Question

Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3, 5).

Answer 1

TO FIND The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(x, y) be any point on the perpendicular bisector of AB. Then,

PA = PB

`=> sqrt((x - 7)^2 + (y - 1)^2) = sqrt((x - 3)^2 + (y - 5)^2)`

`=> (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2`

`=> x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25`

`=> -14x + 6x + 10y - 2y + 49 + 1 - 9 - 25 = 0`

`=> -8x + 8y + 16 = 0`

`=> x - y = 2`

Hence the equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5) is x - y = 2.