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7.6 Equation of a Locus


7.6 Equation of a Locus
1. The equation of the locus of a moving point P(x, y) which is always at a constant distance (r) from a fixed point (x1, y1) is:


2. The equation of the locus of a moving point P(x, y) which is always at a constant distance from two fixed points (x1, y1) and  (x1, y1) with a ratio is:



3. The equation of the locus of a moving point P (x, y) which is always equidistant from two fixed points A and B is the perpendicular bisector of the straight line AB.



Example 1
Find the equation of the locus of a moving point P(x, y) which is always at a distance of 5 units from a fixed point Q (2, 4).

Solution:
(x – x1)2+ (y – y1)2 = r2
(x – 2)2 + (y – 4)2 = 52
x2 – 4x + 4 + y2 – 8y + 16 = 25
x2 + y2– 4x – 8y – 5 = 0


Example 2
Find the equation of the locus of a moving point P(x, y) which is always equidistant from points A (-2, 3) and B (4, -1).

Solution:
Given  P A = P B ( x − ( − 2 ) ) 2 + ( y − 3 ) 2 = ( x − 4 ) 2 + ( y − ( − 1 ) ) 2 Square both sides to eliminate the square roots . ( x + 2 ) 2 + ( y − 3 ) 2 = ( x − 4 ) 2 + ( y + 1 ) 2 x 2 + 2 x + 4 + y 2 − 6 y + 9 = x 2 − 8 x + 16 + y 2 + 2 y + 1 10 x − 8 y − 4 = 0 Hence, the equation of the locus of point  P  is 10 x − 8 y − 4 = 0

Example 3
A (2, 0) and B (0, -2) are two fixed points. Point P moves with a ratio so that AP: PB = 1: 2.  Find the equation of the locus of point P.

Solution:
A P : P B = 1 : 2 A P P B = 1 2 2 A P = P B 2 ( x − 2 ) 2 + ( y − 0 ) 2 = ( x − 0 ) 2 + ( y − ( − 2 ) ) 2 Square both sides to eliminate the square roots . 4 [ ( x − 2 ) 2 + y 2 ] = x 2 + ( y + 2 ) 2 4 ( x 2 − 4 x + 4 + y 2 ) = x 2 + y 2 + 4 y + 4 4 x 2 − 16 x + 16 + 4 y 2 = x 2 + y 2 + 4 y + 4 3 x 2 + 3 y 2 − 16 x − 4 y + 12 = 0 Hence, the equation of the locus of point  P  is 3 x 2 + 3 y 2 − 16 x − 4 y + 12 = 0

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