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I

I

~

A COMPREHENSIVE TEXT BOOK

OF APPLIED MATHEMATICS·

Rakesh Gupta

(M.Sc. B.Ed.)

ABHISHEK PUBLICATIONS CHANDIGARH (INDIA)

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All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without permission in writing from the publishers/copyright owner.

ISBN Copyright First Edition

Publislied by

: 978-81-8247-225-9 : Author : 2009

Abhishek Publications SCO 57-59, Sector 17-C, Chandiharh. Phone: 0172-5003768 Telefax: 0172-2707562 e-mail: abhpub@ yahoo.com, www.abhishekpublications.com

Printed at : Shiva Offset. Naveen Shahadra, Delhi

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PREFACE ______________________________________ __

This book" A Compresensive Text Book of Applied Mathematics-II" has been written for the polytechnic diploma level students preparing for a carrier in Engineering. It covers the latest syllabi of various state boards of technical education in North Region. The subject has been developed in a systematic, logical and concise form to meet the requirements of all types of students. The book has been written in a style which makes it extremely useful for self study. The examples are so arranged that the easier problem come first and the difficult ones later. At the end of chapter a large number of multiple choice questions have been provided. This book covers the complete syllabus but some typical examples in some of the topics have been included to maintain the flow of ideas. The author lay no claim to the original research in preparing this book. Available sources on the subject has been used frequently. The subject matter has been arranged in a proper sequence and presented in a simplified manner.

I feel that no work is perfect and there is always a scope for further improvement Errors might have been crept in inspite of utmost care. Corrections and suggestions for further improvement will be thankfully acknowledged and will be implemented in the subsequent editions.

In last but not least, my thanks are due to my publisher" Abhishek Publications, Chandigarh", for bringing out this book in such a short period of time. Suggestions for further improvement of text are again welcome.

Rakesh Gupta (M.5c. B.Ed.) Lecturer, S.C.N. Polytechnic, Chourmastpur, Ambala

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SYLLABUS ____________________________________ __

APPLIED MATHEMATICS-II

RATIONALE Applied Mathematics forms the backbone of engineering discipline. Basic elements

of permutations and combinations trigonometry, vector, complex number and statistics have been included in the curriculum as foundation course and to provide bases for continuing education to the students.

DET AILED CONTENTS

1. Co-ordinate Geometry (25 hrs)

1.1 Area of triangle, centroid and incentre of triangle (given the vertices of a triangle), simple problems on locus.

1.2 Equation of straight lines in various forms (without proof)with their transformatio~ from to another angle between two lines and perpendicular distance formula(without proof)

1.3 Circle: General equation and its characteristic's given: The centre and radius Three points on it The Co-ordinates of the end's of the diameter

1.4 Conics (parabola, ellipse and hyperbola), standard equation of conics (without proof) , given the equation of conics to calculate foci, directrix, eccentricity, laws rectum, vertices and axis related to different conics Differential Calculus.

2. Differential Calculus (30 hrs)

2.1 Concept of function Four standard limits

Lt x n _ an L t sin x Lt a x_I Lt 1 --------- -- - - ---- (1 + x) x

x~a x a 'x->O x 'x->O x 'x~O

2.2 Concepts of differentiation and its physical interpretation

Differential by first principle of xn, (ax + b)n, Sin x ,Cos x, tan x, cosec x and x x cot x, e ,a log x

Differentiation of sum, product and quotient of different functions

Logarithmic differentiation, Successive differentiation excluding nth order

Application of derivatives for (a) rate measure, (b) errors, (c) real root by Newton's method, (d) equation of tangent and normal, (e) finding the maxima and minima of a function (simple engineering problems)

3. Integral Calculus (20 hrs)

3.1 Integration as inverse operation of differentiation

3.2 Simple integration by substitution by parts and by partial fractions

3.3 Evaluation of definite integrals(simple problems) by using the general properties of definite integrals

3.4 Application of integration for Simple problem on evaluation of area under a curve where limits are prescribed for circle, ellipse, parabola and straight line calculation of volume of a solid formed by revolution of an area about axis (simple problems) where limits are prescribed for sphere and cylinder to calculate average and root mean square of a function area by Trapezoidal Rule and Simpson's Rule

4. Differential Equations

Solution of first order and first degree differential equation by variable separation and their simple numerical problem.

CONTENTS __________________________________ __

Co-ordinate Geometry

1. The Point 2. The Straight Line 3 The Circle 4. The Conic Section

Differential Calculus

1. Functions and Limits 2. Derivation of Functions 3. Tangents and Normals 4. Rate of Change of Quantities 5. Maxima and Minima

Integral Calculus

1. Indefinite Integral 2. Definite Integral 3. Application of Integration and Numerical Integration

Defferential Equations

3 44 84 97

137 162 195 209 224

241 321 376

398

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Section I

Co-ordinate Geometry

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Chapter 1 THE POINT

Co-ordinate geometry is the branch of mathematics, which deals geometry algebraically.

Let X'OX and YaY' be two mutually perpendicular straight lines intersecting at a point '0' is called as origin. The horizontal line X'OX is called X - axis and vertical line YaY' is called Y-axis. These two axes divide the plane into four parts which are called quadrants as shown in the following fig.

Y

lInd Quadrant

X'

lllrd Quadrant

1st Quadrant

O(Origin)

TVth Quadrant

Y' Y-axis

x

X-axis

The X- axis and Y-axis taken together are called as rectangular axes or co -ordinate axes

Rectangular Co - ordinates of a point

Let P be any point in the plane of rectangular axes. Through P draw PL perpendicular on X - axis and PM perpendicular on Y-axis as shown in fig.

Y

M I--~x,---..P(x,y)

y

X' o L X

Y'

/4/

The horizontal distance (PM = OL) of point P from Y- axis denoted by 'x' is called abscissa of point P.

The vertical distance (PL = OM) of point P from X-axis denoted by 'y' is called ordinate of point P

The abscissa and ordinate taken together (x,y) is called co-ordinates (or rectangular co-ordinates) of point P.

Sign conventions for rectangular co-ordinates

1. All horizontal distances (i.e. x-eo-ordinates) to the right of Y-axis are taken as positive and left of Y-axis are taken as negative.

2. All vertical distances (i.e. y-co-ordinates) above the X-axis are taken as positive and below the X- axis are taken as negative

y

lInd Ist

~ y+ve I x +ve I y +ve X' 0 X

IIIrd IVth

~ y -ve I x +ve I y -ve Y'

Note: (i) Co-ordinates of origin are taken as (0,0). (ii) For any point on X-axis, ordinates (i.e. y-coordinate) is zero. (iii) For any point on Y-axis, abscissa (i.e. x-coordinate) is zero.

POLAR CO- ORDINATES OF A POINT P(r,O)

O~~---------------------------~X

Let 'a' be the origin and OX be the fixed straight line. Take any point 'P' join OP. The distance of point P from 'a', denoted by 'r' is called radius vector. The anglf XOP denoted by '8' is called vectorial angle.

The ordered pair (r,8) is called the polar co-ordinates of point P.

Note: The vectorical angle '8' is taken as positive in anticlock wise direction and as negative in clockwise direction.

/5/

RELATION BETWEEN RECTANGULAR AND POLAR CO - ORDINATE

Let (x, y) be the rectangular co-ordinates and (r, 8) be the poldf co-ordinates of the point 'I" y

p(x,y)

X'

Draw PL.l OX, then OL = x, PL = Y

Join OP, then OP = rand LXOP = 8

Now in the right angled triangle OLP,

i.e.

OL x -- = cos e or - = cos 8 OP r x = r cos e PL -= sin 8 OP

y . 8 or - = SIn r

Y'

----(i)

i. e. y = r sin e ---(ii) 'Squaring and adding (i) and (ii) we get

x2 + y2 = r2 cos8 + r2 sin2 e = r2 (sin2 8 +cos2+ e) = r2 (1) (','sin2 e + cos2 8 = 1)

or x2 + y2 = r2 or r = ± ~r-X-2 -+-y-2

(Rejecting -ve sign because r is positive)

Hence r = ~X2 + y2 Dividing (ii) by (i), we get

y rsin8 -=-- = tan8 x rcos8

or 8 = tan-l(~)

x

Note: The value of '8' depends upon the quadrant in which the point lies i.e. signs (+ve or -ve) of the values of 'x' and 'y' which will be clear from the following examples.

/6/

p(~3,l)

X'

Y'

Here x= J3 y=l

r = J( J3) 2 -~ -(-1)-; c= 2

and 8 = tan-1 (~)

= tan-1 ( 53) 8 = 30°

x X'

Here

and

Casell

y

e

Y'

x = -J?, Y =-1

r = [(-J3)2 ~(--l;; = 2

8 = tan-1 (~33)

(-1 ) = tan-1 -=J3 8 = 30° or 210°

x

(As the point P lies in the third quadrant therefore angle '8' will be (180 + 30°) i.e. 210° not 30°) Hence 8 = 2 I 0°

Method to change Cartesian (rectangular) form to Polar form

Put x = r cos8 , y = r sin8 and simplify

Method to change polar form to Cartesian form

x Put cos8 = -

r

sin8 = y and r

r = ~X2 + y2 and then simplify to remove fractional powers.

fJI DISTANcr; FORMULA

To find the distance between two points whose co-ordinates arc given.

Let P(Xl'Yl) and Q(x2 , Y2) be two given points. From P and Q, draw PL and QM perpendiculars on x - axis. From P draw PR perpendicular to QM.

Let'd' be the distance between the points P & Q.

In right angle ~PQR, PQ! = PR2 + RQ2 - - -(i)

y

I.?-'"

Q (X"Y2) (Using Pythagorous theorem)

Now PQ= d PR = LM = OM-OL = X2-X1 RQ = QM-RM = Y -y

~~ p r R 2 • 1

From (i) , d2 = (X2-X1)2 + (Y2-Yl)2

or d= ±~(X2-Xl)2+(Y2-Yl)2 Rejective -ve sign as d is the distance o

d= ~(X2-Xl)2+(Y2-Ylf Hence, distance between two points

= ~(Difference of abscissa)2 + (Difference of ordinate)2 Some Important points to Remember

) L \1 x

(i) When three points A, B, C are given and we have to prove that A, B, C dre collinear (i.e. A, B, C lie on the same straight line) then we will show that

(ii) (a)

AB +BC = AC or AC +BC = AB or AB +AC = BC

When three points are given and we have to prove that an isosceles triangle show that any two sides are equal

(b) an equilateral triangle show that all three sides are equal

(c) right angled triangle show that sum of the squares of any two sides is equal to square of the third side.

(iii) When four points are given and we have to prove that (a) a square

show that all four sides are equal and diagonals are equal. (b) a rectangle

show that pair of opposite sides are equal and diagonals are also equal. (c) a parallelogram

/8/

show that pair of opposite sides are equal but diagonals are not equal (d) a rhombus

show that all four sides are equal but diagonals are unequal.

Example 1 Plot the following points and find the quadrants in which they lie. (i) P (3, 7) (ii) Q (3, -7) (iii) R (-3, -7) (iv) 5 (-3, 7)

Solution S(-3,7) 8 R(3,7)

$------- r------~

6

5

4

3

2

1

-7 -6 -5 -4 -~ -2 -1 1 2 -1

-2

-3

-:1

-5

-6

®------- -7------®

4 5 6 7

R(-3,-7) -8 Q(3,-7)

(i) P(3, 7) lies in the 1st quadrant. (ii) Q(3, -7) lies the IVth quadrant. (iii) R(-3, -7) lies in the IIIrd quadrant. (iv) 5(-3, 7) lies in the Ilnd quadrant.

Example 2 (i) (iii)

Find the polar co-ordinates of the following:-(1, 1) (ii) (1,-1) (-1,-1) (iv) (-1,1)

Solution (i) Given Cartesian coordinate is (1,1)

Here x = 1, Y = 1

X'

y

P(l,l)

x

1t or 8 =:

4 Because x and y both arc positive, therefore, poinllies in the 1st quadrant.

I-Ience polar coordinate is ( .J2,~-) (ii) Given cartesian coordinate is (1, -1)

Here x = 1, Y = -1

r = ~X2--; y2 ~ J(1)2-~-(::-il ~ .J2-

Y

x

/9/

x

and 8 == tan-l(~-) = tan-1 ( i~) = tan-1(-I) P(l,-l)

8 -= tan-1 (-1) Y' Because x is positive and y is negative, therefore point lies in the ivth quadrant

71t Y 8=4

Hence polar co-ordinate is ( .J2, 7: )

(iii) Given Cartesian coordinate is (-1,-1) X' Here x = -1, Y = -1

r = ~~-:£-;y-i cc J(_~)2 + (~I)i C~ Ji P(-l,-l)

and 8=tan-l(~-) =tan-l(=~) = tan-l (1) Y'

5 -IT 4

Because x and y both are negative, therefore, point lies in the I1Ird quadrant

8 = 51t 4

Hence polar coordinate is

(iv) Given Cartesian coordinate is (-1, 1) Here x = -1, Y = 1

r = ~X2 +-y£ --= J( _1)2 ~.(1)2 =--li

x

/10/

and e:::tan-l(~) =tan-l(~l-) =tan-1 (-1) Because x is negative and y is positive, therefore point lies in the lInd quadrant

e = 31t 4

Hence polar coordinate is

P(-l,l)

X'

Example 3. . Change the following into Cartesian co ordinates:-

(i) (ii)

Solution

(i) Given polar co-ordinate is ( 2,- ~ )

Here 1t

r = 2 and e =--4

x = r cose = 2 cos ( - ~) = 2 cos ~ 1

=2xTz=-J'i

and y = r sine = 2sinC- ~) = -2 sin ~

Hence Cartesion co-ordinate is (-J2,--J2)

(ii) Given polar co-ordinate is ( 4, 231t

)

Here 21t

r = 4 and e = ---3

x = r cose = 4 cos ( .?31t

) = 4 cos ( 1t - ;)

Y

X

Y'

1t = -4 cos-

3

1 == -4 x - =-2

2

[.: cos (1t-8) = -cos8]

and y = r sin8 ,= 4 sin ( 231t

) = 4 sin ( 1t - ;)

1t = 4 sin-

3

./3 = 4x-

2

= 2./3

[.: sin (1t-8) == sin8]

Hence Cartesian co-ordinate is (-2,2./3)

Example 4 Solution

or or

Example 5 Solution

or

Transform the cartesian equation X2+y2 = 2ax into polar equation. The given equation is

X2+y2 = 2ax ---- (i) Putting x = r cos 8 and y = r sin 8 in equation(i) we get, r2 cos28 + r2 sin28 = 2ar cos8 r2(sin~8 + cos28) = 2ar cos8 ( .: sin28 + cos28 = 1) r = 2a cosO is the required equation in polar form.

Transform the polar equation r2 = a2 cos28 into cartensian form. The given equation is r2 = a2 cos28 r2 = a2 (cos28 - sin28)

Putting cos8 = ~, sin8 = y and r = ~X2 + y2 , we get r r

or or or

Example 6 Solution

r2(x2+y2) = a2(x2_y2) (X2+y2)(X2+y2) = a2(x2_y2)

(X2+y2)2 = a2(x2_y2) is the required equation in cartesian form. Find the distance between the points (5,7) and (-3,1) Let A = (5,7) and B(-3,1) be the given points.

/11/

/12/

Then AB = ~-;:;)2--:0r~-~Yl-)2

= ~(-3-5)2 +(1_7)2

= .J64 + 36 = .J100 = 10

Example 7 If the points (x,y) be equidistant from the points [(a+b),(b-a)] and [(a-b), (b+a)] show that bx = ay.

Solution

or or or

01'

or or or

Example 8 Solution

A [(a+b),(b-a)]

P(x.y)

B [(a-b),(b+a)]

Let P(x,y) be the point which is equidistant from the points A((a+b), (b-a)) and B((a-b),(b+a)). PA=PB PA2 = PB2 . [x-(a+b)F +[y-(b-a)F = [x-(a-b)F +[y-(b+a)]2 x2+(a+bf -2x(a+b) +y2 +(b-a)2 -2y(b-a)

= xZ+(a-b)2 -2x(a-b) +y2 +(a+b)2 - 2y(a+b) -2x(a+b)-2y(b-a) = -2x(a-b)-2y(b+a) ax+bx+by-ay = ax-bx+ay+by bx+bx = ay+ay 2bx = 2ay or bx = ay is the required result.

Show that the points (-2,3), (1,2), (7,0) are collinear. Let A( -2,3), B(l,2), C(7,O) be the given points

AB = .J(1 + 2)2 + (2 -- 3)2 = /3-z t- (_1)2 = .J9--:;-i = .Jio AC = J(7t-il ~.~ 3)2 =~ /92-~-(~3)-~- -.CO Jsi +-9 =0 -.190 ~ 3./i6 BC = ~ (7 - 1):2 + (0 -- 2):2 -'" ~ b 2. + ( - 2):2 =.J 36 + 4 =- .J 40 = 2 JiO

Here AB+BC = .Jio ·f 2-.1i-o = 3·J10-= AC

/13/

• • ~ A B ,10

I~ ~I~ 2, 10

~I

I~ 3-10

~I

Given points lie on the same line and hence they are collinear.

Example 9 If P(at2, 2at) and Q( t~ , -~a) be the two given points and S(a,O) is another 1 1

point. Prove that SP + SQ is constant for all values of t.

Solution The given points are

(a -2a)

P(at2, 2at), t2 '-t- and S(a,O)

111 1 - + - = + ---;========

.. SP SQ ~(a - at2)2 + (0 - 2at)2

/14j

- --- ---- 1- -----------1 [1 t2 1 - a (1+t2 ) (t2 +1)

1 1

1 = -- which is constant for all values of t.

a

lll'nn' ST; 1 5Q is independent of 't' and so is constant for all values of t. l:x

/15/

Section-Formula

(a) IEapo:intP (x,y) divrlesthe line EEgffi entj):in:ing thepo:intBA ~'Yl) and B(X2'Y2) internally in the ratio m:n, then the co-ordinates of point P are given by

mX2 + nX 1 x= m+n

mY2 +nYl (Xlly,) m:n (x"y,) and • • • y=

m+n A P(x,y) B

[~~ = :] (b) If a point P(x,y) divides the line segment joining the points A(X1'Yl) and B(X2'Y2)

externally in the ratio m:n, then the co-ordinates of points P are given by

and

Proof (a)

In

or

or

or

mX2 - nX 1 x= m-n

mY2 -nYl y=

m-n

Internal Division

~ADP and ~BCP LADP = LBCP = 90°

LAPD = LBPC

By A.A. similarity ~ADP is similar to ~BCP

AD DP AP --=-=-Be CP BP

• P(x,y)

• A(xlly,)

[V erticall Y opposite angles]

X'

A (Xlly,)

Y

o

Y'

• B(x"y,)

• • P(x,y)

B(x"y,)

X

[ ",' corresponding sides of similar ~s are proportional]

DL-AL LM m BN-CN MN n

Y -Yl OM-OL- m = = Y2 -Y ON-OM n

L-:-11.. _~ =_xl m Y2 -Y x2 _.- x n

/16/

(b)

or X-Xl m ---=-- and x2 -x n

Y-]l =E:. Y2 -Y n

or nx-nx1 =mx2-mx and ny-nYl=mY2-my or (m+n)x = mX2 + nXl and (m+n)y = mY2+nYl

or

(mX2 + nXl mY2 + ny1 )

Hence co-ordinates of point P are I m+n m+n

External Division Point P divides AB externally in the ratio m:n

I.e.

In

AP m -=-PB n

flAEP and flPCB LAEP = LPCB = 90° LBPC= LPAE

By AA similarly

- -(1)

--- [Alternate angles]

flAEP is similar to flPCB

AE EP AP -=-=-PC CB PB

X'

F y

o L M N

or EL-AL LN m PN--CN MN n

I-=:.Xi = ON - OL m Y-Y2 ON -OM n

[.: of (1) and corresponding sides

of similar fls are proportional

or

or

or

Y -.- Y 1 = X - Xl = ~ Y-Y2 X-X2 n

Y-Yl = m and X-Xl = m Y-Y2 n X-X2 n

or ny - nYl = my - mY2 and nx - nX1 = mx - mX2 or (n-m)y = nYl + mY2 and (n-m)x = nx1-mx2

ny 1 - my 2 ~n_x~I_-_m_x=-2 Y = --=--=----=-=- and X = -n-m n-m

or

x

/17/

mY2 - nYI mX2 - nX 1 or Y = -- --- -- and x = -- --------m-n m n

" . . (mX2 -nxi mY2 -ny l ) Ihe co-ordmates of pomt Pare ----,-----m--n m--n

COROLLARY Of SECTION FORMULA

Mid Point Formula -

The co-ordinates of point M(x,y) bisecting the line segment joining the points .\(X1'Yl) and B(X2'Y2) are given by

Proof

1 : 1 • •

M(x,y)

Point M bisects the line segment joining the points A(X1'Yl) and B(X2'Y2)'

M is the mid-point of line segment AB, i.e. AM = MB thus M divides the line

segment AB in the ration 1:1 internally. [AM 1] .: AM -= MB => MS- == 1"

By section formula, co-ordinates of point P are given by

mX 2 +nxl X== m+n

Here m: n : : 1 : 1

d mY2 +nYI an Y= m+n

(Xl +X2 YI +Y2)

Here Co-ordinates of mid point M of line segment AB are 2' 2

lixample 12 Find the co-ordinates of a point dividing the line segment joining the points (2,3) and (5,-7)

(i) Internally in the ratio 3:4 (ii) Externally in the ratio 2:1

Solution (i) Let A(Xl'Yl) = (2,3)

B(X2'Y2) = (5,-7) and m:n = 3:4

3:4 • • •

A(2,3) P(x,y) B(5,-7)

/18/

i.e.

or

or

(ii)

i.e

or

or

By internal section formula, co-ordinates of point P are given by

mx') + nXl d mY2 + nYl X= - an Y=

m+n m+n

x = (3)(5) + 4(2) and = (3)(-7) + 4(3) 3+4 Y 3+4

15+8 X=--

7 d

-21+12 an y=

7

23 -9 X=- and y=-

7 7

(23 -9)

Co-ordinates of point Pare -;;-'7 Let A(X.l'Yl) = (2,3)

B(X2'Y2) = (5,-7)

and m:n = 2:1

• A(x"y,) B(X",Y2)

By external section formula, co-ordinates of point P are given by

mX2 - nXl d mY2 - nYl X= an Y=

m -n m-n

x = (2)_(5)(-1)(2) and Y = (2)(-7) - (1)(3) 2-1 2-1

10-2 -14-13 X=-- andy=---

1 1 x = 8 and Y = -17 Co-ordinates of point Pare (8,-17)

• P(x,y)

Example 13 Find the ratio in which the point (2,5) divides the line segment joining the points (2,4) and (2,8).

Solution

i.e

Let the point (2,5) divides 'the line segment joining the points(2,4) and (2,8) in the ratio K:1

By internal section formula ,

mX2 + nX l d mY2 + nYl X= an Y= m+n m+n

2 = K(2) + 1(2) and 5 = K(8) + 1(4) K+1 K+1

2K+2 = 2K+2 and 5K+5 = 8K+4 -3K = 4-5

[

m:n 1 or ::1 = k:1(say)

-3K =-1 Which is true for every value of K.

1 or K="3

The required ratio is K:l

1 =3:1 =1:3

/19/

Hence the point (2,5) divides the line segment joining the given points internally in the ratio 1:3

Example 14 Find the ratio in which the point (16,-15) divides the line segment having end points (1,3) and (6,-3).

Solution

i.e

or

Let P(16,-15) divides the line segment AB having end points A(l,3) and B(6,-3) in the ratio K:1

By internal section formula

mX2 +nxl x=--=--~

m+n d mY2 +nYl an Y=

m+n

16 = K(16) + 1(16) K+l

16K+16 = 6K+l 10K = -15

-15 K= 10

-3 K=-

2

and -15 = K(-3) + 1(3) K+l

and -15K-IS = -3K+3 => -12K = 18

18 K= -12

-3 K=T

The point P divides AB internally in the ratio -3:2

or Externally in the ratio 3:2

Remark: Initially we suppose that the point P divides in the ratio K:l internally. If thw value of K comes out to be +ve, then the division is actually internal. However, if the value of K comes out to be -ve, then the division is external and we say that the given point divides in the ratio k:1 externally.

Example 15 Find the ratio in which the line segment joining the points (-3,4) and (4,-6) is divided by (i) x-axis, (ii) y-axis.

Solution (i) Let the co-ordinates of point P at which the x-axis cuts the line joining

the points A(-3,4) and B(4,-6) in the ratio K:l be (x,D). By internal section formula

/20/

(ii)

(-6)+1(4) Y - co-ordinates of P = 0 = k + 1

or or

or

0= -6K +4 6K = 4

2 K=3 The reqd. ratio is 2:3 Hence x-axis divides AB internally in the ratio 2:3

X' X

Let the co-ordinates of point Q at which y-axis cut the line joining the points A(-3,4) and B(4,-6) in the ratio K:1 be (o,y) [i.e. AQ : QB = K: 1]

x-co-ordinates of Q = 0

i.e.

or or

or

K(4) + 1(-3) = o

K+1 o = 4K-3 4K = 3

3 K =-

4 . . The required ratio is 3:4 Hence, y-axis divides the line segment AB internally in the ratio 3:4.

Example 16 The co-ordinates of three vertices of a parallelogram are (2,3), (-5,-7) and (2,-4), Find the co-ordinates of fourth vertex.

Solution

or

or

or or

Let the parallelogram be ABCD, having co-ordinates of A, Band C as (2,3), (-5-7) and (2,-4) respectively and let (x,y) be the co-ordinates of fourth vertex D. We know that the diagonals of parallelogram bisect each other (say at point P)

Therefore P is the mid point of AC as well as that of BD.

(~ 3 + ( -4) ) = ( -5 + x -7 + y) 2' 2 2 ' 2 (2 -1)=(-5+X -7+ Y)

, 2 2' 2

-5+x -7+y 1 --=2 and ---=--

2 2 2 -5+x = 4 and -7+y = -1 x = 9 and y = 6 Co-ordinates of Dare (9,6)

[By mid-point formula]

~ ____ D-2(X,y)

/21/

Example 17 The co-ordinates of midpoint of a line segment AB are (2,5) and that of A are (-4,-6). Find the co-ordinates of point B.

Solution Let M be the mid-point of AS and let (x,y) be the co-ordinates of point B.

or or

Example 18

Solution

or

By mid point formula

-4 +x -6+y 2=-- and 5=--

2 2 4 = -4 +x and 10 = -6 +y x = 8 and y = 16

• A(-4,-6)

The co-ordinates of other end point B will be (8,16)

M • •

(2,5) B(x,y)

Find a point on the line through the points P(-8,-4) and Q(4,6) which is twice as far from P as from Q.

Let R(x,y) be the required point on the line joining the points P and Q. According to Question, 2(PR) = RQ

M PR 1 ---RQ 2

• • • R(x,y) Q(4,6) P(-8,-4)

or PR : RQ = 1 : 2 i.e. the point R divides the line segment PQ internally in the ratio 1:2

By internal section formula, co-ordinates of point R are given by

or

or

x = (1)(4)+(2)(-8) and = 1(6)+2(-4) 1+2 Y 1+2

4 -16 6-8 and y = ------

3 3 x=

-12 -2 x= -- and y=-

3 3

x =-4 -2

and y=-3

Hence co-ordinates of point Rare (-4, -32 ) Example 19 The co-ordinates of points A and S of a ~ABC are (2,4) and (6,10)

respectively and lengths of sides AC and BC are 3cm. and 6cm. respectively. If the angle bisector of angle C meets the side AB at point P, find the co-ordinates of point P.

Solution Let Co-ordinates of point P be (x,y) We know by Angle Bisector Theorem "The bisector of an angle of a ~ divides the opposite side of ~ in the ratio of the sides containing the angle."

/22/

or

Example 20 Solution

Let

and

or

and

or

AP :.PB = 3: 6 = 1: 2

By internal section formula, co-ordinates of point P are given by

x = 1(6) + 2(2) and y = 1(10) + 2(4) 1+2 1+2

10 x = - and y=6

3

Co-ordinates of point Pare (13° ,6)

A (2,4) p (x,y) B(6,10)

Find the points of trisection of the line joining the points (1,-2) & (-3,4).

P and Q be the points of trisection of line segment AB such that AP: AB = 1:2 AQ: QB = 2:1 By internal section formula, co-ordinates of P are given by

x = 1(-3)+2(1) and y= 1(4)+2(-2) 1+2 1+2

-1 • x = - and y = ° 3 A(l,-2)

Co-ordinates of Pare ( -; ,0 )

Co-ordinates of Q are given by

2(-3)+1(1) and y= 2(4)+1(-2) 2+1 2+1

-5 x= -

3 and y = 2

Co-ordinates of Q are (-: ,2)

1:2 • p • Q

2:1

• B(-3,4)

Reqd. co-ordinates are (-;,0) and ( -: ,2)

MEDIAN AND CENTROID OF A TRIANGLE:

MEDIAN:

The median of a triangle is the line segment joining the vertex of a triangle to the mid-point of the opposite side of triangle.

For example in L\ABC, if D is the mid point of side BC, then AD is a median of L\ABC.

Similary BE and CF are also two medians of L\ABC, where E and Fare mid- points of sides AC and AB respectively.

Thus every triangle has three medians.

Centroid Of A Triangle:

/23/

A

c

The points of concurrency of the three medians of a triangle is called centroid of triangle and is generally denoted by 'G'.

Co-ordinates of centroid of a Triangle:-

Let A (Xl' Yl)' B (X2'Y2) and C(X3'Y3) be the vertices of a L\ABC. Let G be the centroid of L\ABC. It is a

property of centroid of triangle that it divides each median in the ratio 2:1 (internally), 2 towards vertex and 1 towards side.

By mid point formula, Co-ordinates of point D are given by

B(x"y,)

Now by internal-section formula, co-ordinates of point G (centroid) are given by

or

2(X2 ;X3 )+l(Xl ) x = ---'-----'----

2+1

and Y = Y 2 + Y 3 + Y 1 3

Thus Co-ordinates of centroid of L\ABC are given by

(Xl +X2+X 3 Yl+Y2+Y3)

G 3 ' 3

/24/

Inccntre of a triangle:

The point of concurrency of the internal bisector of angles of a triangle is called inc entre of triangle.

To find the co-ordinates of inc entre of a triangle

Let A (xl'YD, B(X2'Y2) and C(X3'Y3) be the vertices of a ~ABC.

Let the internal bisectors AD and BE of angles A and B respectively meet at point 1.

Let BC= a, AC = band AB = c

We know that the angle bisector of a triangle divides the opposite side in the ratio of sides of triangle containing the angle. [Angle Bisector Theorem]

BD AB c -=-=-DC AC b

BD c or -=-

DC b

B(X2'Y2) 14

- -(1)

I.e. D divides BC internally in the ratio c: b Co-ordinates of D are given by

( CX 3 + bx 2 cy 3 + by 2 )

" c+b ' c+b Now, in ~ABD ( By Angle Bisector Theorem)

DI BD BD -=-=-IA AB c

also from (1)

BD c -=-DC b

BD DC or -= --

c b

BD DC BD+DC or -=-=---

c b c+b

BD BC or -=--

c c+b

BD a or -=--

c c+b

- -(2)

------ (3)

a .1

/25/

From (2) and (3)

DI a =

lA c+ b

c, . • (CX 3 + bX 2 CY3 + bY2)

lhus I dIvIdes DA where D -c+b-- ,- c + b

and A(XI' YI) in the ratio a : c + b By internal section formula, co-ordinates of I are given by

or I[aXI + bx'J +cx, ;~L~~J1_~_cr2_l

a+b+c a+b+c J

By symmetry, it is clear that the co-ordinates of points of intersection of aIlble bisectors of angles A and C will also be same as those of point 1.

Thus all the three angles bisectors of ~ABC meet at 1. Ilence I is the point of concurrency of angles bisectors of triangle ABC i.c. the Incentre of Traingle.

I-Ience, co -ordinates of inc entre of triangle ABC are given by

I (aXI + bX2 +cx 3 ; aYI + bY2 +cY3 ) a+b+c a+b+c

Remark 1. In ~ABC, length of sides opposite to vertices A, Band C are taken to be a, band c rl'spectively, i.e. BC= a, AC = bAnd AB = c.

Remark 2 If from incentre I of ~ABC a perpendicular is drawn on anyone side (say I L on BC) and by taking I as center and IL as radius a circle is drawn, then this circle is called incircle of c\ABC. Thus I being the center of incircle of ~ABC, is called inc entre of ~ABC.

A

B c

l~xample 21: Find the centroid of ~ABC having its co-ordinates A(2,4), B(-3,1) and C(5,-2)

Solution Let A (Xl' YI) = (2,4) B (X2' Y2) = (-3, 1)

and C (X3' Y3) = (5, -2) The Co-ordinates of Centroid of ~ABC are given by

/1.6/

Example 22

Solution

(Xl +X2 +X3 .YI +Y2 +Y3)

G 3 ' 3

(2-3+5 4+1-2)

G 3 ' 3

( : ' 1) is required centroid.

The Co-ordinatres of centroid of a ~ABC are (5,4) and that of two of its vertices are (1,2) and (9,7). Find the co-ordinates of third vertex. Let (Xl' Yl) = (1,2)

(x2, Y2) = (9,7) and let the co-ordinates of third vertex be (X3'Y3)' Co-ordinates of centroid are given by

= (Xl +X2+ X3 .YI +Y2+Y3) (5,4) 3' 3

or (5,4) = (1+93+X3 ,2+7

3+ Y3)

or

or

Example 23

Solution

10+x3 =5 and 9+Y2 = 4 3 3

X3 = 5 and Y 3 = 3

The co-ordintes of third vertex are (5,3)

Find the co-ordinates of the center of the circle inscribed in a triangle having its angular points (4,-2), (-2,4) and (5,5)

Let the given ~ be ABC having its angular points

A(4,-2) = (X1'Yl)' A(x"y,) B (-2, 4) = (x2, Y2)'

and C(5,5) = (x3, Y3) say.

We are to find the center of incircle of ~ABC i.e. incenter of ~ABC, whose co-ordinates are given by

I (aX1 + bX2 +cx3 ; aYI + bY2 +cY3) a+b+c a+b+c

Now

,. B(X:VY2)

a

[By distance formula]

= [(5+2)2 +(S-4)2

= .J49+1 =.J50 = s.fi. units

b = AC =~(X3 -X1 )2 +(Y3 - Yl)2

= ~(S-4)2+(S+2)2

= -/i + 49 = .J5O = s.J2 units

and C=AB= ~(X2-Xlf+(Y2-Yl)2

= ~(-2-4)2+(4+2)2

= .J36+36 =m = 6.J2 units

Co-ordinates of center of incircle of ~ABC (incentre ')are given by

I (aXl + bX2 +cx3 ; aYl + bY2 +cY3) a+b+c a+b+c

_ (S.fi(4) + S.fi(-2) + 6.J2(S) ,S.fi(-2)+S.J2(4) + 6J2(S») - s.fi + sJ2 + 6.J2 s.fi + s.fi + 6.Ji

= (20.fi - 10.fi + 30.fi , -10/i + 20.fi + 30/i ) 16Ji 16.fi

= (40.fi , 10.fi) 16.fi 16Ji

AREA OF TRIANGLE

Let A(Xl'Yl) ; B (X2'Y2) and C(X3'Y3) be the vertices of a ~ABC, then its area is given by

ar. ~ ABC

= ar. Trapezium (ABLM) + ar. Trapezium (AMNC) - ar. Trapezium (BLNC)

Y

X' o L

Y'

M

Qx3,v,) i ' N

/27/

X

/28/

1 1 1 = "2 (BL+ AM) x LM +"2(AM + CN) x MN - "2(BL + CN) x LN

1 [.: Area of trapezium = "2 (sum pf parallel sides)

x (Distance between the parallel sides)]

1 1 1 = "2(Y2 + Yl)'(X1-X2) +"2(Yl + Y3)'(X3-X1) -"2(Y2 + y:J(x3- x2)

1 = "2 r Y2Xl- Y2X2 + X1YI-X2Yl +X3Yl -X1Yl+ X3Y3 - X1Y3 - X3Y2 +xzY2 -X3Y3 + x2y31

1 = "2 [(XtY2-X2Yl) +(X3Yl -X1Y3) - X,-\Y2 +X2Y3]

1 = "2 [(XlY2 -X2Yl) + (~Y3- X3 Y2) + (X3YI-XIY3)] Because area of a ~ is always taken to be positive. Thus we take the above value in magnitude. Hence area of ~ ABC is given by

1 ar. ~ABC = "2 I (XlY2- X2Yl) + (X2Y3 - X3Y2) + (X3Yl - X1Y3) I sq. units

Aid to Memory: The above formula is difficult to remember. For the sake of convenient:e We have an aid to memory as given below:-

1) write down the co -ordinates (say) (X1Yl); (X2Y2); (X3'Y3) and (Xl' Yl) vertices A, 13, C and again A respectively of ~ABC vertically £tom top to bottom as shown below.

2) To obtain the area of ~ABC, add the products of pair of terms indicated by arrows in downward direction.

3) Now from the above sum subtract the products of pair of terms indicated by arrows in upwards direction.

4) Take the modulus of the value calculated in above two steps.

5) Multiply the above calculated value by 1/2

6) The value calculated in step 5 will be the magnitude of area of ~ABC.

1 Area of ~ ABC = "2 I (X1Y2 + X2Y3 + X3Yl) - ( X2Y1 + X3Y2 + X1Y3) I

1 = "2 I (X1Y2 -X2Yl) + (X2Y3 - X3 Y2) + (X3YCXlY3) I

Square units.

Example 24 Solution

Find the area of triangle whose vertices are (3,4); (7,2) and (-5,6). Let A (3,4) = (xl' Yt)

B (7,2) = (X2'Y2) and C (-5,6) = (X3'Y3) be the vertices of ~ABC.

Then the area of ~ABC will be

1 = "2 1 (6+42-20) - (28 +(-10) +18) 1

1 = "2 1 (48-20) -(36) 1

1 = -128-361 2

1 1 = "2 1-81 = "2 (8) = 4 Square units.

/29/

Examples 25 The co-ordinates of points A and Bare (3,4) and (5,-2) respectively. Find the co-ordinates of points P such that ar. ~PAB = 5 and PA = PB.

Solution Let the co-ordinates of point P be (x,y).

Now area of ~PAB is given by

f---------~n (5.-2)

1 or "2 1 (4x-6+5y) - (3y+20-2x) 1 = 5

1 or "2 14x-6+5y-3y-20 +2x 1 = 5

1 or "2 16x+2y-26I = 5 or 16x+2y-26I = 10

/30/

or 6x+2y -26 = ± 10

either 6x + 2y = 36 [Taking +ve sign]

i.e. 3x + y = 18 and 3x + y = 8

NowPA = PB

[':1 xl = a or x = ±a] or .. 6x + 2y = 16

[Taking -ve sign] - -(1) - -(2)

=> ~(X_3)2+(Y_4)2 =~(x-5/+(y+2)2 => (X-3)2 +(y_4)2 = (x-5)2 + (y +2)2 => x2-6x +9 +y2 -8y +16 = x2 -19x +25 +y2+ 4y +4 => -6x -8y +25 = -lOx +4y +29 => 4x -12y -4 = 0 => 4x -12y = 4 => x -3y = 1 --(3) on solving (1) and (3), we get

11 3 x="2,y="2

on solving (2) and (3), we get

5 1 x="2,y="2

(11 3) (5 1) Thus the co-ordinates of point Pare "2'"2 or "2'"2 Remark :- When the area and two vertices of a i\ are given and we are to find third vertex there will be two points as the third vertex, one on each side of the line joining the two given vertices of triangles as shown in adjacent figure.

A

p

Example 26 For what value of K the points (4,7), (2,K) and (8,5) will be collinear.

Solution . Let A( 4,7); B(2,K) and C(8,5) be the three vertices of i\ABC.

or

A, Band C will be collinear (lie on a st. line)

if ar. i\ABC = 0

4

1 2 28

4

7

K =0

5

7

B

or

or or or

Example 27

Solution

or

or

or or

or

or

Example 28

Solution

i. e.

or

or or

/31/

1 2."1(4K-14) + (l0-8K) + (56-20)1 = 0

1-4K + 321 = 0 -4K + 32 = 0 K= 8

Find the condition for which the points (at/, 2at1), (atl, 2at2) and (a,O) are collinear. (tl :;et2)

The Points P(at/, 2at1), Q(atl, 2at2) and R(a,O) will be collinear if ar. ~PQR = 0

1(2a2 t/ t2-2a2 tl t1) + (0-2a

2 t2) +(2a2 teO) I =0

2a2 tl t2 (~-t2) +2a2 (t1-t2) = 0 (1 +t1 t2) [tl +t2] = 0

either tl ~ + 1 = 0 or tet2 = 0

either tl t2 =-1 or ~ = t2 [ But tl * t2 given] tl ~ = -1 Which is the required condition.

Find the value of A. for which the area of triangle having vertices (2,5); (-7,A.), (3,8) is 10 square units.

Let P (2,5); Q (-7,A.) and R (3,8) be.the vertices of ~PQR.

According to question, Area ~PQR = 10

2X 5 1 -7'Jf A. - ,Ai = 10 23'Jf8

2 ,Ai 5

1 2." 1(2A. + 35) +(-56 -3A.) + (15 -16) I = 10 I-A. -221 = 20 -A. -22 = ±20

/'J2/

Taking +ve sign, -A -22 = 20

:.:> A = -42 Taking -ve sign,

-A -22 = -20 or A =-2 .. A = -42 or-2

I~xample 29 Find the area of Quadrilateral having vertices (2,6); (4,8); (5,10) and (3 ,8) taken in order.

Solution Let A(2,6); B (4,8); C(5,10) and D(3,8) be the vertices of a quadrilateral ABCD. Area of Quadrilateral ABCD = ar ~ABD + ar. ~BCD

12 X 6 4 8 1 4 X 8 1 5 10 + -= 2 3

X8 2 3 8

2 6 4 8

1 1 = 2 / (16-24) +(32-24) +(18-16) + 2 / (40-40)+(40-30)+(24-32) /

1 1 = 2/-8 +8 +2/ +2/+10 -8/

C(5,lO)

1 1 = -/2/ +-/2/ 2 2

1 1 = 2 (2) + 2 (2) = 2 Square units

Area of quadrilateral ABCD = 2sq. units

Aliter :- Area of Quadrilateral ABCD can also be calculated as follows:-

ar quadrilateral ABCD =

1 = 2 / (16-24) +(40 -40) +(40 -30) +(18 -16) /

/33/

1 = 2" 1-8 +0 +10 +21

1 1 = 2" 141 = 2" (4) = 2 Square units.

LOCUS OF A POINT

Locus of a point moving on a plane is the path traced out by it under the given geometrical conditions.

For Example

1) Circle is the locus of a point moving on a plane such that its distance from a fixed point is always constant. The fixed point is called center of circle and the constant distance is called radius of circle and is denoted by I r' .

i.e. CP = r (constant) is the geometrical condition.

2) Perpendicular bisector of a line segment AB is the locus of a point moving on the plane of AB such that its distance from two end points A and B are always same.

i.e. AP = BP is the geometrical condition.

3) Angle bisector of an angle is the locus of a point moving on the plane containing the angle such that its perpendicular distance from two arms of angle are always same.

i.e. PL = PM is the geometrical condition. x

Example 30 Find the locus of a point P moving on x-y plane whose distance from the point (2,5) is always equal to 4.

Solution Let P(x,y) be the moving point and C(2,5) be the fixed point.

or or

Example 31

Solution

or

or or or or

According to Question, PC = 4

~(x - 2)2 + {y - 5)2 = 4 [By distance formula] (x-2)2 + (y_5)2 = 16 x2 +y2 -4x -lOy +13 = 0, is required locus of point P.

Find the equation of Perpendicular bisector of the line segment AB having end points A (3,5) and B (6,8).

We know that perpendicular bisector of a line segment AB is the locus of a moving point whose distance from the end points A and B are always equal.

Let P(x,y) be the moving point (arbitrary point on perpendicular bisector) Thus AP = PB

~(x - 3)2 + {y _ 5)2 = ~(x _ 6)2 + {y _.8)2 (X-3)2 +(y-5)2 = (X-6)2 + (y -8)2 x2 +y2-6x -lOy +34 = x2 +y2 -12x -16y +100 6x +6y -66 = 0 x +y -11 = 0, is the required locus of point P.

L B(6,8)

Example 32 Find the locus of a point P such that

(i)

Solution

(i)

or

PA 4 -=-PB 3

2PA2 + PB2 = 3 (ii)

where co-ordinates of points A and Bare (-1,2) and (3,4) respectively.

Let (x,y) be the co-ordinates of point P and we are given the co-ordinates of A(-1,2) and B(3,4). By given condition,

PA 4 -=-PB 3

~(X+l)2+{y_2)2 4 =

~(X-3)2+{y-4/ 3 B(3,4)

/35/

(x + 1)2 + (y _ 2)2 16 or (X_3)2 +(y_4)2

=-9

or 9[x2 +y2+2x -4y +5] = 16 [X2 +y2-6x -By +25] or 9x2 +9y2 +1Bx -36y +45 = 16x2 +16y2 -96x -12By +400 or 7x2 +7y2 -114x -92y +355 = 0, is the required locus of point P.

(ii) By given condition, 2PA2 + PB2 = 3

or 2[~(X + 1)2 + (y _2)2 r +[~(X - 3)2 +(y - 4)2 r = 3 or 2[(x+1)2 + (y-2)2] + [(x-3)2 + (y_4)2] = 3 or 2[X2 +y2+2x -4y +5] + [X2+y2 -6x -By +25] =3 or 3x2 +3y2 -2x -16y +35 -3 = 0 or 3x2 + 3y2 -2x -16y +32 = 0, is the required locus of point P.

Example 33 Find the locus of a point so that the join of point (-3,1) and (4,2) subtends a right angle at the moving point.

Solution Let the co-ordinates of moving point P be (x,y).

or or or or

Example 34

Solution

or

We are given the co-ordinates of points A(-3,l) and B (4,2).

(using pythagorous theorem) (x +3)2 +(y_1)2 +(X-4)2 +(y_2)2 = (4+3)2+(2-1)2

A(-3,1)

x2 +9 +6x +y2 +1 -2y +x2 +16 -Bx +y2 +4 -4y = 49+1 2X2 +2y2 -2x -6y -20 = 0 x2 +y2 -x -3y -10 = 0 is the required locus.

B(4,2)

Find the locus of apoint whose distance from x-axis is always equal to three times its distance from y- axis.

Let P(x,y) be the point whose locus is to be found.

Now, distance of point P from x-axis = y (=PL)

And distance of point P from y-axis = x (=PM)

By given condition, PL = 3PM i.e. y = 3 (x) 3x -y =0, is the required locus of point P.

y

o

P(x,y) I-"..t~------'

x

y

r L x

/36/

Example 35 A point moves so that the sum of its distances from the points (-ae, 0) and (ae, 0) is always equal to 2a, show that its locus is

x2 y2 -+ =1 «1) a2 a2(1_e2) ,e

Solution Let p (x,y) be the point whose locus is to be determined. and let A(-ae, 0) and B(ae, 0) be the given points. By given condition,

PA + PB = 2a

~(x + ae)2 + (y - 0)2 + ~(x - ae)2 + (y - 0)2 = 2a

=> ~(x+ae)2+y£+~(x-ae)2+y2 = 2a using identity, [(x +ae)2 +y2] - [(x -ae)2 +y2] = 4aex Dividing (2) by (1) we get

or

[(x + ae)2 + y2] - [(x - ae)2 + y2] 4aex

~(x+ae)2 +y2 +~(x-ae)2 +y2 =~

~(x+ae)2+y2 -~(x-ae)2+y2 = 2ex (1) + (3) gives

2~(x + ae)2 + y2 = 2a + 2ex

or ~(x+ae)2 +y2 = a +ex Squaring both sides we get

---- (1)

---- (2)

---- (3)

(x +ae)2 +y2 = a2 +e2x2 +2aex A (-ae,O) B(ae,O) or x2 +a2e2 +2aex +y2 = a2 +e2x2 +2aex or . x2 _e2x2 +y2 = a2 _a2e2 [on canceling 2aex from both sides] or x2(1 _e2) +y2 = a2 (1-e2)

or

or

x2(1_e2) y2 --'-----'- + = 1 a2(1- e2) a2(1- e2)

x2 2 2+ 2(; 2)- = I, is the required locus of point P. a a -e

/37/

MULTIPLE CHOICE QUESTIONS

1. The point having its co-ordinates (0,0) is called a. Abscissa b. Ordinate c. Origin d. Critical point

2. For a point lying on x-axis a. x co-ordinate = ° b. Y co-ordinate = ° c. both co-ordinates are zeros d. none of these

3. For a point lying on y-axis a. x co-ordinates = ° b. Y co-ordinate = ° c. both co-ordinates are zeros d. none of these

4. The point (2,4) lies in which quadrant? a. 1st b. 2nd c. 3rd d. 4th

5. The point (-3, 4) lies in which quadrant? a. 1st b. 2nd c. 3rd d. 4th

6. The point (-3,-4) lies in which quadrant? a. 1st b. 2nd c. 3rd d. 4th

7. The point (7,-5) lies in which quadrant? a.l st . b.2nd c.3rd d.4th

8. x-co-ordinate of a point is called a. Abscissa b. Ordinate c. Origin d. Critical point

9. y-co-ordinate of a point is called a. Abscissa b. Ordinate c. Origin d. Critical point

10. For all the points lying on a line II to x-axis a. abscissas are same b. ordinates are same c. both co-ordinates are same d. none of these

11. For al the points lying on a line II to yaxis a. Abscissas are same b. Ordinates are same c. Both co-ordinates are same d. None of these

12. The point (2,0) lies on a. X-axis b. Y-axis c. Both axes d. None of these

13. The point (0,3) lies on a. X-axis b. Y-axis c. Both axes d. None of these

14. The only point lying on both the axis is a. (1,0) b. (0,1) c. (-1,0) d. (0,0)

15. 1be point (2,5) represent the co-ordinates of which system? a. Cartasian or rectangular system b. Polar-co-ordinate system c. Both a and b. d. None of these

16. The point (4,30°) represent the co-ordinates of which system? a. Cartasian or rectangular system b. Polar co-ordinate system c. Pedal co-ordiante system d. None of these

/38/

17. The point (6 A ~) represent the co-ordinates of which system? a. Cartasian or rectangular system b. Polar co-ordinate system c. Pedal CO-

/39/

23. For what values of K the distance between the points (2,k) and (6,.J5) is 6 units a. 3$ b. -$ c. both (a) and (b) d. 2$

24. The distance between (-l,a) and (5,5) is 10 unit, the value of a is a. -3 b. 13 c. both (a) and (b) d. 4

25. The points (6,1); (-2,1) and (2,-4) are the vertices of a triangle which is a. Isosceles b. Scalene c. Equilateral d. not form a triangle

26. The points (-.J3,2); (2.j3,-1) and (2.J3,5) are the vertices of a triangle which is a. Isosceles b. Scalene c. Equilateral d. not form a triangle

27. The points (-1,-3); (5,5); (-3,1) are the vertices of a triangle which is a. Isosceles b. Equilateral c. Right angles d. Scalene

28. The points (2,4); (8,4) and (2,1) are the vertices of a triangle which is a. Isosceles b. Equilateral c. Right angles d. Scalene

29. The points (4,2); (7,5) and (9,7) are a. vertices of an isosceles triangle b. vertices of an equilateral triangle c. vertices of a right angled triangle d. collinear (or not form a triangle)

30. The points (2,-2); (8,4); (5,7) are the vertices of a rectangle, then the fourth vertex will be a. (-1,-1) b. (-1,0) c. (0,-1) d. (7/2,5/2)

31. The points (7,3),(3,0),(0,4),(4,-1) are the vertices of a a. Square b. Rhombus c. Rectangle d. None of these

32. The points (1.2); (4,2); (4,5); (1,5) are the vertices of a a. Square b. Rhombus c. Rectangle d. None of these

33. The points (1,1); (5,1); (5,3); (1,3) are the vertices of a a. Square b. Rhombus c. Rectangle d. None of these

34. The points (1,1); (5,1); (7,5); (3,5) are the vertices of a a. Square b. Rhombus c. Rectangle d. Parallelogram

35. The point which divides the join of A (-2,-4) and B (8,11) in the ratio 2:3 (internally) is a. (2,3) b. (2,2) c. (3,2) d. (1,1)

36. The point which divides the joint of A (-2,-4) and B (2,2) externally in the ratio 5:3 is a. (2,3) b. (4,2) c. (8,5) d. (8,11)

37. The ratio in which the join of A (1,4); B (9,-12) is divided by P (6,-6) a. 3:5 internally b. 5:3 externally c. 5:3 internally d. 3:5 externally

/40/

38. The ratio in which x-axis divides the line segment joining the points (3,1) and (7,5) is a. 5:1 internally b. 1:5 externally c. 1:4 internally d. 4:1 externally

39. The ratio in which y-axis divides the line segment joining the points (-3,-4) and (1,-2) is a. 1:3internally b. 1:3 externally c. 3:1 internally d. 3:1 externally

40. The mid point of line segment joining the points (3,5) and (11,3) is a. (8,3) b. (4,1) c. (6,4) d. (7,4)

41 The mid point and one end of a line segment are (3,7) and (4,2) are respectively. The other end point is a. (2,12) b. (2,11) c. (3,12) d. (4,10)

42 Three vertices of a parallelogram taken in same order are (4,-11), (5,3) and C(2,15). The fourth vertex is a. (1,1) b. (2,2) c. (3,3) d. (4,4)

43 The vertices of a triangle are (1,2);(2,6) and (3,4). Its centroid will be a. (2,3) b. (4,2) c. (2,4) d. (5,1)

44. Two vertices of a triangle are (6,4) and (3,2) if the centroid is (4,0) then, its third vertex will be a. (3,6) b. (6,3) c. (3,-6) d. (-6,3)

45 The polar co-ordiantes of point (1,J3) are

a. (2,:) b. (2, :) c. (2, ~) d. (2, ;) 46 The polar co-ordinates of a point (-2,2.J3) are

a. (4,:) b. (4, ;) c. (4, 231t

) d. (4, 431t

)

47. The polar co-ordinates of point (-1,-1) are

(J2 ~~) a. , 4 b. ( J2, 341t) C. ( ~i,~l 2) d. (J2,:) 48. The polar co-ordinates of point (4.j3,--4) are

l( 8 .!.11t)

c. , 6 d. both (b) and (c)

49. The cartasian co-ordinates of point (3,300 )are

[ 3~3 lJ a. 2'2 [l 3.J3] b. 2' 2

50.

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

/41/

The cartasian co-ordinates of point (2,1200 )are

a. (1,F3) b. (-1,F3) c. (F3,1) d. ( -F3,1)

The cartasian co-ordinates of point (4, -3rt ) are

a (2,-F3) b. (2,-2F3) c. (2,2F3) d (2,F3)

The cartasian co-ordinates of point ( J2, 54rt) are

a (1,1) b. (-1,1) c. (1,-1) d (-1,-1)

The transfunn ed equat::bn ofX2 +y2 = 2ax into polar co-ordinate system is a. r = 2a cos 8 b. r2 = 2acos 8 c. r = 2asin 8 d. r2 =2asin 8

The transformed equation of y2=4ax into polar co-ordinate system is a. r sin 8 = 4a cos8 b. r sin28 = 4a cos8 c. r cos 8 = 4a sin8 d. r cos28 = 4a sin8

The equation of circle X2+y2 = 16 in polar co-ordinates system is a. r = 16 b. r = ± 4 c. r = -4 d. r = 4

The area of triangle having its vertices (6,3); (-3,5) and (4,-2) is a. 49 sq. units b. 24 sq. units c. 24.5 sq. units d. 25.5 sq. units

The area of quadrilateral whose vertices taken in order are (1,1) ;(3,4),(5,-2) and (4,-7) is

41 a. 41 sq. units b. 82 sq. units c. 2 sq, units d. None of these. For what value of A, the area of triangle having vertices (1, A); (4,3), and (9,7) is 15 square units?

18 -12 18 -12 5 -5 a. 7'7- b. 5'5 c. 18'U d. None of these Points (a, 0); (0, b) and (1,1) are collinear if

1 1 1 1 a - + - = 1 b. -a - b = 1 . a b. -.:!.+1.=1 c. a b If the area of quadrilateral whose vertices taken in order are (1, 2); (-5, 6); (7, -4) and (K, -2) be zero, then the value of K will be a. 1 b. 2 c. 3 d. 4

Which point lies on the locus of a point whose equation is given by x2 +2y2 = 9 a. (1,2) b. (2.1) c. (0,3) d. Both (b) and (c)

The locus of a point which lies on the line joining the point (3A) and (2,2) is a. x = 2y + 2 b. 2x = Y + 2 c. 2x = y-2 d. x = 2y- 2

/42/

63.

64.

65.

66.

67.

The locus of a point P moving on a plane such that PA = PB; where A(4,1) and B(3,5) is a. 2x -8y +17 = 0 b. x -8y +17 = 0 c. 2x +8y -17 = 0 d. x +8y +17 = 0

If A(1,3) and B(5,7) are two fixed points, then the locus of point P(x, y) for which 2PA2 +3PB2 = 10 is a. x2 +5y2 -34x -54y +232 = 0 c. 5x2 +5y2_34x -54y +232 = 0

. b. 5x2+ y2-34x -54y +232 = 0 d. None of these.

For the points A(a, 0) and B(-a, 0) the locus of a point C(x, y) such that LACB is right Angle is given by a. x2_y2 = a2 b. _x2 + y2 =a2 c. x2+ y2 = _a2 d. x2 + y2 = a2

x2 y2 The point lying on the curve ~ - 11 = 1 is a. (a secS , b tanS) b. (a sinS, b cosS) c. (a sinS, -b cose) d. None of these.

x2 y2 The point lying on the curve 2"" + -2 = 1

a. (a sece, b tane) c. (-a sinS, -b cosS)

a b b. (a sine, b cosS) d. Both (b) and (c)

68. Centroid divides a medium in ratio (From vertex towards side) is a. (1:2) b. (2:1) c. (3:1) d. (1:3)

69. The co-ordinates of the incentre of the triangle whose vertices are (2,-2) (8,6) and (8,-2) are a. (0,6) b. (-6,0) c. (6,0) d. (0,-6)

70. If (Xl' Yl); (x2, y2J and (x3, Y3) are the vertices of a triangle. The its centroid is given by

(axl + bX2 +cx3 , aYl + bY2 +CY3) (Xl +x2 +x3 , Yl +Y2 +Y3)

c. a+b+c a+b+c d. 3 3

71. If A (Xl' Yl); B(x2, Y2) and C(x3' Y3) are the vertices of a triangle ~ ABC, BC = a, AC = b, AB = c, then incentre of triangle ABC is given by.

(axl + bX2 + cx3 , aYl + bY2 +cY3)

c. a+b+c a+b+c

/43/ - ANSWERS-

(1) c (2) b (3) a (4) a (5) b (6) c (7) d (8) a (9) b (10) b (11) a . (12) a (13) b (14) d (15) a (16) b (17) b (18) c (19) d (20) d (21) b (22) c (23) c (24) c (25) a (26) c (27) c (28) d (29) d (30) a (31) b (32) a (33) c (34) d (35) b (36) d (37) c (38) b (39) c (40) d (41) a (42) a (43) c (44) c (45) d (46) c (47) a (48) d (49) a (50) b (51) b (52) d (53) a (54) b (55) d (56) c (57) c (58) b (59) a (60) c (61) d (62) b (63) a (64) c (65) d (66) a (67) d (68) b (69) c (70) d (71) c

Chapter 2 THE STRAIGHT LINE

INTRODUCTION

In last chapter we studied about the point. The co-ordinates of a given point determines the exact location of that point. Now, in the following chapter we shall study about the straight line, equation of a straight line in its various forms, perpendicular distance between two given straight lines, slope of straight line, condition for the concurrency of three straight lines etc.

LINE SEGMENT - Line segment is the shortest distance between two given points. Thus every line segment has a definite length equal to the shortest distance between its two end points. However, a straight line is infinite in length, i.e. neither there is any initial point nor there is an end point of a straight line.

A ... ---------. B Fig. (a)

( )

Fig (b)

Figure (a) represents a line segment AB. where as, figure (b) represent a straight line.

STRAIGHT LINE - The straight line is the locus of a point moving in a fixed direction. At this level we shall restrict ourselves to two dimensional geometry only. Thus the straight line, about which we are going to study is supposed to be lying in xy plane. Thus every point lying on the given straight line would have only two co-ordinates x-eo-ordinate and y-co-ordinate.

SLOPE OF A STRAIGHT LINE- The tangent of the angle subtended by a straight line with the +ve direction of x-axis measured anti clockwise is called slope of straight line. Y

Some times, slope of a straight line is also called gradient of the straight line.

In above figure(c), e is the inclination of straight line AB and its slope is tane. X' X

Slope of a straight line is denoted by 'm'

Thus slope =m=tane B

Y' Fig. (c)

Remark 1 The inclination of a straight line varies from 0° to 180°.

Remark 2 The inclination of a straight line parallel to x-axis is 0° or 180°. Thus the slope of any line parallel to x-axis is tan 0° or tan 180° i.e. O.

Remark 3 The inclination of a straight line parallel to y-axis is 90°. The slope of any line parallel to y- axis is tan 90°. I.e. 00 (infinite)

INTERCEPTS OF A GIVEN STRAIGHT LINE

If a straight line AB cuts x-axis and y-axis at points P and Q respectively. Then OP = a = x -intercept of AB and OQ = b = Y -intercept of AB and the pair (a,b) is called intercept made by AB on x-axis and y -axis respectively.

Remark 1 (a) If PQ is in 1st Quadrant, then OP = + ve, OQ = +ve i.e. both intercepts are +ve.

A

y

8=00 or 1800

1800

y

8=900

900

o I+-~----::'a-----+t

(b) If PQ is in IInd Quadrant, then OP= -ve, OQ = +ve,

/45/

x

x

x

B

(c) If PQ is in lIIrd quadrant then OP= -ve and OQ = -ve, i.e. both the intercepts are -yeo

(d) If PQ is in IVth quadrant, then OP = +ve and OQ = -ve

Remark 2 (a) Fora stra.:ghtline II to x-axis, x-intercept = 0 (b) For a straight line II to y-axis, y-intercept = 0 (d) For a straight line (inclined) passing through the origin, x- intercept and

y-intercept both are zeros. y

Condition for two lines to be parallel:-

Two lines are parallel if 81 = 82 => tan 81 = tan 81 i.e. m1 =m2

Thus for two lines to be parallel their inclinations and their slopes should be equal.

o

Conditions for two lines to be perpendicular to each other.

Q

x

s

Let two straight lines PQ and RS intersects each other at point T such that PQ and RS are mutually perpendicular to each other.

/46/

or or

or

or

or

or

in ,1ABT, 82 = 81 + 90°

82 -8 1 = 90° tan (82 -8 1) = tan90°.

tan82 - tan8} ---=---~ = 00 1 + tan8 2 . tan8}

m2 -m} --"----"-- = 00 1+m2 ·m1 mm =-1 } 2

-1 m2 =-m}

[.,' Exterior angle of a triangle is equal to the sum of interior opposite angles]

y S Q

o x

p R

Thus ,two straight lines are .1 to each other if and only if, the product of their slopes is -l.or their slopes are -ve reciprocal of each other.

V ARIOUS FORMS OF STRAIGHT LINES

1) Straight Line Passing Through The Origin

Let P(x,y) be any point lying on a straight line AB inclined by an angle 8 with the +ve direction of x- axis. Y

B

or

or

Example 1

Solution

Here

tan 8 = y x

y -=m x

[.,' m = slope = tan 8 ]

y=mx A x

Thus y = mx is the equation of straight line passing through the origin and having slope m.

Find the equation of straight line passing through the origin and having

1 slope "2' The equation of straight line passing through origin and having slope' m' given by y= mx

1 m=-

2 the required equation of straight line will be

1 y ="2 x

or or or

2y = x 2y -x = 0 x -2y = 0

/47/

Example 2 Find the equation of straight line passing through origin and having an inclination of (i) 45° (ii) 135°.

Solution (i)

i.e. or or

(ii)

8 = 45° m = tan 8 = tan 45°

=1 The equation of straight line will be y =mx y = (1)x Y = x x-y = 0 8 = 135° m=tan8

= tan 135° = tan(1800-45°) = -tan45° = -1

The equation of straight line will be, y =mx

or y = (-l)x i.e. or

y =-x x+y = 0

2) Slope-intercept From of Straight Line

Let AB be a straight line which intersects y-axis at point M such that OM = c(y-intercept)

Let P(x,y) be an arbitrary point on AB.

Now, PL = PR+RL

= x tan8 +OM

= x tan8 +c

y = x(m) +c

or y = mx +c; where c is the y-intercept

B

[

. PR PRJ .,' ill i1PRM tan 8 = --=--, MR x

=> PR = xtan8

Thus, y = mx +c is the slope-intercept form of straight line.

x

/48/

Example 3 Solution

Find the equation of straight line having inclination 60°& y-intercept 5. flere e = 60° m = tan 8

= tan60°

=J3 c =5

The equation of straight line in slope-intercept from is, y = mx +c

i.e. y = J3 x+5

Example 4 Find the equation of straight line having

1 (i) m = J3' c = 4 (ii) m = -I, c = -3

Solution

3)

1 (i) Here, m = J3' c = 4

The equation of straight line in slope-intercept form is y = mx +c

1 or y = J3 x +4

or J3y = x +4J3

or x - J3y +4 J3 = 0 is required equation. (ii) Here, m = -I, c = -3

The equation of straight line in slope-intercept form is y = mx +c

or y = (-I)x +(-3) or y = -x-3 or x +y+3 = 0 isthe equation of straight line.

Intercept Form of Straight Line Let AB be the straight line which cuts x-axis and y-axis at points Rand Q

respectively B Y

Let OR = a and OQ = b Let L'QRO = LQPL = a and P(x,y) be an arbitrary point on the straight line

In ~QPL,

tan a QL LP

L ______ P(x,y,)

b ! Mi

o 14---"'a---+t x A

= OQ-OL = b-y OM x

In ~PRM,

tan a PM MR

PM =----

OR-OM From (1) and (2)

b-y Y --=--

x a-x or ba - bx - ay +xy = xy or ba - bx - ay = ° or bx +ay = ab

=-y-a-x

------ (1)

------ (2)

or -; + ~ = 1 is the required equation of line in intercept form.

/49/

Example 5 Find the equation of straight line having x-intercept 4 and y-intercept 5.

Solution Equation of straight line in intercept form is given by

~+y =1 a b

Given a = 4 and b = 5

or or

x y . -+-=1 4 5

5x +4y = 20 5x +4y -20 = 0 is the required equation of line

Example 6 A straight line cuts x-axis and y-axis at points (3,0) and (0,5) respectively. Find the equation of straight line.

Solution Since, the straight line cuts x-axis and y-axis at points (3,0) and (5,0).

.. x-intercept (a) = 3

and y-intercept (b) = 5 We know, that equation of straight line in

intercept form is given by -; + ~ = 1

Q Y

x

p

/50/

or

or or

~+I.=1 3 5

5x + 3y = 1 5x + 3y -1 = 0 is the required equation of line.

4) One Point Form or Slope Point Form

line.

Let a given straight line cuts x-axis and y-axis at points Rand S respectively.

Let P(x,y) be an arbitrary point and A(xl, Yl) be a fixed point on the given straight

Let LSRO = LSAN = a In L\ PAk

or

or

tan a PK AK PM-KM

=----

tan a =

AN-NK OQ-ON OL-OM

Y-YI

Y-YI tan(1800-e) = x I-X

Y-YI -tane =

Xl -x

(X-Xl) tane = (Y-Yl)

y

x

or or (Y-Yl) = m(x-x1); where tane = m, the slope of straight line

x

Thus Y-Yl = m(x-x1) is the equation of straight line in one-point form or slope point form.

Example 7 Find the equation of straight line at an inclination of 120° with the +ve direction of x-axis and having a point (-2,4) on it.

Solution

or or

Given e =120° and (Xl'Yl) - (-2,4) The equation of straight line in slope point form is given by, Y -Yl = m(x-x1) Y -4 = tan 120° (x-(-2)) Y -4 = tan(1800-600)[x+2]

or

or

or

or

Example 8

Solution

or or or or

Example 9

Solution

or or or or

Example 10

Solution

/51/ y-4 = -tan600(x+2)

y-4 = -J3(x+2)

y-4 = -J3x - 2J3

.J3x + y + 2.J3 - 4 = 0 is the required equation of straight line.

Find the equation of straight line passing through the point (3,-5) and having slope 3. Given (X1'Yl) = 3,-5) and m= 3 The equation of straight line in one-point form is given by Y-Yl = m(x-x1) y-(-5) = 3(x-3) Y +5 = 3x-9 3x-y -9 -5 = 0 3x -y -14 = 0 is the required equation of straight line.

Find the equation of straight line passing through the point (-2,4) and parallel to line 2x+y-7 = 0 Equation of give line is 2x+y-7 = 0 (1) slope of (1) is given

-Coefficient of x (

2x+y-7=O

m = Coefficient of y

( (-2,4)

2 • - = -2 A

1 Slope of required line is -2 as lines are parallel. By slope-point form the equation of required straight line is given by, y -Yl = m(x-x1) Y -4 = -2[x-(-2)] Y -4 = -2x-4 2x +y -4 +4 = 0 2x +y = 0 is required line

~

~

Find the equation of straight line passing through the point (4,7) and perpendicular to line x-2y +5 = 0

Equation of given line is x-2y+5 = 0

Slope of line(1) is given by

= -Coefficient of x Coefficient of y

1 (--2) =

1

2

- - - (1) (4,7) A

111,

x-2y+5=O m,

/52/

or i.e.

Let the slope of required line be m 2. Then by the condition of perpendicularity

1 =

1 - 1/2 =-2

By slope-point form, the equation of required straight line will be (Y-Yl) = m2(x-x1) (y-7) = (-2)(x-4) y-7 = -2x+8 2x+y-7-8 = 0 2x+y-15 = 0 is the required line.

5) Two Point Form of Straight Line

Let the given straight line is inclined at an angle 8 to the +ve direction of x-axis.

Let A(Xl'Yl) and B(X2'Y2) be two fixed points and P(x,y) be an arbitrary point on that straight line. y

In ~APS,

tan 8

In ~ABT,

tan 8

PS PN -SN =-=

=

=

AS LN

PN-SN ON-OL

Y-Yl

BT AT

RS RN-SN =-=---

1M OM-OL

= x2 -Xl

From (1) and (2)

Y-Yl Y2 -Yl --=---

---- (1)

---- (2)

y

y,

M N x

x

or

Example 11

Solution

or

or

or

/53/

( ) Y2 - Yl ( )' h 'f' h 1" 'f Y-Y 1 = X-Xl ; IS t e equation 0 straIg t me m two-pomt orm~ X2 -Xl

Find the equation of straight line passing through the points (2,3) and (-7,8) Given (Xl'Yl) = (2,3) and (X2'Y2) = (-7,8) By two point form, equation of required straight line will be

Y2 -Yl (Y-Yl) = X2 - Xl (X-Xl)

8-3 (y-3) = --(x-2)

-7-2

5 (y-3) = -9 (x-2)

-9y +27 = 5x -10 5x +9y -10 -27 = 0 5x +9y -37 = 0

Example 12 Find the equation of straight line passing through the point (-1,4) and through the point of intersection of straight lines 2x-y+2 = 0 and 3x-2y-18 = 0,

Solution The equation of given lines are 2x-y+2 = 0 - - (1)

and 3x+2y-18 = 0 - - (2) The point of intersection of lines can be obtained by solving (1) and (2) for X and y (1) x 2 + (2) gives

4x -2y =-4 3x +2y= 18 7x = 14

X = 2 Substituting the value of X in equation (1) we get

2(2)-y =-2 = y-6

Y = 6 The point of intersection of (1) and (2) is (2,6) Thus the required line passes through the points B(-I,4) and (2,6), By two-point form equation of required straight line will be

f54/

i.e.

Y2 -YI (Y-YI) = X2 - Xl (X-Xl)

(y-4)

y-4

6-4 = 2+1 (x+l)

2 = - (x+l)

3 or 3y -12 = 2x +2 or 2x -3y +14 = 0 is the required line.

Example 13 The vertices of a triangle ABC are A(l,2); B(3,-2) and C(4,3). Find the equations of its sides.

Solution Since we are to find the equations of sides AB, BC and AC of ~ABC, we shall use two point form of straight line.

i.e. Y2 -YI

Y-YI = x2 - Xl (X-Xl)

Equation of side AB

or

-2-2 (y-2) = -- (x-I)

3-1

4 y-2 = -"2 (x-1) y-2 = -2(x-l) 2x+y-4 = 0

Equation of side BC

3+2 (y +2) = 4 _ 3 (x-3)

Y +2 = 5(x-3) or 5x-y-17 = 0

Equation of side AC

3-2 (y-2) = -(x-I) 4-1

y-2 1

= - (x-I) 3

or x-3y+5 = 0

A(1,2)

B(3,-2) C(4,3)

6) Straight Lines Parallel to Axes

a) Straight Lines Parallel to x-axis

Let AB and CD be two straight lines parallel to x-axis, such that AB is above x-axis at a distance 'a' unit from x-axis and CD is below x-axis at a distance' a' units from x-axis. X'

b)

Equation of AB

i.e. or

Slope of AB = m = 0 [.: tane = tanO° =0]

and AB passes through the point P(O,a) By one point form, equation of AB will be (y-a) = O(x-O) y-a = 0

y = a, is a line parallel to x-axis

Equation of CD

I.e. or

Slope of CD = m = 0 [.: tane = tanO° =0] and CD passes through the point Q(O, a) By one point form, equation of CD will be (y+a) = O(x-O)

y+a = 0 y = -a is a line parallel x-axis

Straight Lines parallel to Y-axis Let AB and A'B' be two straight lines parallel

to y-axis at a distance 'b' units to the right and left of y-axis respectively.

Equation of AB X'

i.e.

i.e. or

Slope of AB = m = 00 [.: tane = tan90° = 00]

and AB passes through the point P(b,O) By one point form, equation of AB will be (y-O) = 00 (x-b)

y-O

x-b x

1 = - (x-b) o =0 = b is a line parallel to y-axis.

Equation of A'B'

Y

P(O a) A

a

0 a

c Q

Y'

Y A'

b b

Q 0

B'

Y'

Slope of A'B' = m = 00 [.: tane = tan90° = 00] and A'B' passes through the point Q(-b,O) By one point form, equation of A'B' will be

/55/

B

X

D

P X

B

/56/

i.e. or

(y-O)

y-O

x+b x

= 00 (x+b) 1

= -(x+b) o =0 = -b is a line parallel to y-axis.

Example 14 Find the equation of straight line (i) Parallel to x-axis at a distance 4 units above it. (ii) Parallel to x-axis at a distance 4 units below it. (iii) Perpendicular to x-axis (parallel to y-axis) at a distance 5 units to the right of y-

axis. (iv) Perpendicular to x-axis (parallel to y-axis) at a distance 5 units to the left of y-

axis.

Solution (i) Equation of straight line parallel to x-axis at a

distance of I a' units above it is gives by y=a

Here a = 4 The equation of required straight line is

y=4 or y - 4 = 0

(ii) Equation of straight line parallel to x-axis at a distance of I a' units below it is gives by

y =-a Here a = 4 The equation of required straight line is

y =-4 or y+4 = 0

(iii) Equation of straight line perpendicular to x-axis (or parallel to y-axis) at a distance of b units to the to the right of y-axis is given by

x=b Here b = 5 The equation of required straight line is

x=5 or x - 5 = 0

X'

X'

X'

Y

4

o X ,

Y'

Y

4 o X

Y'

b

o X

(iv) Equation of straight line perpendicular to x-axis (or parallel to y-axis) at a distance of 'b' units to the to the left of y-axis is given by

x =-b Here b = 5 The equation of required straight line is

x =-5 or x + 5 = 0

7) Normal or Perpendicular form of straight line -

Let AB be any straight line meeting x-axis and y-axis at A and B respectively. Y

Let the perpendicular OL=p drawn on the straight line AB be inclined to the +ve direction of x-axis at an angle a. measured anticlockwise.

The co-ord,nates of L foot of perpendicular drawn from the origin on

/57/

b

-'" X' o X

straight line AB are x,~--:~"="'--"::-----"~~~

OK OK = -.OL = pcosa.

OL Y'

KL = KL .OL = psina. OL

Also LNMB = 900+a. [':LNMB = LMLO+LLOM = 900+a.1 Slope of~traight line AB is m = tan(900+a.) = -cota.

Thus by slope - point form, equation of straight line AB will be

(Y -psin~) = -cota. (x-pcosa.)

or cos a.

(y-psina.) = -.- (x-pcosa.) sma.

ysina. - psin2a.= -x cosa. + pcos2a. or xcosa. + ysina. = p cos2a. + psin2a. or xcosa. + ysina. = p (cos2a. + sin2a.) or xcosa. + ysina. = p, is equation of straight line in normal form

Example 15 Find the equation of straight line at a perpendicular distance of 5 units from the origin such that the perpendicular drawn from the origin makes an angle of 30° with the +ve direction of x-axis measured anticlockwise.

/58/

Solution

or

or

or

or

Example 16

Solution

or

or

Given p = 5 a = 30°

By the normal form of straight line, the equation of required line is x cosa + ysina = p x cos30° +ysin30° = 5

.J3x +y = 10

.J3x +y-l0 = 0

Find the equation of straight line at a perpendicular distance of 4 units from the origin and making an angle of 120° with the +ve direction of x-axis measured anticlockwise, Given p = 4

e = 120° also e = 90° +a 120° = 900 +a a = 30°

By normal form of straight line, the equation of required straight line will

y

be x cos30° +ysin30° = 4 x,~-~-='------~-'--~

J3 1 x'T+y'2"=4

or .J3x +y, 8

or .J3 x +y -8 = 0

Example 17 Find the equation of straight line by using perpendicular form for which p = 3 and a = 60°

Solution

or

or

or

Given p = 3 and a = 60° By normal form x cos a + ysin a = p x cos60° + ysin60° = 3

x+J3y=6

x +.J3y -6 = 0

/59/

Example 18 Find the equation of straight line for which p = 7..fi and ex = 135°

Solution

or

or

or

or or

Given p = 7.fi and ex = 135° By normal form, x cos ex + ysin ex = p

x cos135° + ysin135° = 7.fi x cos(1800-45°) + ysin(1800-45°) = 7.fi

x[ -cos45°] + y[ sin45°] = 7.fi

-x +y = 14 x -y +14 = 0

8) Symmetric form of straight line (Distance form)

Let RS be any straight line inclined at an angle e(measured antic1ockwise) to the +ve direction of x-axis.

Let P(x,y) be any arbitrary point and A(Xl'Yl) be a fixed point on straight line RS such that distance between A and P is J r'

In ~APK,

cose AK

-AP

y

LM --AP

OM-OL y = r

cose = X-Xl y,

r M X

or r X - Xl

----- (1) = cose R

PK and sine -

AP

PM-KM = r

sine y - Yl

= r

/60/

or r = sinO

----- (2)

From (1) and (2)

X-Xl Y-YI --=--=r cosO sinO

is the symmetric form of straight line.

Example 19 Find the equation of straight line in symmetric form having point (2,5) on it and inclined at an angle of 45° with the +ve direction of x-axis.

Solution Given (Xl'Yl) = (2,5) and 0 = 45°

Equation of straight line in symmetric form is

X-Xl Y-YI --=--=r cosO sinO

x-2 y-2 ---=---=r cos 45° sin 45°

x-2 y-5 or -1- = -1- = r, is the equation of reqd. straight line in symmetric form.

h h Example 20 Find two point at distance of 4 units on either side of a point (2,3) lying

on a straight line making an angle of 60° with the +ve direction of x-axis( measured anticlockwise)

Solution Given (Xl'Yl) = (2,3) r = ±4

Equation of straight line in symmetric form is

or

or

X-Xl Y-YI --=--=r cose sinO

_x_-_2_= y-3 =±4 cos 60° sin 60°

x-2 y-3 -1-=±4 and J3 =±4

2 2

/61/

or 2x-4 = ±4 and 2y-6 = ±4.J3

or 4±4 d 6±4.J3 x=-- an y=

2 2

i.e. x = 4, Y = 3 + 2.J3 (taking +ve sign)

and x = 0, y = 3 - 2.J3 (taking -ve sign) The required two points are (4, 3 + 2../3) and (0, 3 - 2../3)

General Equation of Straight Line

Proof

To show that every first degree equation in x and y represents a straight line

Let Ax+By+C = 0 be any first degree equation in x and y. - - - (1) Let P(Xl'Y1)' Q(X2'Y2) and R(X3'Y3) by any three points lying on the locus of by

equation (1), therefore these points would satisfy equation (1).

i.e. AX1 + BY1 + C = 0 - - - (2) AX2 + BY2 + C = 0 - - - (3)

and Ax + By + C = 0 - - - (4) 3 3 (3) - (2) gives

A(X2-X1) + B(Y2-Yl) = 0 or A(X2-X1) = B(YI-Y2) and (4) - (3) gives

A(X3-X2) + B(Y3-Y2) = 0 or A(X3-X2) = B(Y2-Y3) Dividing (5) by (6) we get

or or

or or

or

X3- X2 Y2-Y3

(X2-X1) (Y2-Y3) = (YCY2) (X3-X2) (X1Y2-X2Yl) +(X2Y3 -X3Y2)

+(X3YCXIY3) = 0 2(Area of dPQR) = 0 Area of dPQR = 0 Points P,Q,R are collinear

- - - (5)

- - - (6)

.,' Aera of dPQR, where P(x1y 1)'

Q(X2'Y2) and R(x3Y3) is given by

Three points lying on the locus represented by (1) are collinear

Locus represented by (1) is a straight line. i.e. equation (1) represents a straight line. Hence every equation of first degree in x and Y represents a straight line.

/62/

Reduction of General Equation of straight Line To Various Standard Forms.

a) Reduction of general equation of straight line to slope-intercept form.

Let Ax + By + C = 0 by any general equation of a straight line - - - (1) By = -Ax - C

A C or y= -Bx-"B

or - - - (2)

which is the required slope intercept form, comparing (2) with y = mx+c

A -Coefficient of x we get, slope(m) = -13 = Coefficient of y

C and y-intercept(c)

B

-Coefficient of x Coefficient of y

Example 21 Reduce the equation 2x -5y +10 = 0 to slope intercept form. Hence find its slope(m) and y-intercept(c).

Solution Given equation of straight line is 2x -5y + 10 = 0 or -5y = -2x -10

2 or y = 5 x +2 -- - - (1)

Which is the required equation of straight line in SlOpE intercept form. Comparing equation (1) with y = mx+c we get,

2 Slope (m) = 5

and y-intercept (x) = 2

(b) Reduction of general equation of straight line to intercept form. Let Ax + By +C = 0 be the equation in general form of any straight line.

Ax +By =-C

A B or - C x- C y =l

or x y

--+--=1

(-~) (-~~) - - - (1)

Which is the required straight line in intercept form, Comparing (1) with intercept form

x Y -+-=1 we get a b

C -Constant term x-intercept (a) = - A = C ff" f oe lClent 0 x

C -Constant term y-intercept (b) = -13 = Coefficient of y

/63/

Example 22 Reduce the equation /3x - Y + 6 = 0 to intercept form. Hence find its

x-intercept and y-intercept.

Solution Given equation of straight line in general form is

(c)

J3x - y + 6 = 0 - - - (1)

or J3x-y=-6

or

or - - -(2)

is the required equation of straight line in intercept from.

Comparing (2) with ~ + bY = 1 we get, a '

a = -2/3, b = 6 x - intercept (a) = -2./3 and y - intercept (b) ~': 6

Reduction of general equation of straight line to normal (perpendicular) form Let Ax + By +C = 0 be the equation of straight line in general form. - - (1) Let xcosa. +y sin a. = p - - (2) or xcosa. +y sin a.-p = 0 be the required Normal form of eq.(1)

or

ABC --=--=-cosa sina -p

Cos a. = -~p and Sin a. = -~p} Squaring and adding we g;et,

- -(3)

/64/

or

or

or

or

and

2 C2 P = A2+B2

+ C P = - ~A2+B2 P is the length of perpendicular drawn from the origin to the straight line, therefore p is always positive.

C Thus if C is +ve, then p = ~ A 2 + B2

-C and if C is -ve, then p = ~A2 + B2

-A -B From (3) cosa = ~ A 2 + B2 , sina = ~ A 2 + B2 for c > a

A B

Equation (2) becomes

-Ax By C -;====== = -;====

.JA2+B2 ~A2+B2 ~A2+B2 for c > a

Ax + By = -C ~A2 + B2 ~A2 + B2 ~A2 + B2 for c < a

/65/

Thus during reduction of general form of straight line to Normal form we must take following steps-

Step 1: Take the constant to R.H.S.

Step 2:

Step 3:

Make R.H.s. +ve if required by multiplying both sides with -ve sign.

Divide both sides by J(Coef. of X)2 + (Coe£. of y)2

Resulting equation obtained will be in normal form.

Example 23 Reduce the equation 3x - 4y + 10 = 0 into normal form and hence find p and a..

Solution Equation of given st. line in general form is 3x-4+10 = 0 or 3x -4y = -10 or -3x +4y = 10

dividing both sides by ~ (_3)2 + (4)2 = 5

3 4 we get --x+-y=2

5 5 which is the required equation of straight line in Normal form. Comparing it with x cos a. + y sin a. = p we get,

3 4 cos a. = -5 and sin a. 5"' p =2

4/5 tan a. =---3/5

4 tan-1 ( - !) tan a. - or

3

Thus P =2 and a. = tan-1 ( - !). Example 24 Reduce the equation /3x + y - 7 = 0 to normal form and hence find the

value of p and a..

Solution Equation of given straight line in general form is /3x + y - 7 = 0

or /3x+y=7 - - -(1)

Dividing both sides by ~(/3)2 +(1)2 = 2, we get

/3 1 7 -x+-y=-

2 2 2 - - -(2)

/66/

or

Which is the required equation of straight line in normal form. Comparing (2) with x cos a + y sin a = p, we get

7 ..[3-. 1 P = 2' cos a = 2-' sm a = 2

1 2 1

tan a =../3 =../3 = tan30° 2

a = 30° 7 300 7t Thus P = - ,a = = - radians 2 6

Point of intersection of two straight lines

Let Alx + Bly +CI = 0

A2x + B2y +C2 =0

(1)

(2)

be two non parallel straight lines

Their point of intersection P can be obtained

by solving equation (1) and (2) for x and y.

By Cross- multiplication method

or

and

___ x __ = __ -"y ___ = ___ 1 __ _ B1C2 - B2C1 C1A2 -C2A1

B1C2 - B2C1 x= A1B2 -A2B1

C1A 2 -C2A 1 y= A1B2 -A2B1

Cross - multiplication

2 3 1 2

B1XC1XA1XB1 B2 C2 A2 132

[ B1C2 - B2C1 C1A2 - C2A1]

P A B - A B 'A B _ A B is the point of intersection of lines (1) and 12211221

Remark:

(2).

The point of intersection of st. lines (1) and (2) can also be obtained by method of elimination.

Example 25. Find the equation of the straight line passing through the point of intersection of lines x +2y +3 = 0 and 3x +4y +7 = 0 and perpendicular to the line x -y +9 = 0

/67/

Solution: The given straight lines are x + 2y + 3 = 0 - -(1)

and 3x + 4y + 7 = 0 - - (2)

or

Example 26

Solution:

By cross- multiplication method

x y 1 -_.------ = --.. -~- - ---:=: -------(2)(7) - (4)(3) (3)(3) - (7)(1) (1)(4) _. (3)(2)

_x_~..::~_=_1_ 14 -12 9 - 7 4 -- 6

x Y 1 --=-=-2 2 -2

.. x = -1, Y = -1 P (-1,-1) is the point of intersection of st. lines (1) and (2). The required st. lilne is perpendicular to the straight line

x - y + 9 = 0 - - (3) Slope of line (3) is given by

-Coefficient of x m =

1 Coefficient of y

1 = ---=1 (-1)

slope of required line will be

-1 -1 m= -=-=-1

ill! 1

By slope - point form, the equation of required line is

(y - y 1) = m( x - Xl) [y - (-1)] = (-1)[x-(-1)] Y +1 = -(x +1)

X'

x +y + 2 = 0 is the required line.

x

Find the equation of straight line through the point (2,5) and through the point of intersection of line x -2y +1 = 0 and 2x -5y +3 = O. The given straight lines are

x -2y +1 = 0 and 2x -5y +3 = 0 By cross- multiplication method

x _ y _ 1 - -

- - (1) - - (2)

(-2)(3) - (-5)(1) (1)(2) - (1)(3) (1)( -5) - (2)(2)

/68/

x Y 1 --=--=---6 + 5 2 - 3 -5 + 4

x Y 1 -=-=--1 -1 -1

or

x = 1, Y = 1 P(I,I) is the point of intersection of lines (1) and (2) The required line passes through the points P(I,I) and Q(2,5) equation of required st. line will be

Y2 -YI (y - Yl) = x2 - Xl (X - Xl)

(2,5) 5-1

(y - 1) = 2 -1 (x-I) Q

4 or y-l = 1 (x-I) or 4x -Y -3 = a is the required line.

Example 27 Find the equation of straight line parallel to 6x -7y +8 and through the point of intersection of straight lines X +2y +3 = a and 3x +4y +7 = O.

Solution The given straight lines are 6x -7y +8 = 9 - - (1) X +2y +3 = a - - (2)

and 3x +4y +7 = a - -(3) The required st. line passes through the point of intersection of st. lines (2) and (3)

Solving (2) and (3) by cross multiplication method, we have P (-1,-1) , the point of their intersection.

Since the required equation is parallel to line (1) their slopes will be same. Now, slope of eq. (1) is given by

-Coefficient of X

Example 28

Solution

6 or [y-(-I)] = -;; [x -(-1)]

or 7(y +1) = 6( x+l) 6x -7y -1 = 0 is the required line.

For what value of K, are the three lines x -2y +1= 0; 2x -5y +3 = 0 and 5x -9y +K = 0 are concurrent? The given equation of straight lines are

x-2y+l=0 --(1) 2x -5y +3 = 0 - - (2)

and 5x -9y + K = 0 - - (3)

/69/

Straight lines (1), (2) and (3) will be concurrent if they pass through same point

The point of intersection of (1) & (2) is given by

x _ Y = 1 (-2)(3) - (-5)(1) (1)(2) = (1)(3) (1)( -5) - (2)( -2)

x Y 1 --=--= -6 + 5 2 - 3 -5 + 4

or x=l,y=1 Le. P (1,1).

Lines (1), (2) and (3) will be concurrent if point P(I,I) of intersection of (1) and (2) would lie on (3)

5(1) -9 (1) + K = 0 or K=4

Angle between two lines Let the given two lines make angles 81 and

82 (measured anticlockwise ) with the +ve direction of x-axis.

Let 8 and ~ be two angles between two given straight lines as shown in figure.

Now, 81 = 8 + 82

or 8 = 8 - 8 1 2

[Exterior angle = sum of interior opposite angles1

tan 8 = tan (8 - 8)

y

o

flO/

Now or

m l -m2 = 1+m l ·m2 ;

m l -m2 Hence tan 8 = 1 +m l .m2 8+~=1t

~= 1t-8 tan ~ = tan(1t - 8)

-tan8

=

where m l = tan8 l and m2 = tan82 are the

slope of given lines respectively.

- - - (1)

-- - (2)

From (1) and (2) , tangent of the angle between two st. lines is given by

tan8 = ±( 1:~~J - - - (3) Positive value of tan8 implies 8 is the acute angle between two given lines, whereas

Negative value of tan8 implies 8 is the obtuse angle between two given lines. So, the acute angle '8' between the lines is given by

i.e.

m l -m2 tan 8 = 1 + m

l . m2

m l -m2 8 = tan-l 1 + m

l . m 2

- - - (4)

CONDITIONS FOR TWO STRAIGHT LINES TO BE PARALLEL OR PERPENDICULAR.

(a) Condition for two lines to be parallcl-If two lines are parallel. Then angle between them is 00 or 1800

or tan 8 = 0

i.e. =0

or ~ - m2 = 0 or m l = m 2 i.e. Slopes should be same, is the required condition for two st. lines to be

parallel.

/11/ (b) Condition for two st. lines to be perpendicular: .

If two straight lines are perpendicular to each other then angle between them is 90° or tan e = tan90° = 00

or ( ml -m2 ) + -00

- 1+ml· ffi2 -

or 1+m1 +m2 =0

i.e.

-1 Hence m1. m2 = -1 or m2 = ml i.e. product of slopes is -1 or slope of one is -ve

reciprocal of other is the required condition for two st. lines to be mutually perpendicular.

Example 29 Find the angle between the straight lines .J3x - y + 7 = 0 and x -y +2 :: 0 Solution

and

and

Equation of given lines are

.J3x-y+7 = 0

x - Y +2 = 0

slope of line (1) is m1

= -.J3 =.J3 -1

-(-1) slope of line (2) is m2 = ~ = 1

-- (1)

-- (2)

Angle e between lines (1) and (2) are given by

( .J3-1) = ± 1 +.J3.1

= ±(.J3 -1) .J3 +1

[ 1 J

1--+ .J3

- - 1 - 1+ J3 [

.: tan(450-e) = 1- tane] 1 + tane

/72/ = ± tan [45° -30°] = ± tan 15° = tan15°, -tan 15° = tan (180° -15°) = tan 165°

tan e = tan 15°, tan 165° e = 15° , 165°

Example 30 Find the acute angle between the St. lines 13x - y + 4 = 0 and

x-13y+5 =0

Solution