Find the equation to the locus of the perpendicular bisector of the li

1.	Find the equation to the locus of the perpendicular bisector of the line joining A(3, 2) and B(4,1).
Answer» The perpendicular bisector of the line joining A and B is the locus of the point which moves such that it is equidistant from A and B. By data we have A = (3, -2), B = (4, 1) Let P(x, y) be any point on the perpendicular bisector Thus we have PA = PB = PA² = PB² => (x – 3)² + (y + 2)² = (x – 4)² + (y – 1 )² ⇒ x² + 9 - 6x + y² + 4 + 4y = x² + 16 - 8x + y² + 1 - 2y ⇒ -6x + 4y + 13 = -8x – 2y + 17 ⇒ 2x + 6y – 4 = 0. ⇒ x + 3y – 2 = 0 is the equation of the locus of a point.

Find the equation to the locus of the perpendicular bisector of the line joining A(3, 2) and B(4,1).

Answer»

The perpendicular bisector of the line joining A and B is the locus of the point which moves such that it is equidistant from A and B.

By data we have A = (3, -2), B = (4, 1)

Let P(x, y) be any point on the perpendicular bisector Thus we have PA = PB = PA² = PB²

=> (x – 3)² + (y + 2)² = (x – 4)² + (y – 1 )²

⇒ x² + 9 - 6x + y² + 4 + 4y = x² + 16 - 8x + y² + 1 - 2y

⇒ -6x + 4y + 13 = -8x – 2y + 17

⇒ 2x + 6y – 4 = 0.

⇒ x + 3y – 2 = 0 is the equation of the locus of a point.