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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
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In Mathematics:
Locus = Place
Definition:
The locus of a moving point P(x,y) is the pathtaken or traced out by it.
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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
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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
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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
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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.
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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.
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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
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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
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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
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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 !
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HOMEWORK
Exercise 6.6.1:
Do the number 4, 5 and 6.