustrative Examples

Example

Find the equation of the parabola with focus at (-2, 0) and whose directrix is the line x +2y -3 = 0.

Solution

The focus of the parabola is at F(-2, 0) and directrix is the line x +2y -3 = 0.Let P(x, y) be any point on the parabola and |MP| be the perpendicular distance from P to the directrix, then by def. of parabola   |FP| = |MP|         (As e = 1 for parabola)So ((x +2)² +y²) = |x +2y -3|/ (1² +2²)On squaring,=> 5 ((x +2)² +y²) = (x +2 y -3)²          (since |x|² = x²)=> 5 (x² +4 +4 x +y²) = x² +4 y² +9 +4 x y -6 x -12 y=> 4 x² -4 xy +y² +26 x +12 y +11 = 0,which is the required equation of the parabola.

Example

Find the equation of the ellipse whose focus is (1, -2), the corresponding directrix x -y +1 = 0 and eccentricity is 2/3.

Solution

The focus of the ellipse is at F(1, -2), the corresponding directrix is the line x -y +1 = 0 and e = 2/3.Let P (x, y) be any point on the ellipse and | MP | be the perpendicular distance from P to the directrix, then by def. of ellipse |FP| = e |MP|

=> => 9 [(x -1)² +(y +2)²] = 2 (x -y +1)²=> 9 [x² -2 x +1 +y +1 +y² +4 y +4] = 2 [x² +y² +1 -2 x y +2 x -2 y]=> 7 x² +4 x y +7 y² -22 x +40 y +43 = 0,which is the required equation of the ellipse.

Example

Find the focus, directrix and eccentricity of the conic represented by the equation 3y² = 8x.

Solution

The given equation is 3 y² = 8 x     i.e. y² = (8/3) x ...(i)which is the same as y² = 4ax, so (i) represents a standard (right hand) parabola, and hence its eccentricity is 1, as e = 1 for parabola.

Also 4a = 8/3 => a = 2/3, therefore, focus is (a, 0) i.e. and the equation of directrix isx +2/3 = 0 i.e. 3 x +2 = 0.

Example

Find the locus of a point P, the sum of whose distances from the points F(-2, 3) and F'(2, 0) is constant equal to 4 units.

Solution

Here, we note that |FF'| = [(2+2)² +(y -2)²] = 5and |PF| +|PF'| = 4 (given)=> |PF| +|PF'| < |FF'|, which is not possible wherever P may be (since sum of two sides of a triangle cannot be less than third side)Therefore, the locus of P is the empty set.

Exercise

1. Write the equation of a conic with eccentricity e, focus (a, 0) and directrix y-axis. 2. Write the equation of a conic with eccentricity e, focus (0, a) and directrix x-axis. 3. Write the equation of the parabola with the line x +y = 0 as directrix and the point (1,

0) as focus. 4. Find the equation of the parabola whose focus is (2, -1) and directrix is x +2 y -1 = 0. 5. Find the equation of the ellipse whose focus is (1, -1), the corresponding directrix is x

-y +3 = 0 and e = 1/2. 6. Find the equation of the hyperbola with directrix x + 2 y = 1, focus at (0, 0) and

eccentricity 2. 7. Find the equation to the parabola with the focus (a, b) and directrix x/a + y/b = 1. 8. Find the equation to the parabola whose focus is (-2, 1) and directrix is 6 x -3 y = 8. 9. Find the equation of the parabola whose focus is (5, 3) and the directrix is given by

3x -4 y +1 = 0.also find the equation of axis. 10. Find the equation to the conic section whose focus is (1, -1), eccentricity is 1/2 and

the directrix is the line x -y = 3. Is the conic section an ellipse? 11. Find the eccentricity of the ellipse if:

(i) the latus-rectum is one half of its minor axis.(ii) the latus rectum is one half of its major axis.(iii) the distance between foci is equal to the length of latus-rectum.

12. Find the eccentricity, co-ordinates of the foci and the length of the latus-rectum of the ellipse 4 x² +3 y² = 36.

13. Find the co-ordinates of the foci and the ends of the latera-recta of the ellipse 16x² +9y² = 144.

14. Find the vertices, eccentricity, foci and the equations of the directrices of the hyperbola x² -y² = 1.

15. Find the lengths of axes, co-ordinates of foci, the eccentricity and the length of latus-rectum of the hyperbola 25 x² -9 y² = 225.

16. In each of the parabolas(i) y² = 3 x    (ii) y² = -4 x    (iii) 3 x² = 4 y     (iv) x² = -12 y,find the length of latus-rectum, coordinates of focus and the equation of directrix.

17. If the parabola y² = 4 p x passes through the point (3, -2), find the length of latus-rectum and the co-ordinates of the focus.

18. Find the equation of the parabolas with vertices at the origin and satisfying the following conditions:(i) Focus at (-4, 0)       (ii) Directrix y -2 = 0(iii) Passing through (2, 3) and axis along x-axis.

19. Find the equation to the ellipse referred to its axes as co-ordinates axes(i) whose major axis = 8, eccentricity = 1/2(ii) which passes through the points (-3, 1) and (2, -2).

(iii) which passes through the point (-3, 1) and has eccentricity .(iv) whose minor axis is equal to the distance between foci and whose latus-rectum is 10.

20. Find the equation of the ellipse whose eccentricity is 1/2 and whose foci are (±2, 0).

Answers

1. (1 -e²) x² +y² -2 a x +a² = 02. x² +(1 -e²) y² -2 ay +a² = 03.(x -y)² -4 x +2 = 04. 4 x² -4x y +y² -18 x +14 y +24 = 05. 7 x² +2 x y +7 y² -22 x +22y +7 = 06. x² +8 2x y +5 y² -8 x -8y +4 = 07. a² x² -2 a b x y +b² y² -2 a³ x -2 b³ y +(a 4 - a² b² +b4) = 08. 9 x² +36 x y +36 y² +276 x -138 y -169 = 09.16 x² +24 x y +9 y² -256 x -142 y +849 = 0; 4x +3 y -29 = 010. 7 x² +2 x y +7 y² -10 x +10 y +7 = 0; Yes11. (i) ( 3)/2    (ii) 1/ 2      (iii) ( 5 1)/212. e = 1/2, foci are (0, 3), (0, - 3), length of latus-rectum = 3 3

13. (0, 7), (0, - 7); 14. (1, 0), (-1, 0); 2  ; ( 2, 0), (- 2 , 0); 2 x -1 = 0, 2x +1 = 015. 6, 10; ( 34 , 0), (- 34, 0); ( 34)/3; 1016. (i) 2 3 (( 3)/2, 0) ; 2 x + 3 = 0 (ii) 4; (-1, 0); x -1 = 0     (iii) 4/3 ; (0, 1/3) ; 3 y +1 = 0        (iv) 12; (0, -3); y -3 = 017. 4/3, (1/3, 0)18. (i) y² = -16 x    (ii) x² = -8y   (iii) 2 y² = 9 x19. (i) 3 x² +4 y² = 48  (ii) 3x² +5 y² = 32     (iii) 3 x² +5 y² = 32  (iv) x² +2y² = 10020. 3 x² +4 y² = 48

 Factorial Notation

The continued product of first n natural numbers is called n factorial or factorial n.It is denoted by | n or n!Thus, | n or n! = 1. 2. 3. 4. .... (n -1) . n                       = n (n -1)(n -2). .... 3.2.1          (in reverse order)

Meaning of zero factorial

According to the above definition, makes no sense. However we define 0! = 1Note. When n is a negative integer or a fraction, n factorial is not defined.

Illustrative Examples

Example

Find the value ofi. 25! / 23! ii. 9! / [6!.3!] iii. (7 -4)! iv. 5! -4! v. 2! +3!

Solution

i. 25!/23! = 25. 24 23!/ 23! = 25. 24 = 600 ii. 9! /[6! 3!] = 9.8. 7. 6! /[6! 3!] = 9. 8. 7/3! = 84 iii. (7 -4)! = 3! = 3. 2. 1 = 6 iv. 5! - 4! = (5. 4. 3. 2. 1) - (4. 3. 2. 1) = 120 -24 = 96 v. 2! + 3! = (2. 1) + (3 . 2. 1) = 2 +6 = 8

Example

Find n, if n! /[2! (n -2)!] and n! /[4! (n -4)!] are in the ratio 2 : 1.

Solution

Given n! /[2! (n -2)!] : n!/[4!(n -4)!] = 2:1=> [n! / 2! (| n -2)!].[4! (n -4)! /n!] = 2/1=> 4.3.2!.(n -4)! /[2!.(n -2).(n -3).(n -4)!] = 2/1=> 4.3 /[(n -2)(n -3)] = 2/1

=> (n -2)(n -3) = 6=> n² -5 n = 0      =>    n (n -5) = 0=>   n = 0 or n = 5But, for n = 0, (n -2)! and (n -4)! are not meaningful, therefore, n = 5.

Example

Prove that 33! is divisible by 216. What is the largest integer n such that 33! is divisible by 2n?

Solution

33! = 1. 2. 3. 4. 5. 6. ... . 29. 30. 31. 32. 33= [2. 4. 6. ... . 30. 32] [1. 3. 5. ... . 29. 31. 33]= [(2. 1) (2. 2) (2. 3) ... (2. 15) (2. 16)] [1. 3. 5. ... . 31. 33]= 216. (1. 2. 3. 4. ... . 15. 16). 1. 3. 5. ... . 31. 33 ...(i)=>   33! is divisible by 216.Further, 1. 2. 3. 4. ... . 15. 16 = (2. 4. 6...16) (1. 3. 5. ... . 15)= (2. 1) (2. 2) (2. 3) ... (2. 8) (1. 3. 5. ... . 15)= 28.(1. 2. 3. ... . 8) (1. 3. 5. ... . 15)= 28 (2. 4. 6. 8) (1. 3. 5. 7) (1. 3. 5. ... . 15)= 28. 24. (1. 2. 3. 4) (1. 3. 5. 7) (1. 3. 5. ... . 15)= 212. (2. 4) (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15)= 212. 2³. (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15)= 215. (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15) ...(ii)From (i) and (ii), we get33!    = 216. 215. (1.3) (1. 3. 5. 7) (1. 3. 5. ... . 15) (1. 3. 5. ... . 33)         = 231. (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15) (1. 3. 5. ... . 33).Hence, 31 is the largest value of n for which is divisible by 2 n.

Exercise

1. Find the values of(i) 6!    (ii) 4! +3!    (iii) (9 -6)!(iv) 6! -4!    (v) 3!  . 5!

2. Convert into factorials:(i) 3. 6. 9. 12. 15. 18(ii) 6. 7. 8. 9. 10. 11. 12(iii) 1. 3. 5. 7. ... (2 n -1)(iv) (n +1)(n +2) ... 2 n.

3. Which of the following are true?(i) 4(3!) = 4!(ii) 3(4!) = (3 × 4)!(iii) 3! + 4! = (3 +4)!(iv) (4 -3)! = 4! - 3!(v) 4!. 3! = (4. 3)!

(vi) = 4. Find the L.C.M. and H.C.F. of 10!, 11!, 15!. 5. Prove that 310 divides 30!. Which is highest natural number n such that 3n divides 30!

>

Answers

1. (i) 720      (ii) 30      (iii) 6     (iv) 696     (v) 7202. (i) 36. 6!    (ii) 12!/5!      (iii) 2n! / [2n. n!]       (iv) 2n! / n!3. Only (i) is true.4. H.C.F. is 10! and L.C.M. is 15!5. 14  

Permutations and Combinations

Combination

Each of the groups or selections which can be made by taking some or all of a number of things without reference to the order of things in each group is called a combination.For example, choosing two fruits out of a basket containing 4 oranges and 5 bananas - three combinations are possible - two oranges, two bananas or one orange and one banana.

Permutation

Each of the arrangements which can be made by taking some or all of a number of things is called permutation.For example, the letters of word CAT can be arranged in six different ways     CAT, CTA, ACT, ATC, TAC, TCASometimes, students may be confused whether combination or permutation is being talked about. Certain key words/phrases can be helpful in deciding.Combinations: Selection, choose, make group, distribute, committee, geometric problems.Permutations: Arrangements, standing in a line, seated around a table, problems on digits or letters of a word.

Fundamental Theorem

1. Multiplication Principle (Principle of Association).If a certain thing can be done in m ways, and if when it has been done, a second thing can be done in n ways, then the total number of ways in which two things can be done is mn.For example, if there are eight top heroes and 4 top heroines, a film producer can choose from 8 x 4 = 32 pairs.2. Addition Principle:If there are two jobs which can be done in m and n ways respectively, then either of two jobs can be done in m + n ways.For example, if the film producer has limited budget and he can afford either a top hero or a top heroine but not both, then he can choose in 8 +4 = 12 ways.

Illustrative Examples

Example

Three persons enter a railway carriage, where there are 5 vacant seats. In how many ways can they seat themselves?

Solution

First man can sit on any of 5 vacant seats. Then the second can sit on any of 4 vacant seats left. And the third can sit on any of 3 vacant seats left. Hence by fundamental principle of counting, the required number of ways is 5 x 4 x 3 = 60.

Example

How many numbers are there between 100 and 1000 such thati. every digit is either 2 or 5 ii. there is no restriction iii. the digit in the hundreds place is 5 iv. atleast one of the digits is 5 v. exactly one of the digits is 5 vi. no digit is repeated vii. atleast one digit is repeated viii. the digit in units place is 5?

Solution

The numbers between 100 and 1000 consist of 3 digits - 100 is included and 1000 is excluded.

i. Since every digit is 5 or 2, there are 2 ways of filling up of each of three digits (places). Thus the three digits can be filled in 2 x 2 x 2 = 8 ways. It is easy to see that the eight numbers satisfying giving condition are 222, 225, 252, 255, 522, 525, 552, 555.

ii. The first digit has to be non-zero so there are 9 choices. Second and third digits can be any of ten digits 0 to 9. Hence required number of numbers is 9 x 10 x 10 = 900 (including 100 but excluding 1000).

iii. Digit in hundreds place is fixed (5). The other two places can each be filled in 10 ways. Hence possible ways are 1 x 10 x 10 = 100. It is easy to see that numbers between 500 and 599 (both inclusive) fulfill given criteria.

iv. First, let us find all numbers between 100 and 1000 which do not have digit 5 in any place. The number of such numbers is 8 x 9 x 9 = 648 (including 100). We have also seen in part (ii) that there are 900 numbers between 100 and 100 (including 100). Hence there are 900 -648 = 252 numbers between 100 and 1000 which have at least one digit as 5.

v. The numbers with exactly one digit as 5 are:(a) of type 5 x x = 1 x 9 x 9 = 81 (where x is non-five digit)(b) of type x 5 x= 8 x 1 x 9 = 72 (first digit is non-zero, non five)(c) of the x x 5 = 8 x 9 x 1 = 72Thus there are 81 +72 +72 = 225 numbers between 100 and 1000 which have exactly one digit as 5

vi. First digit (i.e. hundreds digit) can be anything from 1 to 9; second digit can be then filled in 9 ways, and third digit in 8 ways. Thus number of numbers with no digit repeated is 9 x 9 x 8 = 648.

vii. As there are 900 numbers between 100 and 1000 (including 100) and 648 of them have no digit repeated, there are 900 -648 = 252 numbers (including 100) which have at least one digit repeated.

viii. The digit in hundreds place can be filled in 9 ways (1 to 9); the digit in tens place can be filled in 10 ways (0 to 9), and digit in units place can be filled in one way only as it is fixed (5). Hence required number of numbers is 9 x 10 = 90.

Exercise

1. There are 12 buses running between Delhi and Agra. In how many ways can a man plan his journey from Delhi to Agra and back? In how many ways can he go from Delhi to Agra and return by a different bus?

2. How many words (with or without meaning) of three English alphabets can be formed? How many of these have all distinct alphabets?

3. In how many ways can 2 prizes (in Science and Maths) be awarded to 15 students? In how many ways can the first and second prize in History be awarded to 15 students?

4. How many different numbers of 3 digits can be formed using only odd digits (i.e. 1, 3, 5, 7, 9)? Using only even digits (0, 2, 4, 6, 8)?

5. There are 4 doors in Lotus temple. In how many ways can a person enter the temple and leave by a different door?

6. There are 20 boys and 20 girls in a class. For staging the play Romeo and Juliet, how many lead pairs are possible? If a shy girl refuses to go on stage, how many pairs are possible?

7. Sandy has 5 shirts, 4 pants, 3 pairs of socks and 2 kinds of shoes. In how many different ways can he dress himself? (He does not know about colour combinations!)

8. Two cards are drawn, one at a time, and without replacement, from a deck of 52 cards. Determine the number of ways in which cards can be drawn. What will be the number of ways if the first card is replaced before the second is drawn?

9. How many orderings of the letters in the word SMILE are possible? 10. For a set of six true or false statements, no student in a class has written all correct

answers and no two students in the class have written the same sequence of answers. What is the maximum number of students in the class, for this to be possible?

Answers

1. 144, 132                 2. 17576, 15600             3. 225, 2104. 125, 100                 5. 12                               6. 400, 3807. 120                         8. 2652, 2704                 9. 12010. 63  

Formula for P(n,r)

The number of permutations of n different things taken r at a time is given byP(n, r) = n. (n -1). (n -2) ... (n -r +1) = n! / (n -r)!Note. n Pn = n (n -1) ... (n - n +1) = n (n -1) ... 1 = n!

Illustrative Examples

Example

Find the values of   (i)5P5     (ii) 5P3     (iii)60P48     (iv)60P2.

Solution

i. 5P5 = 5. 4. 3. 2. 1 = 120 ii. 5P3 = 5. 4. 3 = 60 iii. 60P48 = 60! / (60 -48)! = 60! / 12! iv. 60P2 = 60. 59 = 3540

Example

Find n ifi. 5P4 :nP5 = 1 : 2 ii. nP4 :n -1P3 = 9 : 1 iii. P (n, 4) = 2 P (5, 3)

Solution

i. nP4 :nP5 = 1 : 2=> [n (n -1)(n -2)(n -3)] / [n (n -1)(n -2)(n -3)(n -4)] = 1/2

ii. nP4 :n -1P3 = 9 : 1=> [n (n -1)(n -2)(n -3)]/[(n -1)(n -2)(n -3)] = 9/1=> n = 9

iii. P (n, 4) = 2 P (5, 3) = 2. 5. 4. 3 = 5. 4. 3 . 2=> n (n -1)(n -2)(n -3) = 5 . 4. 3. 2=> n = 5

Example

i. In how many ways can the letters of the word WONDERFUL be arranged? ii. Ten students participate in a debate. In how many ways can the first three prizes be

won? iii. Find the number of signals which can be given with 5 flags of different colours

hoisted one above the other, by using any number of flags.

Solution

i. Since there are 9 distinct letters in WONDERFUL, the required number of permutations is   9P9 = 9! = 9. 8. 7. 6. 5. 4. 3. 2. 1 = 362880

ii. The given problem is similar to one of filling 3 places with any 3 of 10 distinct objects. This can be done in 10P3 ways. Hence required number is   10P3 = 10. 9. 8 = 720

iii. When 1 flag is used, number of signals = 5P1 = 5When 2 flags are used, number of signals = 5P2 = 5. 4 = 20When 3 flags are used, number of signals = 5P3 = 5. 4. 3 = 60When 4 flags are used, number of signals = 5P4 = 5. 4. 3. 2 = 120When 5 flags are used, number of signals = 5P5 = 5. 4. 3. 2. 1 = 120Hence total number of signals = 5 +20 +60 +120 +120 = 325

Example

i. Find the number of permutations of n different things taken r at a time when each thing may be repeated any number of times in any permutation.

ii. In how many ways can three prizes be given to 20 boys when a boy may receive any number of prizes?

iii. Find the sum of all three digit numbers formed by using odd digits 1, 3, 5, 7, 9 with repetition of digits allowed.

Solution

i. The first place may be filled with any of n things. After the first place has been filled up, the second place can also be filled in n ways since we are not prevented from repeating the same thing. When the first two places have been filled in n x n ways,

the third place can also be filled up in n ways and so on.By fundamental principle of association, the r places can be filled in(n x n x ... r times) ways, i.e. nr ways.Hence the required number of permutations = nr

ii. The first prize can be given in 20 ways, and then second prize can also be given in 20 ways, since repetition is allowed. Similarly third prize can also be given in 20 ways. Thus by principle of association, required number of ways = 20 x 20 x 20 = 8000

iii. Since any of 5 digits can be placed at any place, there are 5 x 5 x 5 = 125 such numbers.Now consider the digits in units place. 25 numbers end with digit 1; 25 numbers end with digit 3, and so on. Hence sum of all units digits of 125 numbers= 25(1 +3 +5 +7 +9) = 625.Similarly sum of tens digits of 125 numbers is 625, and sum of hundreds digits of 125 numbers is 625.Hence sum total of these 125 numbers is 625 (1 +10 +100) = 625 x 111 = 69375

Example

Find the number of divisors of the number 36000.

Solution

Factorising the given number, we find 36000 = 25 . 3². 5³. This means that any divisor of 36000 is of the type 2a. 3b. 5c where a can take values 0, 1, 2, 3, 4, 5; b can take values 0, 1, 2; c can take values 0, 1,2, 3. Hence number of divisors is 6 x 3 x 4 = 72. Note that both 1 and 36000 are counted among 72 divisors.

Exercise

1. Find the values of(i) P (4, 3)(ii) P (65, 15)(iii)7P5

(iv)10P3 2. Prove that

(i) P (10, 3) = P (9, 3) +3. P (9, 2)(ii) P(n, n) = 2. P (n, n -2)(iii) P (n, n) = P (n, n -1)(iv) P(n, r) = (n -r +1). P (n, r -1)(v) 1 +1. P1 +2. P2 +... + n. Pn = Pn = 1 where Pm = mPm

3. Find n if(i) 5 P (4, n) = 6 P (5, n -1)(ii)n+1P3 = nP4

(iii) P (n -1, 3) : P (n +1, 3) = 5 : 12(iv) P (n, 4) = 5040

4. Find the value of r if(i) P (11, r) = P (12, r -1)(ii) P (9, r) = 3024(iii) P (15, r -1) : P (16, r -2) = 3 : 4

5. If r s n, prove that P (n, s) is divisible by P (n, r).[Hint. P(n, s) / P(n, r) = ... = a whole number]

6. In how many ways can the letters of following words be permuted?(i) CATS(ii) DELHI(iii) JAIPUR(iv) HEXAGON

7. In how many ways can you take 3 letters of above words and arrange into words with no letter being repeated?

8. (i) In how many ways can 5 people be seated on a sofa if there are only 3 seats available?(ii) In how many ways can 3 people sit if there are 5 vacant seats?

9. In how many ways can 5 children stand in a queue? 10. In how many ways can 6 women draw water from 6 taps if no tap remains unused? 11. Seven candidates are contesting an election. In how many ways can their names be

listed on the ballot paper? 12. There are three different rings to be worn in four fingers with atmost one in each

finger. In how many ways can this be done? If the wedding ring is to be worn on ring finger only, then what are the number of ways?

13. In how many ways can captain and vice captain be chosen out of 11 players? 14. In how many ways can 8 boys and 7 girls be photographed if

(i) girls are to sit on chairs in a row and the boys stand in a row behind them(ii) girls sit on even numbered chairs and boys sit on odd numbered chairs and there are 15 chairs in a row(iii) there are 15 chairs in a row and there is no restriction?

15. How many numbers consisting of five different digits can be made from the digits 1, 2, 3, ..., 9 if(i) the number must be odd(ii) the first two digits of each number are even?

16. Solve problem 15 if repetition of digits is allowed. 17. Find the number of all five-digit numbers with distinct digits. 18. How many natural numbers less than 1000 can be formed with the digits 1, 2, 3, 4

and 5 if(i) no digit is repeated(ii) repetition of digits is allowed?

19. There are 3 candidates and 5 voters. In how many ways can the votes be given? If 3 candidates are included in 5 voters and they vote for themselves only, then how many ways of giving votes are possible?

20. There are 8 true-false statements in a question paper. How many sequences of answers are possible?

21. Find the number of ways in which 5 boys and 5 girls can be seated alternately in a row?

22. Find the number of divisors of the number 8800. How many of these are even?

Answers

1. (i) 24  (ii) 65! / 50!     (iii) 2520    (iv) 7203. (i) 3    (ii) 5               (iii) 8         (iv) 104. (i) 9 (ii) 4 (iii) 146. (i) 24 (ii) 120 (iii) 720 (iv) 50407. (i) 24 (ii) 60 (iii) 120 (iv) 2108. (i) 60 (ii) 609. 120     10. 720     11. 5040     12. 24, 613. 110     14. (i) 8!.7!    (ii) 8!.7!   (iii) 15!15. (i) 8400 (ii) 2520     16. (i) 32805 (ii) 1166417. 27216     18. (i) 85 (ii) 155     19. (i) 243 (ii) 920. 256     21. 2880022. 36, 30 [Hint. For even, power of 2 should be minimum 1]  

Permutations under restrictions

Number of permutations of n distinct objects when a particular object is not taken in any arrangement is n-1Pr

Number of permutations of n distinct objects when a particular object is always included in any arrangement is r.n-1Pr-1.

Illustrative Examples

Example

Without using factorials prove that nPr = n-1Pr + r.n-1Pr-1

Solution

nPr indicates permutations of n distinct objects taken r at a time. Alternatively, we have two ways of arranging these :One, in which a particular object is always excluded - there are n-1Pr ways of doing so.Two, in which the particular object is always included - there are r.n-1Pr-1 ways of doing so.Hence nPr = n-1Pr + r.n-1Pr-1

Example

In how many ways can 4 books on Mathematics and 3 books on English be placed on a shelf so that books on the same subject always remain together?

Solution

Consider the 4 books on Mathematics as one big book and 3 books on English as another big book. These two can be arranged in 2! ways. In each of these arrangements, 4 books on Mathematics can be arranged among themselves in 4! ways and 3 books on English can be arranged among themselves in 3! ways.Hence, the required number of ways = 4! 3! 2! = 288.

Example

How many different (eight letter) words can be formed out of the letters of the word DAUGHTER so that

i. the word starts with D and ends with R ii. position of letter H remains unchanged iii. relative position of vowels and consonants remains unaltered iv. no two vowels are together?

Solution

The given word consists of 8 different letters out of which 3 are vowels and 5 are consonants.

i. If the words have to start with D and end with R, then we can arrange remaining 6 letters at 6 places in 6P6 = 6! = 720 ways.

ii. If position of H remains unchanged, the remaining 7 letters can be arranged in 7 places in 7P7 = 7! = 5040 ways.

iii. The relative position of vowels and consonants remains unaltered means that vowel can take the place of vowel and consonant can take place of consonant. Now the 3 vowels can be arranged among themselves in 3! = 6 ways and the 5 consonants can be arranged among themselves in 5! = 120 ways. Thus the total number of words that can be formed = 6.120 = 720

iv. First let us arrange the consonants in a row. This can be done in 5P5 = 5! = 120 ways     C x C x C x C x C xNow no two vowels are together if they are put at places marked . The 3 vowels can fill up these 6 places in 6P3 = 6.5.4 = 120 ways. Hence, the total number of words = 120.120 = 14400

Exercise

1. In how many ways can 7 books be arranged on a shelf if(i) any arrangement is possible(ii) 3 particular books must always stand together(iii) two particular books must occupy the ends?

2. Four different mathematics books, six different physics books, and two different chemistry books are to be arranged on a shelf. How many different arrangements are possible if(i) the books in each particular subject must all stand together(ii) only the mathematics books must stand together?

3. In how many ways can 5 children be arranged in a row such that two boys Ajay and Sachin(i) always sit together(ii) never sit together?

4. When a group photograph is taken, all the nine teachers should be seated in the first row and all the eighteen students should be in second row. If the two corners are reserved for the two tallest students (interchangeable only between them), and if the middle seat of the first row is reserved for the principal, how many arrangements are possible?

5. Find the number of ways in which 5 boys and 5 girls may be seated in a row so that no two girls are together.

6. If ten students appear in an examination and 4 of them are appearing for mathematics and rest for 6 different subjects, in how many ways can they be seated in a row so that no copying is possible?

7. Find the number of ways in which the candidates A1, A2, A3, ..., A10 can be ranked if(i) A1 and A2 are next to each other(ii) A1 is always above A2.

8. How many words can be formed out of the letters of the word ARTICLE so that vowels occupy even places?

9. How many words can be formed out of the letters of the word ORIENTAL so that A and E occupy odd places?

10. How many words beginning and ending with a consonant can be formed by using the letters of the word EQUATION?

11. How many different five letter words starting with a vowel can be formed from the letters of the word EQUATION?

12. How many arrangements can be formed by the letters of the word VOWELS if(i) there is no restriction(ii) each word begins with S(iii) each word begins with S and ends with E(iv) all vowels come together(v) all consonants come together?

13. The letters of the word TRIANGLE are arranged in such a way that vowels and consonants remain together. How many different arrangements will be obtained?

14. How many 6 digit telephone numbers can be constructed with the digits 0, 1, 2, ..., 9 if each number starts with 35 and no digit appears more than once?

15. How many four digit numbers divisible by 4 can be made with the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?

Answers

1. (i) 5040           (ii) 720         (iii) 240        2. (i) 207360     (ii) 87091203. (i) 48               (ii) 72           4. 2. 6! 8!5. 86400                                  6. 6048007. (i) 2! 9! = 725760              (ii) 10!/2 = 18144008. 144                                       9. 864010. 4320                                   11. 420012. (i) 720          (ii) 120          (iii) 24          (iv) 240             (v) 14413. 1440            14. 1680                           15. 24  

Permutations of objects which are not all different

The number of permutations of n things taken all together, when p of the things are alike of one kind, q of them alike of another kind, r of them alike of a third kind and the remaining all different is n! / [p! q! r!]

The number of permutations of n objects, of which m are of one kind and the rest n -m of another kind, taken all at a time is n! / [m! (n-m)!]

Illustrative Examples

Example

In how many ways can the letters of the word permutations be arranged such thati. there is no restriction ii. P comes before S iii. words start with P and end with S iv. T's are together v. vowels are together vi. order of vowels remains unchanged?

Solution

The given word has 12 letters -two Ts and 10 different letters.i. Total number of arrangements is12!/2! = 6. 11! ii. Out of above, P comes before S in half the arrangements. Hence required number of

arrangements = 3. 11! iii. As position of P and S is fixed, remaining 10 letters (that is, two T's and eight other

different letters) can be arranged in 10!/2! = 5. 9! ways. iv. Considering two T's as a block, we have to arrange 11 different things, which can be

done in 11! ways. v. Considering the five vowels in given letter -E, U, A, I, O as a block, we have 8 things

having 2 alike things (T's). So this can be arranged in 8!/2! = 4. 7! ways. Now within the block, 5 different vowels can be arranged in 5! ways. Hence required number of arrangements is 4. 7!5! ways.

vi. If order of five vowels has to remain unchanged, we can consider them like five alike things, so only one ordering is possible. Thus we have 12 things of which 2 are alike and 5 are alike. Hence required number of arrangements is 12!/[2! 5!]

Exercise

1. How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used?

2. In how many ways can five red marbles, two white marbles and three blue balls be arranged in a row?

3. In how many ways can you arrange six identical coins in a row so that you get exactly 2 heads?

4. Find the number of arrangements that can be made out of the following words:(i) BANANA(ii) APPLE(iii) PINEAPPLE(iv) INDEPENDENCE(v) ASSASSINATION

5. How many arrangements can be made with the letters of the word MATHEMATICS if(i) there is no restriction(ii) vowels occur together(iii) all vowels don't occur together(iv) consonants occur together(v) M is at both extremes(vi) order of vowels remains unchanged?

6. How many different numbers can be formed out of all the digits of 111223? How many of these are greater than 300000?

7. How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4? 8. How many numbers greater than 100000 can be formed by using the digits 2, 4, 2, 3,

0, 2 taken all together? 9. How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits

always occupy odd places? 10. If permutations of all the letters of the word AGAIN are arranged in dictionary order,

which is the fiftieth word? 11. In how many ways can the letters of the word ASSASSINATION be arranged so that

all the S's occur together?

Answers

1. 1260                       2. 2520            3. 154. (i) 60                      (ii) 60                (iii) 30240        (iv) 1663200        (v) 108108005. (i) 4989600            (ii) 120960        (iii) 4868640     (iv) 75600    (v) 90720               (vi) 415800        6. 60, 10          7. 3608. 100                        9. 18                10. NAAIG      11. 151200  

Circular Permutations

In general, the number of ways of arranging n objects around a round table is (n-1)!An easier way of thinking is that we "fix" the position of a particular person at the table. Then the remaining n -1 persons can be seated in (n-1)! ways. Done!Thus the number of ways of arranging n persons along a round table so that no person has the same two neighbours is(n-1)!/2Similarly in forming a necklace or a garland there is no distinction between a clockwise and anti clockwise direction because we can simply turn it over so that clockwise becomes anti clockwise and vice versa. Hence the number of necklaces formed with n beads of different colours = (n-1)!/2

Illustrative Examples

Example

In how many ways can 3 men and 3 women be seated at a round table ifi. no restriction is imposed ii. each woman is to be between two men iii. two particular women must sit together iv. two particular women must not sit together v. all women must sit together vi. there is exactly one person between two particular women?

Solution

i. Total six persons can be seated at a round table in 5! = 120 ways.

ii. Three men can be seated first at the round table in 2! = 2 ways.Then the three women can be seated in 3 gaps in 3! = 6 ways.Hence the required number of ways = 2 x 6 = 12

iii. Temporarily treating two particular women as one big fat woman, five persons can be seated at a round table in 4! = 24 ways. However these two women can be arranged within themselves in 2! = 2 ways.Hence the required number of arrangements = 24 x 2 = 48

iv. As out of total 120 arrangements, there are 48 ways in which these two women sit together, the required number of arrangements = 120 -48 = 72

v. Temporarily treating three women as one person, four persons can be arranged at round table in 3! = 6 ways. Further, these 3 women can be arranged among themselves in 3! = 6 ways.Hence the required number of arrangements is 6 x 6 = 36

vi. Temporarily leave aside two particular women. The remaining 4 persons can be seated in 3! = 6 ways. Now these two particular women may be seated "around" any of 4 persons, and further the two can be arranged within themselves in 2 ways.Hence the required number of arrangements is 24 x 6 = 48

Example

i. A cat invites 3 rats and 4 cockroaches for dinner. How many seating arrangements are possible along a round table? Assume that animals of a species all look alike, though they will be deeply offended at this assumption.

ii. If m indistinguishable men from Mars and n indistinguishable women from Venus sit around a round table, how many possible seating arrangements are there?

Solution

i. "Fix" the position of the cat. Now remaining 3 rats and 4 cockroaches can be seated in 7!/(3! 4!) = 35 ways.

ii. Important. You may think that the formula (m -n -1)!/[m! n!] should work in such cases. Try putting m = 3, n = 3, you get 5!/[3! 3!] = 10/3, which is a fraction! In general, there is no formula for circular permutations where all items are repeated. However, even if a single item is there which is not repeated, we can "fix" its position and then find permutations of all remaining items.

Exercise

1. In how many ways can 7 boys be seated at a round table so that two particular boys are(i) next to each other(ii) separated?

2. In how many ways can 4 ladies and 4 gentlemen be seated at a round table so that all ladies sit together?

3. 6 person sit around a table. In how many ways can they sit so that no person has the same neighbors?

4. How many different necklaces can be made with 6 beads(i) of different colors(ii) of same color?

5. Find the number of ways in which 5 men and 4 women can be seated round a table so that no two women are together.

6. In how many ways can 7 men and 7 women be seated round a table so that no two women are together?

7. In how many ways may six Hindus and six Muslims sit round a table so that no two Hindus sit together?

8. Three boys and three girls go out for dinner. A shy boy does not want to sit with any girl and a shy girl does not want any boy as a neighbour. How many seating arrangements are possible?

9. A round table conference is to be held between 20 delegates. How many seating arrangements are possible if two particular delegates are(i) always to sit together(ii) never to sit together(iii) always separated by exactly one person?

10. Indian cricket team sits down for dinner at a round table. In how many arrangements is Saurav flanked by Sachin and Dravid?

11. (i) How many ways can a necklace be formed from 2 red and 2 blue beads?(ii) Two twin brothers are married to two twin sisters. In how many ways can they sit at a round table?

12. How many different garlands can be made from 6 marigolds and 2 roses?

Answers

1. (i) 240     (ii) 480         2. 576         3. 604. (i) 60       (ii) 1             5. 2880       6. 36288007. 86400                         8. 49. (i) 2(18!) (ii) 17(18!) (iii) 2(18!)10. 8064011. (i) 2        (ii) 2            12. 4  

Combinations - Formula for nCr

The number of combinations of n different things taken r at a time is                   nCr = n!/[ r! (n -r)!]

Corollaries:

i. nCr = [n (n -1)(n -2).....upto r factors]/r! ii. nCn = 1 iii. nC0 = 1 iv. nCr = nCn -r

Illustrative Examples

Example

Prove that nCr +nCr -1 =n +1Cr.

Solution

nCr +nCr -1 = n!/[ (n -r)! r!] + n!/[(n -(r -1))! (r -1)!]

= = [n!/[(n -r)!(r -1)!]].[(n -r +1 +r)/[r. (n -r +1)]]= [(n +1) n!]/[r (r -1)!. (n -r +1).(n -r)!]= (n +1)!/[r! (n +1 -r)!] = n +1Cr

Example

Evaluatei. C(12, 5) ii. 10C8 iii. 10C7 +10C6 iv. 15C8 +15C9 -15C6 -15C7

Solution

i. C(12, 5) = 12!/(5! 7!) = 792 ii. 10C8 = 10C2 = (10.9)/(1.2) = 45 iii. 10C7 +10C6 = 11C7       (using nCr +nCr -1 =n +1Cr)

      = 11C4 = (11. 10. 9. 8)/(1. 2. 3. 4) = 330      (using nCr = nCn -r) iv. 15C8 +15C9 -15C6 -15C7  = (15C8 +15C9) -(15C6 +15C7 )

        =16C9 -16C7 = 0    (because16C9 =16C7)

Exercise

1. Evaluate the following:(i) C(15, 11)(ii) 100C98

(iii) 52C52

(iv) C(11, 7) -C(10, 6)(v) 7C4 +7C5 +8C6

(vi) 5Cr.

2. Prove that (i) nCr +2.nCr -1 +nCr -2 = n +2Cr

(ii) (iii) n.n -1Cr-1 = (n -r +1).nCr-1

3. (i) If 16Cr = 16Cr +2 find rC4

(ii) If nC2 = nC3 find nC5.(iii) If 2nC3 : nC3 = 11 : 1 find n.(iv) If nPr = 720 and nCr = 120 find r.

4. Find the value of 47C4 + (52 -r)C3 5. If nCr -1 = 36, nCr = 84 and nCr +1 = 126, find the values of n and r.

Answers

1. (i) 1365  (ii) 4950    (iii) 1   (iv) 120  (v) 84    (vi) 313. (i) 35      (ii) 1          (iii) 6   (iv) 3.      4. 52C4

5. n = 9, r = 3  

Combinations - Some important results

1. If out of n different things, any number of items can be chosen, it can be done in nC0 +nC1 +nC2 +... +nCn ways. Alternatively, any of n items may or may not be chosen. Hence number of selections = 2×2×...n times = 2n

=> nC0 +nC1 +... +nCn = 2n.It can be shown that nC0 +nC2 +nC4 +... = nC1 +nC3 +nC4 +... = 2n-1.

2. Out of n different things, at least one (or more) can be chosen in 2n -1 ways. 3. Number of combinations of n different things taken r at a time when p particular

things always occur is n -pCr -p. Number of permutations of these is n -pCr -p.r! 4. Number of combinations of n different things taken r at a time when p particular

things never occur is n-pCr. Number of permutations of these is n -pCr.r! 5. The number of ways in which m +n different things can be divided into two groups

containing m and n things respectively is (m +n)!/[m!n!]. The reason is that whenever you choose a group of m out of m +n, a group of n is automatically left behind. Number of combinations of m +n things taken m at a time is          n +mCm = (m +n)!/[m! (m +n -1)!] = (m + n)!/[m! n!]

6. If subgroups are equal i.e. n = m, then 2m things can be divided into two groups of m each in (2m)!/(m!)² ways. If you distribute things to two persons, then this formula gives number of subdivisions. If it is possible to interchange the two groups then number of divisions is (2m)!/[(m!)².2!]

7. Number of division of m +n +p things into groups of m, n, p things respectively is (m +n +p)!/[m!.n!.p!]

8. If 3m things are divided into 3 equal groups, then number of divisions is (3m)!/(m!)³ and if the groups are interchangeable, the number of divisions is (3m)!/[(m!)³.3!]

9. If there are p +q +r things, where p things are alike, q things are alike and r things are alike, a non-empty selection can be made in (p +1)(q +1)(r +1) -1 ways as 0, 1, 2, ..., p items of p may be chosen; 0, 1, 2, ..., q items of q may be chosen etc.

10. If there are p +q +r things, where p things are alike, q things are alike and remaining r are all different, then a non-empty selection can be made in (p +1)(q +1). 2r -1 ways. (Prove it!)

Illustrative Examples

Example

In how many ways can final eleven be selected from 15 cricket players ifi. there is no restriction ii. one of them must be included iii. one of them, who is in bad form, must always be excluded iv. two of them being leg spinners, one and only one leg spinner must be included?

Solution

i. 11 players can be selected out of 15 in 15C11 ways= 15C4 = (15.14.13.12) / (1.2.3.4) = 1365 ways.

ii. Since a particular player must be included, we have to select 10 more out of remaining 14 players. This can be done in14C10 = 14C4 = (14.13.12.11)/(1.2.3.4) = 1001 ways.

iii. Since a particular player must be always excluded, we have to choose 11 players out of remaining 14. This can be done in14C11 ways = 14C3 = (14.13.12)/(1.2.3) = 364 ways.

iv. One leg spinner can be chosen out of 2 in² C1 = 2 ways. Then we have to select 10 more players out of 13 (because second leg spinner can't be included). This can be done in13C10 ways = 13C3 = (13.12.11)/(1.2.3) = 286 ways.Thus required number of combinations = 2×286 = 572

Example

Out of 5 boys and 4 girls, a committee of 5 is to be formed so as to include no more than 2 girls. In how many ways can it be done?

Solution

We may choose 5 boys and no girl in 5C5.4C0 = 1×1 = 1 way only.We may choose 4 boys and 1 girl in 5C4 .4C1 = 5C1.4C1 = 5×4 = 20 ways.We may choose 3 boys and 2 girls in 5C3.4C2 = 10×6 = 60 ways.Hence required number of combinations = 1 +20 +60 = 81.

Example

i. Find the number of ways in which 18 different objects can be divided into three groups each containing 6 objects.

ii. Find the number of ways in which 18 different objects can be distributed equally among 3 persons.

Solution

i. If 18 different objects are to be divided into 3 groups of 6 each and the order of groups is not important, number of ways of doing so = 18!/[(6!)³.3!]

ii. If 18 different objects are to be distributed 6 each to 3 persons, it can be done in 18!/(6!)³ ways.

Example

There are 15 points in a plane, no three of which are in the same straight line except 4 which are collinear. Find the number of

i. straight lines ii. triangles formed by joining them.

Solution

i. If no three points out of 15 points lie on a line, then the number of lines is 15C2 as exactly one line is drawn through two points. Since 4 out of 15 points are collinear, they from only one line instead of 4C2 lines. Hence the number of lines gets decreased by 4C2 -1. Hence the required number of lines= 15C2 -(4C2 -1) = 105 -6 +1 = 100

ii. A triangle is formed by joining 3 non-collinear points. So if the 15 points are all non-collinear, 15C3 triangles can be formed. But here 4 points lie on a line and form no triangle instead of 4C3. Thus the required number of triangles= 15C3 -4C3 = 455 -4 = 451

Example

Find the number of (i) combinations (ii) permutations of four letters taken from the word EXAMINATION.

Solution

The word examination consists of 11 letters -    (AA), (II), (NN), E, X, M, T, O.The following combinations are possible:

(a) 2 alike, 2 alike:³C2 = 3 ways(b) 2 alike, 2 different:³C1×7C2 = 63 ways(c) all 4 different: 8C4 = 70 waysHence required number of combinations = 3 +63 +70 = 136.To find the number of permutations:In (a), the number of permutations = 3×4!/[2!2!] = 18In (b), the number of permutations = 63×4!/2! = 756In (c), the number of permutations = 70×4!= 1680Hence required number of permutations = 18 +756 +1680 = 2454.

Exercise

1. A roller coaster has 3 seats and 4 children want to ride. How many ride combinations are possible? (Make an impossible assumption that little children don't fight over who sits where!)

2. A baseball team has 13 members. How many lineups of 9 players are possible? The position of each member in the lineup is not important.

3. In an examination, a candidate is required to answer 6 out of 10 questions which are divided into two groups, each containing 5 questions, and he is not permitted to attempt more than 4 questions from each group. In how many ways can he make up his choice?

4. A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

5. Out of 3 books on Economics, 4 books on Political Science and 5 books on Geography, how many collections can be made if each collection consists of (i) exactly one book on each subject (ii) at least one book on each subject?

6. Find the number of three digit numbers (repetitions allowed) such that at least one of the digits is 9.

7. A committee of 6 is chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. In how many different ways can it be done if two particular women refuse to serve on the same committee?

8. There are 6 boys and 3 girls in a class. A committee of 5 is to be selected such that 3 boys and 2 girls are on the committee.(i) In how many ways can the committee be selected?(ii) What is the number of ways if there is at least one girl on the committee?

9. In how many ways can a lawn-tennis mixed doubles be made up from 7 married couples if no husband and wife play in the same set?

10. In how many ways can a pack of 52 cards be divided equally among 4 players in order?

11. Determine the number of ways of obtaining 4 heads and 2 tails in 6 tosses of a coin. 12. From 4 mangoes, 5 oranges and 6 apples, how many selections can be made by

taking at least anyone of them? 13. A committee of 6 is to be formed from 6 boys and 4 girls. In how many ways can this

be done if the committee contains (i) 2 girls (ii) at least two girls? 14. A man has six friends. In how many ways can he invite one or more of them to a tea

party? 15. How many even numbers are there with three digits such that if 5 is one of the digits,

then 7 is the next digit? 16. Find the number of ways in which 5 white balls and 4 black balls can be arranged in

a row so that no two black balls are together. 17. Everybody in a room shakes hands with everybody else. The total number of

handshakes is 66. How many people are there in the room? 18. At a party, each person shook hands with everyone else. Mr. Lee shook hands with 3

times as many men as women. Mrs Lee shook hands with 4 times as many men as women. How many men and women were there at the party?

19. (i) There are 12 points in a plane of which 5 are collinear. These points are joined in pairs. Find the number of straight lines formed.(ii) How many triangles can be formed by joining the vertices of a hexagon?

20. How many diagonals are there in a polygon with n sides? 21. How many words can be formed by taking 4 letters at a time out of the letters of the

word(i) MATHEMATICS(ii) EXAMINATION

Answers

1. 4             2. 715             3. 200           4. 65. (i) 60      (ii) 3255           6. 252           7. 7800

8. (i) 60      (ii) 120             9. 420          10. 52!/(13!)4

11. 15        12. 209            13. (i) 90        (ii) 18514. 63       15. 365            16. 15            17. 1218. 16 men, 5 women        19. (i) 57       (ii) 2020. [n (n -3)]/2                   21. (i) 2454   (ii) 2454  

Compound Interest and Depreciation

Interest: It is the additional money besides the original money paid by the borrower to the money lender in lieu of the money used.Principal: The money borrowed (or the money lent) is called principal.Amount: The sum of the principal and the interest is called amount.Thus, amount = principal +interest.Rate: It is the interest paid on Rs 100 for a specified period.Time: It is the time for which the money is borrowed.Simple Interest: It is the interest calculated on the original money (principal) for any given time and rate.Formula: Simple Interest = (Principal x Rate x time)/100

Compound interest

Compound interest (abbreviated C.I.) can be easily calculated by the following formula:

      A = P   where A is the final amount, P is the principal, r is the rate of interest compounded yearly and n is the number of years.

C.I. = A -P = Remark.Interest may be converted into principal annually, semiannually, quarterly, monthly etc. The number of times interest is converted into principal in a year is called the frequency of conversion, and the period of time between two conversions is called the conversion period or interest period. Thus "rate of 5%" means a rate of 5% compounded annually; 12% compounded semi-annually means that each interest period of 6 months earns an interest of 6%. Thus the rate of interest per interest period is  r = (annual rate of interest) / (frequency of conversion)and the number of interest periods is  n = (given number of years) x (frequency of conversion).In solving problems on compound interest, remember the following:

1. A = P and C.I. = where A is the final amount, P is the principal, r is the rate of interest compounded yearly (or every interest period) and n is the number of years (or terms of the interest period).2. When the interest rates for the successive fixed periods are r1 %, r2 %, r3 %, ..., then the final amount A is given by

    A = 3. S.I. (simple interest) and C.I. are equal for the first year (or the first term of the interest period) on the same sum and at the same rate.4. C.I. of 2nd year (or the second term of the interest period) is more than the C.I. of Ist year (or the first term of the interest period), and C.I. of 2nd year -C.I. of Ist year = S.I. on the interest of the first year.5. Equivalent, nominal and effective rates of interestTwo annual rates of interest with different conversion periods are called equivalent if they yield the same compound amount at the end of the year. For example, consider an amount of Rs 10,000 invested at 4% interest compounded quarterly. So, the amount at the end of one year = 10000(1·01)4 = 10406. This is equivalent to interest of 4·06% compounded annually because 10000(1·0406) = 10406.When interest is compounded more than once in a year, the given annual rate is called

nominal rate or nominal annual rate. The rate of interest actually earned is called effective rate. In the above example, nominal rate is 4% while effective rate is 4·06%.If nominal rate is r% compounded p times in year, then effective rate of interest is

   6. Present value or present worth of a sum of Rs P due n years hence at r% compound interest is

   P.V. = In particular, present value of sum of a Rs P due one year hence (i.e. n = 1) at r% (compound) interest is

    P.V.=7. Equal instalments (with compound interest)

Loan amount = , whereP = each equal instalmentR = rate of interest per annum (or per interest period)T = time, say 3 years (or 3 interest terms).Note. If T = n years (or interest terms), then there will be n brackets.8. Formulae for populationIf the present population of a town is P and its annual increase is r%, the population after n

years will be P , and n years ago, the population was .If, however, there is annual decrease of r% per annum, the population after n years will be

, and n years ago, the population was .

Depreciation

All fixed assets such as machinery, building, furniture etc. gradually diminish in value as they get older and become worn out by constant use in business. Depreciation is the term used to describe this decrease in book value of an asset.There are a number of methods of calculating depreciation. However, the most common method which is also approved by income tax authorities, is the Diminishing Balance Method. Here each years depreciation is calculated on the book value (i.e., depreciated value) of the asset at the beginning of the year rather than original cost. Note that as the book value decreases every year, the amount of depreciation also decreases every year. Therefore, this method is also called Reducing Instalment Method or "Written Down Value Method".If the rate of depreciation is i% per year and the initial value of the asset is P, the depreciated

value at the end of n years is and the amount of depreciation is

. If n is large, log tables should be used for calculation.The number of years a machine can be effectively used is called its life span. After that it is sold as waste or scrap.

Illustrative Examples

Example

Find the compound amount of Rs 1500 for 6 years 7 months, at 5·2% compounded semi annually.

Solution

Using formula, we could find the value of But in these kinds of problems, generally we use compound interest for full interest period and simple interest for fractional interest period. Here we find compound interest for 13 interest periods and simple interest for 1 month.Required compound amount

   A = 1500        =1500(1·026)13(1·0043)Taking logs,   log A = log 1500 +13 log (1·026) +log (1·0043)            = 3·1761 +13(0·0112) +0·0017) = 3·3234.A = antilog (3·3234) = 2144Thus the required compound amount is Rs 2144.

Example

The simple interest in 3 years and the compound interest in 2 years on a certain sum at the same rate are Rs 1200 and Rs 832 respectively. Find (i) the rate of interest. (ii) the principal. (iii) the difference between the C.I. and S.I. for 3 years.

Solution

i. Let the principal be Rs P and rate of interest be R% p.a.According to the first condition of the question,(P x R x 3)/100 = 1200 => P x R = 40000 ... (1)According to the second condition of the question,

  

=>   

=> =>  4[(100)² +R² +2. 100. R -(100)²] = 832 R=> R² +200 R = 208 R  => R² +200 R -208 R = 0=> R² -8R = 0   =>  R(R -8) = 0=> either R = 0 or R -8 = 0=> either R = 0 or R = 8, but R cannot be zero.Hence the rate of interest = 8% p.a.

ii. On using (1), we get P x 8 = 40000, so P = 5000 iii. Rate of compound interest = 8% p.a. and principal = Rs 5000

Amount due after 3 years = Rs 5000 x

                                       = Rs 5000 x = Rs 6298·56Hence C.I. for 3 years = A -P = Rs 6298·56 -Rs 5000 = Rs 1298·56The difference between the C.I. and S.I. for 3 years = Rs 1298·56 -Rs 1200 = Rs 98·56

Example

The population of an industrial town is increasing by 5% every year. If the present population is 1 million, estimate the population five years hence. Also estimate the population three years ago.

Solution

Present population, P = 1 million, rate of increase = 5% per annum

Hence population after 5 years =

                    = 1000000 = 1000000 (1·05)5

                   = 1276280

Population three years ago = = 863838.

Example

Avichal Publishers buy a machine for Rs 20000. The rate of depreciation is 10%. Find the depreciated value of the machine after 3 years. Also find the amount of depreciation. What is the average rate of depreciation?

Solution

Original value of machine = Rs 20000,Rate of depreciation, i = 10%

Hence the book value after 3 years = 20000                  = 20000 (0·9)³ = 20000 (0·729) = Rs 14580.Amount of depreciation in 3 years = Rs 20000 -Rs 14580 = Rs 5420Average rate of depreciation in 3 years          = (5420/20000) x (100/3) = 9·033%

Exercise

1. A person invests Rs 5600 at 14% p.a. compound interest for 2 years. Calculate:(i) the interest for the first year.(ii) the amount at the end of the first year.(iii) the interest for the second year, correct to the nearest rupee.

2. A man saves every year Rs 4000 and invests it at the end of the year at 10% per annum compound interest. Calculate the total amount of his savings at the end of the third year.

3. The simple interest on a certain sum for 3 years is Rs 225 and the compound interest on the same sum at the same rate for 2 years is Rs 153. Find the rate of interest and the principal.

4. A sum of money is lent out at compound interest for two years at 20% p.a., C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at the same rate percent per annum, C.I. being reckoned half-yearly, it would have fetched Rs 482 more by way of interest. Calculate the sum of money lent out.

5. A man borrowed a certain sum of money and paid it back in 2 years in two equal instalments. If the rate of compound interest was 4 percent per annum and if he paid back Rs 676 annually, what sum did he borrow?

6. A sum of Rs 32800 is borrowed to be paid back in 2 years by two equal annual instalments allowing 5% compound interest. Find the annual payment.

7. A loan of Rs 4641 is to be paid back by 4 equal annual instalments. The interest is compounded annually at 10%. Find the value of each instalment.

8. A man borrows Rs 5800 at 12% p.a. compound interest. He repays Rs 1800 at the end of every six months. Calculate the amount outstanding at the end of the third payment. Give your answer to the nearest Re.

9. A man borrows Rs 5000 at 10% p.a. compounded annually. He repays Rs 1000 at the end of each of first three years. Find the amount which he has to pay at the end of the fourth year.

10. Divide Rs 21866 in two parts such that the amount of one in 3 years is same as the amount of the second in 5 years. The rate of compound interest is 5% per annum.

11. Two partners A and B together invest Rs 10000 at 12% per annum compounded annually. After 3 years, A gets the same amount as B gets after 5 years. Find their shares in the sum of Rs 10000.

12. A debtor may discharge a debt by paying (a) Rs 80000 now, or (b) Rs 100000 five years from now. It money is worth 5% compounded semiannually to him, which alternative should he accept?

13. At the birth of a daughter, a father wishes to invest sufficient amount to accumulate at 12% compounded semiannually to Rs 1 lac when the daughter is 16 years old. How much should he invest?

14. In buying a house, X pays Rs 1 lac cash and agrees to pay Rs 75000 two years later. At 6% compounded semiannually, find the cash value of the home.

15. The cost of a refrigerator is Rs 12000. If it depreciates at 10% per annum, find its value 3 years hence.

16. The present value of a machine is Rs 160000. If its value depreciates 6% in the first year, 5% in the second and 4% in the third year, what will be its value after 3 years?

17. I buy a mobike at Rs 20000 cash payment and three annual instalments of Rs 20000 each. If rate of interest is 15% compounded annually, what is the present worth of the mobike? If the rate of depreciation is 10%, what will be the resale value after 7 years?

18. A person buys a land at Rs 30 lacs and a year later constructs a building on it at the cost of Rs 20 lacs. Assuming that land appreciates at 20% annually and building depreciates at 20% for first 2 years and at 10% thereafter, find the total value of property after 5 years from date of purchase of land.

19. A loan of Rs 1 lac is to paid back in 5 equal annual instalments. The rate of interest charged is 20% annually. Find the amount of each instalment.

20. The population of a town increased from 2 lacs to 8 lacs in last 10 years. If the same trend continues, in how many years will it become 1·6 million?

21. Find the nominal rate compounded monthly equivalent to 6% compounded semiannually. Also find the effective rate of interest.[Hint. (1 +r/1200) 12 = (1·03) 2]

22. The machinery of a certain factory is valued at Rs 18400 at the end of 1990. If it is supposed to depreciate each year at 8% of the value at the beginning of the year, calculate the value of the machine at the end of 1989 and 1991.

Answers

1. (i) Rs 784   (ii) Rs 6384   (iii) Rs 894        2. Rs 132403. 4% p.a.; Rs 1875            4. Rs 20000     5. Rs 12756. Rs 17640                        7. Rs 1464·10  8. Rs 11779. Rs 3679·50                     10. Rs 11466 and Rs 1040011. Rs 5565, Rs 4435         [Hint. x (1·12)³ = (10000 - x) (1·12)5]12. He should accept (b)      13. Rs 1549614. Rs 166636·50               15. Rs 8748     16. Rs 13720017. Rs 65664·50, Rs 47869·40                    18. Rs 85 lacs approx.19. Rs 33438                      20. 5 years       21. 5·926%, 6·09%22. Rs 20000; Rs 16928  

Index Numbers

Index number is a specialized average designed to measure the change in the level of an activity or item, either with respect to time or geographic location or some other characteristic. It is described either as a ratio or a percentage. For example, when we say that consumer price index for 1998 is 175 compared to 1991, it means that consumer prices have risen by 75% over these seven years.Wholesale Price Index(WPI) and Consumer Price Index (CPI) are widely used terms. They indicate the inflation rates, and also changes in standard of living. Consumer price index is based on prices of five sets of items - Food, Housing (Rent), Household goods, Fuel and light, and Miscellaneous. Each item is based on study of a number of items - e.g. Food includes Rice, Wheat, Dal, Milk, and so on.

Methods of constructing index numbers

Price Relative means the ratio of price of a certain item in current year to the price of that item in base year, expressed as a percentage (i.e. Price Relative = (p2/p1)×100). For example, if a colour TV cost Rs 12000 in 1981 and Rs. 18000 in 1998, the price relative is       (18000/12000)×100 = 150.

Generally instead of one item, rates of a number of items are given, for current year as well as for base year. Sometimes different weights, or quantities are also given for those items. There are a number of ways to calculate index numbers in such cases.

Unweighted index numbers

1. Simple Aggregative

2. Simple Average of Price Relatives

Weighted (or Arithmetic Mean method)

1. Weighted Aggregative

2. Weighted Average of Price Relatives

where I = (p2/p1)×100

Illustrative Examples

Example

During a certain period, the cost of living index number goes from 110 to 200 and the salary of a worker is also raised from Rs 325 to Rs. 500. Does the worker really gains or loses, and by how much amount in real terms?

Solution

Real wage = [(Actual wage)/(cost of living index)] × 100So real wage of Rs 325 = (325/110) × 100 = Rs 295·45and real wage of Rs 500 = (500/200) × 100 = Rs 250So the worker actually loses, i.e. Rs (295·45 -250)= Rs 45·45 in real terms.

Example

With price index of 1991 as 100, the cost of living index for 1996 is 160 and for 1997 is 180. The salary of an employee increased from Rs 5000 in 1996 to Rs 5500 in 1997. Find out whether the real income in 1997 has increased or decreased as compared to 1996. Also calculate if any extra dearness allowance should be paid to him to compensate for loss. Also calculate purchasing power of rupee in 1996 and 1997.

Solution

The price relative of 1997 compared to 1996 isI = (p2/p1)×100 = (180/160)×100 = 112·50.Hence the wages of Rs 5500 in 1997 are equal to wages of(5500×100)/112.50 = Rs 4889 in 1996.Thus we find that real income in 1997 has decreased compared to 1996.Now let us assume that Rs x extra dearness allowance is paid to compensate for this loss.Hence (5500+x)/180 = 5000/160  => x = Rs 125Now purchasing power of a rupee is defined asPurchasing power = (Index number for base year)/(Index number for current year)So purchasing power of rupee in 1996 compared to 1991 is

Rs 100/160 = paise (100/160) × 100 = 62·5 paise 62 paisePurchasing power of rupee in 1997 compared to 1991 isRs = 100/180 paise = (100 × 100)/180 = 55·55 paise 56 paiseThus we see that due to inflation (price rise), the purchasing power of rupee is falling.

Example

Using 1985 as base year, the index numbers for the price of a commodity in 1986 and 1987 are 118 and 125. Calculate the index numbers for 1985 and 1987 if 1986 is taken as the base year.

Solution

Let p1, p2, p3 be prices in 1985, 1986, 1987.Then p1/p2 × 100 = 118 and p3/p2 ;× 100 = 125Now price index of 1985 with 1986 as base   = p1/p2 × 100 = (100/118) × 100 = 84·75Price index of 1987 with 1986 as base

= This procedure is called shifting of base.

Exercise

1. Find by simple aggregate method, the index number of the following data:

Commodity Base Price Current Price

Rice 140 180

Oil 400 550

Sugar 100 250

wheat 125 150

Fish 200 300

2. Construct index number for above data using price relatives. 3. Calculate a cost of living index from the following table of prices and weights.

  Weight Price index

Food 35 108.5

Rent 9 102.6

Clothes 10 97.0

Fuel 7 100.9

Miscellaneous 39 103.7

4. Construct Index number for following data:

  Butter Bread Tea Bacon

Relative Index 181 116 110 152

Weight 4 12 3 7

5. Construct the consumer price index number for 1998 on the basis of 1988 from the following data, using simple aggregates.

Commodity A B C D E

1988 Price per unit(Rs)

16.00 40.00 0.50 5.12 2.00

1988 Price per unit(Rs)

20.00 60.00 0.50 6.25 1.50

6. Redo above problem using price relatives. 7. Based on year 1988 as base, the index numbers for 1988, 1989, 1990, 1991 and

1992 are 100, 110,120, 200 and 400. Now taking 1992 as base year, calculate index numbers for years 1988, 1989,1990, 1991 and 1992.

8. Taking 1995 as the base year, with an index number 100, calculate an index number for 1999, based on (i) simple aggregate (ii) price relatives derived from the table given below:

Commodity A B C D

Price per unit in 1995

20 10 25 40

Price per unit in 1999

24 20 30 40

9. Taking 1995 as the base year with an index number 100, calculate an index number for 1995 based on the weighted average of price relatives derived from the table given below:

Commodity A B C D

Price per unit in 1995

10 20 5 40

Price per unit in 1995

30 35 10 80

Weight 20 30 10 40

10. Net monthly income of an employee was Rs 3000 per month in 1990 and Rs 5000 per month in 1998. Consumer price index in 1990 was 150. Find the consumer price index for 1998, given that net income of the employee is linked to consumer price index (so that his standard of living does not decrease).

Answers

1. 148·19                               2. 157·21

3. Index = =104.4        4. Index = = 1355. Consumer price index = 138·716. Consumer price index = 114·417. 25·0, 27·5, 30·0, 50·0, 1008. (i) 114·00 (ii) 135·00

9. Index = = 212·5        10. 250  

Shares, Stock and Debentures

To start any big business (company or industry), a large sum of money is needed and, in general, it is not possible for an individual to invest such a large amount. Then some persons, interested in the business, join together and form a company called joint stock company. They divide the estimated money required into small parts. Each such part is called a share. A share may have value Rs 5, Rs 10, Rs 25, Rs 100 etc. Each person who purchases one or more shares is called a share-holder The original value of a share is called its nominal value (abbreviated N.V.) or face value or printed value.

The price of a share at any time is called its market value (abbreviated M.V.) or cash value.

If the market value of a share is the same as its nominal value, the share is called at par.

If the market value of a share is more than its nominal value, the share is called at premium or above par. If a share of (face value) Rs 100 is selling at Rs 135, then it is said to be selling at a premium of Rs 35 or at Rs 35 above par.

If the market value of a share is less than its nominal value, the share is called at discount or below par. If a share of Rs 100 is selling at Rs 88, then it is said to be selling at a discount of Rs 12 or at Rs 12 below par. 6. The profit which a share holder gets for his investment from the company is called dividend. The dividend is always expressed as the percentage of the face value of the share. The dividend is always given (by the company) on the face value of the share irrespective of the market value of the share.

Shares are of two types:(i) Preferred shares(ii) Common shares or ordinary shares.Preferred shares carry a provision that dividend of a specific percent must be paid to preferred shareholders before any dividend is paid to common shareholders. Note that there is no guarantee of any returns; even preferred shareholders will get some dividend only if the company has some profits after paying for all expenses and taxes.Quotations:"Shares of 15% at Rs 145" means that(1) the face value of 1 share is Rs 100.(2) the market value of 1 share is Rs 145.(3) the dividend (profit) on 1 share is Rs 15 per annum.(4) the income on Rs 145 is Rs 15 for one year.STOCKJoint stock companies or the government can also raise loans from the market by issuing bonds or promisory notes. They promise to pay a fixed amount (called redemption value) on a future date and interest payments at fixed periods until that time. The money paid to company or government for buying such bonds is called stock. However, even before redemption date, the stock can be sold and purchased in open market, and the rate varies, just like shares. Note that stock can be bought or sold in fractions, unlike shares.BrokerageStocks and shares are sold and purchased in share market through stockbrokers, and their charges are called brokers commission or brokerage. Brokerage is usually calculated as percentage of face value (not market value or sale/purchase value), unless given otherwise.Note that:(i) Brokerage is added to market value while purchasing stock/shares.(ii) Brokerage is deducted from market value while selling.

Thus a Rs 100 stock at 95 and brokerage ½% means that a buyer will spend Rs to

buy a Rs 100 stock, while a seller will get Rs . The difference between these prices shows profit margin of the broker. In real life, however, brokers use the formula which is beneficial to them. If a share of Rs 100 is selling at Rs 300, they will calculate brokerage as percentage of sale value; if a share of Rs 10 is being sold at Rs 2, they may charge a fixed brokerage of 20 paise per share; if a debenture of Rs 100 is being sold at Rs 80, they will charge brokerage as percentage of face value!DEBENTURESTo meet working expenses, a company may borrow money from the public/shareholders by issuing debentures. They promise a fixed rate of interest for a fixed period. The main difference between stock/shares and debentures is that debentures give a fixed return, whether the company is in losses or profit. Also note that shares form a part of the capital of the company whereas debentures are debt taken by the company.

Illustrative Examples

Example

A man invests Rs 11200 in a company paying 6% dividend when its Rs 100 shares can be bought for Rs 140. Find

i. his annual income ii. his percentage income on his investment.

Solution

i. Since Rs 100 share can be bought for Rs 140,if the investment is Rs 140, income = Rs 6if the investment is Re 1, income = Rs 6/140if the investment is Rs 11200, income = Rs (6/140)×11200 = Rs 480

ii. Rs 480 is the income on Rs 11200,percentage of income =(480/11200)×100% = 30/7 %

         =

Example

Mr. Singh invested Rs 8000 in 7% hundred-rupee shares at Rs 80. After a year he sold these shares at Rs 75 each and invested the proceeds in 18% twenty five-rupee shares at Rs 41 each. Find

i. his gain or loss after a year. ii. his annual income from the second investment. iii. the percentage of increase in return on his original investment.

Solution

i. Since Mr. Singh invested Rs 8000 at Rs 80 each, the number of shares bought by Mr. Singh = 8000/80 = 100Dividend received on one share = 7% of Rs 100 = Rs 7Hence the total dividend received after one year = Rs 7×100 = Rs 700As Mr. Singh sold his shares at Rs 75 each, the sale value of his shares= Rs 75×100 = Rs 7500The amount received (proceeds) after a year= Dividend received on shares +sale value of shares= Rs 700 +Rs 7500 = Rs 8200Gain after a year = proceeds -investment= Rs 8200 -Rs 8000 = Rs 200

ii. Since Mr. Singh invested his proceeds i.e. Rs 8200 in twenty-five rupee shares at Rs 41 each, the number of shares bought = 8200/41 = 200Dividend received on one share = 18% of Rs 25= Rs (18/100)×25 = Rs 4.75Total dividend received on his second investment= Rs 4.75×200 = Rs 900

iii. The increase in return = dividend on second investment -dividend on first investment= Rs 900 -Rs 700 = Rs 200The percentage of increase in return on his original investment= (200/8000)×100% = 2·5%

Example

A company is authorised to issue 1 lac shares of face value Rs 100 each. It decides to issue 70000 shares in first instance. It fixes application money as Rs 40, allotment money as Rs 30 and first and second calls as Rs 15 each. The company receives applications for 50000 shares only. At a given time if investors holding 6000 shares have not paid first call money and investors holding 4000 shares have not paid second call money, describe various types of capital.

Solution

Authorised capital = 1 lac × 100 = Rs 1 croreIssued capital = 70000 × 100 = Rs 70 lacsSubscribed capital = 50000 × 100 = Rs 50 lacsAs both calls have been issued,Called up capital = Rs 50 lacsUncalled capital = 0Paid up capital = 40000×100 + 6000×70 + 4000×85 = Rs 47,60,000Unpaid capital (or calls in arrear) = 6000×30 + 4000×15 = Rs 2,40,000

Example

A man sold Rs 4500 of 3% stock. He sold the stock at 87¾ and invested the proceeds in 4% stock at 98¾. Find the change in his annual income. (Brokerage = ¼%)

Solution

Since the man holds Rs 4500 stock at 3%, his annual income= Rs (3/100)×4500 = Rs 135Since he sells this stock at 87¾, brokerage ¼%, money received by selling this stock= Rs 87.50×(4500/100) = Rs 3937·50Now in new stock, investment of Rs 99 (i.e. 98¾ + ¼)earns a return of Rs 4.Hence investment of Rs 3937·50 will earn an income of

= Rs (4/99)×3937.50 = Rs 159·09Hence, increase in annual income = Rs 159·09 -Rs 135 = Rs 24·09

Example

A small investor holds some 6% preference shares quoting at 112 "cum-div". If he sells just after getting the dividend, what price would he get? What amount would he realise if he holds 50 shares?

Solution

Note that "ex-div" means "without dividend" and "cum-div" means including dividend.Thus cum-div price = ex-div price +dividendAs cum-div price is Rs 112, the ex-div price after claiming the 6% dividend is Rs 112 -Rs 6 = Rs 106.Also, the amount realised by selling 50 shares = Rs 106×50 = Rs 5300

Example

A company has a total capital of Rs 2 crores divided equally into preference shares of 6% and ordinary shares, each of face value Rs 100. It gives an annual dividend of Rs 10 lacs. If a person holds 100 preference shares and 200 ordinary shares, how much dividend does he receive?

Solution

As the total capital is Rs 2 crores, the number of preference shares = (1 crore)/100 = 1 lac, and the number of ordinary shares is also 1 lac.Hence total dividend paid to 6% preference shareholders= 1 lac × 6 = Rs 6 lacDividend paid to ordinary shareholders= Rs 10 lac -Rs 6 lac = Rs 4 lacDividend paid to ordinary shareholders= (4 lac)/(1 lac) % = 4%Thus the dividend received by a person holding 100 preference shares and 200 ordinary shares= Rs (100×6 + 200×4) = Rs 1400

Exercise

1. At what price should a 6·25% Rs 100 share be quoted when the money is worth 5%? 2. At what price should a 6·25% Rs 50 share be quoted when the money is worth 10%? 3. A man buys Rs 40 shares of a company which pays 10% dividend. He buys the

shares at such a price that his profit is 16% on his investment. At what price did he buy each share?

4. A company with 10000 shares of Rs 100 each, declares an annual dividend of 5%.(i) What is the total amount of dividend paid by the company?(ii) What would be the annual income of a man who has 72 shares in the company?(iii) If he received only 4% on his investment, find the price he paid for each share.

5. A man sold some Rs 100 shares paying 10% dividend at a discount of 25% and invested the proceeds in Rs 100 shares paying 16% dividend quoted at Rs 80 and thus increased his income by Rs 2000. Find the number of shares sold by him.

6. A man invests Rs 6750, partly in shares of 6% at Rs 140 and partly in shares of 5% at Rs 125. If his total income is Rs 280, how much has he invested in each?

7. Divide Rs 5300 into two parts such that if one part is invested in 7% hundred-rupee shares at Rs 98 and the other in 8% at par, the resulting incomes are equal.

8. Divide Rs 95680 into two parts such that if one part is invested in 8% Rs 100 shares at 4% discount and the other in 9% Rs 100 shares at 8% premium, the annual incomes are equal.

9. A person invested 20%, 30% and 25% of his savings in buying shares of three different companies A, B and C, which declared dividends of 10%, 12% and 15% respectively. If his total income on account of dividends be Rs 2337·50, find his savings and the amount which he invested in buying shares of each company.

10. A person transfers the shares of face value Rs 11000 from 4% at 92 to 5% at 110. Find the change in his income.

11. A person invests equal amounts in 5% stock at Rs 110 and 6% stock at Rs 126.If his total income from the investments is Rs 82, find the amount invested in each case.

12. Sohan invests Rs 6760 partly in 5% stock at Rs 120 and the rest in 4% stock at Rs 96.If he derives equal income from each investment, how much has he invested in each stock?

13. What amount of money will Ankita get by selling 6% debentures worth Rs 20000 at 15% discount, the face value of each being Rs 100 and brokerage 2% of transaction amount?

14. A company has a total capital of Rs 4 crores divided equally into preferred shares of 6% dividend, and ordinary shares of face value 100 each. The company gives an annual dividend of 20 lacs. Find the dividend received by a shareholder holding 200 preferred shares and 3000 ordinary shares.

15. A company has a capital stock of Rs 200000 divided into 500 shares of 6% preferred stock and 1500 shares of common stock, each with par value of Rs 100. The company declares a dividend of Rs 15500. If Mohan holds 30 shares of preferred stock and 75 shares of common stock, find the amount of dividend he receives.

16. A person invested equal sums of money in 5% debentures and 6% preferred shares. Income was same in both cases. If debentures were available at par, find the issue price of preferred shares.

Answers

1. Rs 125                   2. Rs 31·25     3. Rs 254. (i) Rs 50000           (ii) Rs 360        (iii) Rs 125     5. 4006. Rs 3500, Rs 3250   7. Rs 2500, Rs 28008. Rs 47840 in 8%; Rs 47840 in 9% shares9. Rs 25000; Rs 5000, Rs 7500, Rs 625010. Increase of Rs 20   11. Rs 84012. Rs 3380 in each stock                  13. Rs 1666014. Rs 2400               15. Rs 805      16. Rs 120