A(- 5,2) and B(4,1). Find the equation of the locus of point P, which

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

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

Answer»

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.