presentation topic 6.6

8/9/2019 Presentation TOPIC 6.6

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TOPIC 6 : COORDINATE GEOMETRY

6.6 : THE EQUATION OF LOCUS

INVOLVING DISTANCE

BETWEEN TWO POINTS

6.6.1 : THE EQUATION OF LOCUS



In Mathematics:

Locus = Place

Definition:

The locus of a moving point P(x,y) is the pathtaken or traced out by it.



Determining The Locus of Points

Circle Perpendicular Bisector

The locus of pointsthat are of constant

distance from a fixed

point is a circle

The locus of pointsthat are equidistant

from two fixed points

is the perpendicular

bisector of the line

joining the two points

A B

Locus of P

A.

Locus of P

P



The equation of locus can be determine by using

the distance formula

From Pythagoras¶ Theorem:

c2 = c x c

c2 = 4 x 1 ab + (b ± a)2

2

= 2ab + b2 ± 2ab +a2

c2 = b2 + a2



Distance Between Two Points

Let and be any two points on the

coordinate plane, using Pythagoras¶ Theorem that:

),( 11 y x P ),( 22 y xQ

222

QN PN PQ!

2

12

2

12 )()( y y x x !

2

12

2

12 )()( y y x x PQ !

0

y

x

),( 11 y x

),( 22 y x

12 y y

12 x x N

Hence, the equation of locus can be determine

by using the distance formula



a) The locus of moving point that is of constant

distance from a fixed point.

Example:Find the equation of locus of a point which moves such that its

distance from the point A(3,2) is 5 units.

Solution:

Let the point be P(x,y).

The given condition is PA = 5PA = 5

(x ± x1)2 + (y ± y1)

2 = d

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

(x-3)2

+ (y-2)2

= 25X2 ± 6x + 9 + y2 ± 4y + 4 = 25

X2 ± 6x + y2 ± 4y + 9 + 4 ± 25 = 0

X2 + y2 ± 6x ± 4y ± 12 = 0

Hence, X2 + y2 ± 6x ± 4y ± 12 = 0 is required locus.

Squaring both

sides to

eliminate the

square root.



b) The ratio of the distance of a moving point from the two

fixed points is constant.

Example 1:

Find the equation of the locus of P if P is equidistant from the point

A(-2,0) and B(0,3).

Solution:

Let the point be P(x,y).

The given condition is distance P(x,y) from A(-2,0) and B(0,3)

is equidistant that:

PA = PB

(x ± x1)2 + (y ± y1)

2 = (x ± x2)2 + (y ± y2)

2

(x ± (-2))

2

+ (y ± 0)

2

= (x ± 0)

2

+ (y ± 3)

2

(x + 2)2 + y2 = x2 + (y ± 3)2

(x + 2)2 + y2 = x2 + (y ± 3)2

X2 + 4x + 4 + y2 = x2 + y2 ± 6y + 9

4x + 6y ± 5 = 0

Hence, 4x + 6y ± 5 = 0 is required locus.

Squaring both

sides to

eliminate thesquare root.



Example 2:

A point P moves such that its distances from S(1,2) and T (4,-1) are

in the ratio 2 : 1. Find the equation of locus P.

Solution:

Let P(x,y) be the moving point.

The given condition is PS : PT = 2 : 1

PS = 2PT 1

PS = 2PT

(x ± x1)2 + (y ± y1)

2 = 2 (x ± x2)2 + (y ± y2)

2

(x ± 1)2 + (y ± 2)2 = 2 (x ± 4)2 + (y + 1)2

x2 ± 2x + 1 + y2 ± 4y + 4 = 2 (x2 ± 8x + 16 + y2 + 2y + 1)

Squaring

both

sides to

remove

thesquare

root sign

S T

2 1

P



x2 ± 2x + 1 + y2 ± 4y + 4 = 4 (x2 ± 8x + 16 + y2 + 2y + 1)

x2 ± 2x + y2 ± 4y + 5 = 4x2 ± 32x + 64 + 4y2 + 8y + 4

x2 ± 2x + y2 ± 4y + 5 = 4x2 ± 32x + 4y2 + 8y + 68

3x2 ± 30x + 3y2 + 12y + 63 = 0

x2 ± 10x + y2 + 4y + 21 = 0

x2

+ y

2

± 10x

+ 4y + 21 = 0

Thus, the equation of the locus of P is x2 + y2 ± 10x + 4y + 21 = 0.

Dividing

each term

of the

equation by

3



Exercises:

Find the equation of the locus of the point P(x,y) if it moves:

a) Its distance from the point A (3,5) is 3 cm.

b) It is equidistant from the points A (-1,-2) and B (2,3).

c) Its distance from the points A (2,6) and B (-4,0) is in the

ratio 2 : 1



Summary:

The Equation of Locus

a) The locus of moving point that is of

constant distance from a fixed point.

b) The ratio of the distance of a moving

point from the two fixed points is constant.

The equation of locus can be determine by using thedistance formula.

d2

12

2

12 )()( y y x x !



HOMEWORK

Exercise 6.6.1:

Do the number 4, 5 and 6.

presentation topic 6.6

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