Question

A(– 5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B.

Answer 1

Let P(x, y) be any point on the required locus.
P is equidistant from A(– 5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² – 4y + 4 = x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = – 8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
∴ The required equation of locus is 9x – y + 6 = 0.