Confidential | Copyright

www.testcracker.in

Image s

ourc

e:

Inte

rnet

[Pow

erP

lugs –

Tem

pla

te f

or

Pow

erP

oin

t

Coordinate Geometry – 1

Straight Lines CAT

Coordinates

X X’

Y

Y’

O

Origin

1 2 3 4

+ve direction

-1 -2 -3 -4

-ve direction

-1

-2

-3 -ve d

irection

1

2

3

+ve d

irection

X-axis : X’OX

Y-axis : Y’OY

Coordinates

X X’

Y

Y’

O

1 2 3 4 -1 -2 -3 -4

-1

-2

-3

1

2

3

(2,1)

(-3,-2)

Ordinate

Abcissa

(?,?)

Coordinates

X X’

Y

Y’

O

1 2 3 4 -1 -2 -3 -4

-1

-2

-3

1

2

3

(2,1)

(-3,-2)

Ordinate

Abcissa

(4,?)

Coordinates

X X’

Y

Y’

O

1 2 3 4 -1 -2 -3 -4

-1

-2

-3

1

2

3

(2,1)

(-3,-2)

Ordinate

Abcissa

(4,-2.5)

Quadrants

X X’ O

Y

Y’

I II

III IV

(+,+) (-,+)

(-,-) (+,-)

Distance Formula

x1

X X’

Y’

O

Y

x2

y1

y2

N PQN is a right angled .

PQ2 = PN2 + QN2

2 2

2 1 2 1PQ x x y y

y2-y

1

(x2-x1)

PQ2 = (x2-x1)2+(y2-y1)

2

Distance From Origin

Distance of P(x, y) from the origin is

2 2

x 0 y 0

2 2x y

Applications of Distance Formula

Parallelogram

Applications of Distance Formula

Rhombus

Applications of Distance Formula

Rectangle

Applications of Distance Formula

Square

Area of a Triangle

X X’

Y’

O

Y A(x1, y1)

C(x3, y3)

B(x

2,

y2)

M L N

Area of ABC =

Area of trapezium ABML

+ Area of trapezium ALNC - Area of trapezium BMNC

Area of a Triangle

X X’ Y’

O

Y A(x1, y1)

C(x3, y3)

B(x

2,

y2)

M L N

Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC

1 1 1

BM AL ML AL CN LN BM CN MN2 2 2

2 1 1 2 1 3 3 1 2 3 3 2

1 1 1y y x x y y x x y y x x

2 2 2

1 1

2 2

3 3

x y 11

x y 12

x y 1

Area of Polygons

Area of polygon with points Ai (xi, yi)

where i = 1 to n

2 21 1 n 1 n 1 n n

3 32 2 n n 1 1

x yx y x y x y1. . .

x yx y x y x y2

Can be used to calculate area of Quadrilateral,

Pentagon, Hexagon etc.

Collinearity of Three Points

Method I :

Use Distance Formula

a b

c

Show that a+b = c

Collinearity of Three Points

Method II :

Use Area of Triangle

A (x1, y1)

B (x2, y2)

C (x3, y3)

Show that 1 1

2 2

3 3

x y 1

x y 1 0

x y 1

Section Formula – Internal Division

X X’

Y’

O

Y

L N M

H

K

Clearly AHP ~ PKB

AP AH PH

BP PK BK

1 1

2 2

x x y ym

n x x y y

2 1 2 1mx nx my nyP ,

m n m n

Midpoint

Midpoint of A(x1, y1) and B(x2,y2)

m:n 1:1

1 2 1 2x x y yP ,

2 2

Section Formula – External Division

X X’

Y’

O

Y

L N M

H

K

Clearly PAH ~ PBK

AP AH PH

BP BK PK

1 1

2 2

x x y ym

n x x y y

2 1 2 1mx nx my nyP ,

m n m n

P divides AB externally in ratio m:n

Centroid

Intersection of medians of a triangle is called the centroid.

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,

2 2

1 2 1 2x x y yF ,

2 2

Centroid is always denoted by G.

Centroid

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,

2 2

1 2 1 2x x y yF ,

2 2

Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

2 3 2 31 1

x x y yx 2 y 2

2 2L ,1 2 1 2

Centroid

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,

2 2

1 2 1 2x x y yF ,

2 2

Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

1 3 1 32 2

x x y yx 2 y 2

2 2M ,1 2 1 2

Centroid

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,

2 2

1 2 1 2x x y yF ,

2 2

Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

1 2 1 23 3

x x y yx 2 y 2

2 2N ,1 2 1 2

Centroid

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,

2 2

1 2 1 2x x y yF ,

2 2

1 2 3 1 2 3x x x y y yL ,

3 3

1 2 3 1 2 3x x x y y yM ,

3 3

1 2 3 1 2 3x x x y y yN ,

3 3

We see that L M N G

Medians are concurrent at the centroid, centroid divides medians in

ratio 2:1

Centroid

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F G

2 3 2 3x x y yD ,

2 2

1 3 1 3x x y yE ,

2 2

1 2 1 2x x y yF ,

2 2

1 2 3 1 2 3x x x y y yL ,

3 3

1 2 3 1 2 3x x x y y yM ,

3 3

1 2 3 1 2 3x x x y y yN ,

3 3

We see that L M N G

Centroid

1 2 3 1 2 3x x x y y yG ,

3 3

Incentre

Intersection of angle bisectors of a triangle is called the incentre

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F I

Incentre is the centre of the incircle

Let BC = a, AC = b, AB = c

AD, BE and CF are the angle bisectors of A, B and C respectively.

BD AB b

DC AC c

2 3 2 3bx cx by cyD ,

b c b c

Incentre

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F I

BD AB b

DC AC c

2 3 2 3bx cx by cyD ,

b c b c

AI AB AC AB AC c bNow,

ID BD DC BD DC a

2 3 2 31 1

bx cx by cyax b c ay b c

b c b cI ,a b c a b c

1 2 3ax bx cxI

a b c

Similarly I can be derived using E and F also

Incentre

A(x1, y1)

B(x2, y2) C(x3, y3) D

E F I

BD AB b

DC AC c

2 3 2 3bx cx by cyD ,

b c b c

AI AB AC AB AC c bNow,

ID BD DC BD DC a

2 3 2 31 1

bx cx by cyax b c ay b c

b c b cI ,a b c a b c

1 2 3ax bx cxI

a b c

Angle bisectors are concurrent at the incentre

Cirumcentre

Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre.

OA = OB = OC

= circumradius

A

B

C

O

The above relation gives two simultaneous linear equations. Their solution gives the coordinates of O.

Orthocentre

Intersection of altitudes of a triangle is called the orthocentre.

A

B C

H

Orthocentre is always denoted by H

Cirumcentre, Centroid and Orthocentre

The circumcentre O, Centroid G and Orthocentre H of a triangle are collinear.

O

H

G

G divides OH in the ratio 1:2

Locus – a Definition

The curve described by a point which moves under a given condition or conditions is called its locus

e.g. locus of a point having a constant distance from a fixed point :

Circle!!

Locus – a Definition

The curve described by a point which moves under a given condition or conditions is called its locus

e.g. locus of a point equidistant from two fixed points :

Perpendicular bisector!!

Equation to a Locus

Algorithm to find the equation to a locus :

Step I : Assume the coordinates of the point whose locus is to be found to be (h,k)

Step II : Write the given conditions in mathematical form using h, k

Step III : Eliminate the variables, if any

Step IV : Replace h by x and k by y in Step III. The equation thus obtained is the required equation to locus

Illustrative Example

Find the equation to the locus of the point equidistant from A(1, 3) and B(-2, 1)

Let the point be P(h,k)

PA = PB (given)

PA2 = PB2

(h-1)2+(k-3)2 = (h+2)2+(k-1)2

6h+4k = 5

equation of locus of (h,k) is 6x+4y = 5

Solution :

Illustrative Example

A rod of length l slides with its ends on perpendicular lines. Find the locus of its midpoint.

Let the point be P(h,k)

Let the lines be the axes

Let the rod meet the axes at

A(a,0) and B(0,b)

h = a/2, k = b/2

Also, a2+b2 = l2

4h2+4k2 = l2

equation of locus of (h,k) is 4x2+4y2 = l2

B(0,b)

A(a,0) O

P(h,k)

Solution :

Prove it!

If the line joining the points A(a,b) and B(c,d) subtends an angle at the origin, prove that

2 2 2 2

ac bdcos

a b c d

Solution

On simplifying,

2 2 2 2

ac bdcos

a b c d

Let O be the origin.

OA2 = a2+b2, OB2 = c2+d2, AB2 = (c-a)2+(d-b)2

Using Cosine formula in OAB, we have

AB2 = OA2+OB2-2OA.OBcos

2 2 2 2 2 2 2 2 2 2c a d b a b c d 2 a b c d cos

Solve it.

Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given such that

Find x. DBC 1

ABC 2

Given that ABC = 2DBC

6 3 1 x 3x 1

3 5 1 2 3 5 1

4 2 1 4 2 1

6 5 2 3 4 3 1 6 20 2 x 5 2 3x 4 3 1 6 20

2 28x 14 49

4928x 14

2

11 3x or x

8 8

Solution :

Prove it.

If a b c, prove that (a,a2), (b,b2) and (c,c2) can never be collinear.

Let, if possible, the three points be collinear.

2

2

2

a a 11

b b 1 02

c c 1

R3 R3- R2, R2 R2-R1

2

2 2

2 2

a a 1

b a b a 0 0

c b c b 0

2a a 1

b a c b 1 b a 0 0

1 c b 0

Solution :

Solution Cont.

b a c b c a 0

This is possible only if a = b or b = c or c = a.

But a b c. Thus the points can never be collinear.

Q.E.D.

Solve it.

Three vertices of a parallelogram taken in order are (a+b,a-b), (2a+b,2a-b) and (a-b,a+b). Find the fourth vertex.

Let the fourth vertex be (x,y).

Diagonals bisect each other.

a b a b 2a b x a b a b 2a b yand

2 2 2 2

the required vertex is (-b,b)

Solution :

Solve it.

The locus of the midpoint of the portion intercepted between the axes by the line xcos+ysin = p, where p is a constant, is

2 2 2

2 2 2

2 2

2 2 2 2

1 1 4(a)x y 4p (b)

x y p

4 1 1 2(c)x y (d)

p x y p

Solution

Let the line intercept at the axes at A and B. Let R(h,k) be the midpoint of AB.

p p

R h,k ,2cos 2sin

p psin , cos

2k 2h

2 2

2 2

p p1

4k 4h 2 2 2

1 1 4Locus

x y p

Ans : (b)

Slope of a line

X X’

Y’

O

Y

Slope = tan y

x

1

Steps forward

Ste

ps u

p

is always w.r.t. X’OX

y

x

(x,y)

Slope

slope +ve is acute

slope -ve is obtuse

= 0 slope = 0

= 90 slope = ?

Infinite?

Not infinite. It is not defined.

Slope is usually denoted by m

Slope in terms of points on a line

X X’

Y’

O

Y

P (x1, y1)

Q (x2, y2)

L M

N

2 1

2 1

y yQN difference of ordinatestan

PN x x difference of abcissae

Slope of reflection in either axis

X X’

Y’

O

Y

Slope of a line = m slope of reflection = -m

Angle between two lines

2 1

2 1

2 1tan tan

2 1

2 1

tan tantan

1 tan tan

X X’

Y’

O

Y

1 2

-

2 1

2 1

tan tanAlso tan tan

1 tan tan

2 1

1 2

m mtan

1 m m

Parallel lines

tan 0

2 1

1 2

m m0

1 m m

1 2m m

Perpendicular lines

cot 0

1 2

2 1

1 m m0

m m

1 2m m 1

Intercepts on x axis, y axis

X X’

Y’

O

Y

A

B

x-intercept

Consider a line cutting the axes in A and B

OA = x-intercept OB = y-intercept

Slope intercept form

X X’

Y’

O

Y P(x, y)

c

Q x

y-c

L

M

Consider a line making an angle with the x-axis and an intercept c with the y-axis Consider a point P (x, y) on it

Slope = m = tan = PM

QM

y c

x

y m x c

Coefficient of y = 1

Locus definition of a straight line

Condition 1: A point on the line is given

Any number of lines may pass through

a given point.

Condition2: Direction of the

line is given

Any number of lines can

lie in a certain

direction.

Point slope form

Consider a line passing through P (x1, y1) and having a slope m.

Consider any point Q (x, y) on it.

slope m = 1

1

y y

x x

1 1y y m x x

BUT ONLY ONE straight line can pass through a given point in a given direction

Illustrative example

Slope of AB = 3 5

12 6

Slope of perpendicular = 1

Perpendicular bisector will pass through midpoint of AB which is (2, -1)

the required equation is y+1 = x-2 or y = x-3

Find the equation of the perpendicular bisector of the line segment joining the points A (-2, 3) and B (6, -5)

Solution :

Two point form

Consider a line passing through P (x1, y1) and Q (x2, y2).

slope m = 2 1

2 1

y y

x x

2 11 1

2 1

y yy y x x

x x

Using point slope form,

Illustrative example

By section formula,

D (-3, 4), E (0, 2) and F (-1, 7)

Using two point form,

4 5

AD y 5 x 2 x 5y 23 03 2

Find the equation of the medians of the triangle ABC whose vertices are A (2, 5), B (-4, 9) and C (-2,-1) through A

Let the midpoints of BC, CA and AB be D, E and F respectively

Solution :

Intercepts form

X X’

Y’

O

Y

Consider a line making intercepts a and b on the axes.

Consider a point P (x, y) on it.

Area of OPB

P (x, y)

y

x

a

b

A

B

+ Area of OPA = Area of OAB

1 1 1bx ay ab

2 2 2

x y1

a b

Illustrative example

Let the y-intercept = c.

the x-intercept = c+5

Intercept form of line is given by 22 6

1c 5 c

As this passes through (22,-6)

22 61

c 5 c

Find the equation of a line which passes through (22, -6) and is such that the x-intercept exceeds the y- intercept by 5.

Solution :

Solution Cont.

the required equation is

x y x y1 or 1

10 5 11 6

Rearranging,

x+2y-10 = 0 or 6x+11y-66 = 0

c2-11c+30 = 0

(c-5)(c-6) = 0

c = 5 or c = 6

Solve these. Fast!

(i) Find the equation of the line which passes through (1, 2) and the sum of the intercepts on the axes is 6.

(ii) Find the equation of the line through (3, 2) so that the segment of the line intercepted between the axes is bisected at this point.

Solution (i)

Let the equation of line in intercept form be

x y1,

a b if it passes through (1, 2), then

1 21,

a b also a + b = 6

1 21

a 6 – a 6 – a 2a a 6 – a

26 a 6a – a 2a – 5a 6 0 a – 3 a – 2 0

a 3 or 2 Corresponding b = 3 or 4

Hence, equations become

x y x y1 or 1

3 3 2 4

Solution (ii)

Let x-intercept and y-intercept of the line be a and b respectively i.e. line passes through (a, 0) and (0, b)

As segment joining (a, 0) and (0, b) is bisected by (3, 2)

a 0 0 b3 and 2

2 2

a = 6 and b = 4 Equation of line becomes

x y1

6 4 or 2x + 3y = 12

Solve this. Fast!

Find the equations of the lines which passes through the origin and trisect the portion of the straight line 3x + y = 12 which is intercepted between the axes of coordinates.

Solution :

A

B

P

Q

O

y

x

(0,12)

(4,0)

Let P be the point dividing AB in 2 : 1,

4 2.0 1.0 2.12 4P , , 8

3 3 3

Solution

And Q be the point dividing AB in the ratio 1 : 2, then

1.0 2.4 1.12 2.0 8Q , , 4

3 3 3

Equation of 8

OP y x y 6x4

3

Equation of 4 3

OQ y x y x8 2

3

Now try this.

Line L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angle; keeping the origin fixed, the same line has intercepts p and q, then

(a) 2 2 2 2a b p q (b) 2 2 2 2

1 1 1 1

a b p q

(c) 2 2 2 2a p b q (d)

2 2 2 2

1 1 1 1

a p b q

Solution

X Y1

p q ...(ii)

...(i) x y1

a b

Now both equations represents the same line with different axes.

Hence the distance of origin from both the lines is the same.

2 2 2 2

1 1 1 1or

a b p q

2 2 2 2

1 1

1 1 1 1

a b p q

Thank you…

Email your doubts to: support@testcracker.in

Contact 9035001996 for any query.