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.