Question

Find an equation for the perpendicular bisector of the line segment whose endpoints are (8,9) and (4,−3).

Answer 1

Answer:

that is the answer which you ask me

Answer 2

x + 3y -15 = 0

Step-by-step explanation:

let (x,y) be the coordinate of bisector,

so (x,y) should be equal distance from point (8,9) and (4,-3)

(x-8)^2 + (y-9)^2 = (x-4)^2 + (y-(-3))^2

or, (x^2 - 16x + 64) + (y^2 - 18y + 81 )= (x^2 -8x + 16) + (y^2 + 6y + 9)

After cancelling x^2 and y^2 from both side, we get

-16x - 18y +145 = -8x +6y + 25

or, -16x + 8x - 18y - 6y +145 -25 = 0

or, -8x - 24y + 120 = 0

or -8 ( x + 3y - 15) = 0

or, x + 3y - 15 = 0 ------ this is the equation of the perpendicular bisector of line segment with endpoints (8,9) and (4,-3)