Question

A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)

Answer 1

Let P(x, y) be any point on the required locus.
Given, A(1, 6) and B(3, 5),
∠APB = 90°

∴ ΔAPB is a right angled triangle.

By Pythagoras theorem,

AP² + PB² = AB²

∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 – 5)²

∴ x² – 2x + 1 + y² – 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1

∴ 2x² + 2y² – 8x – 22y + 66 = 0

∴ x² + y² – 4x – 11y + 33 = 0

∴ The required equation of locus is
x² + y² – 4x – 11y + 33 = 0.