Question

Find the point on x-axis which is equidistant from (2 , -5) and (-2 , 9)​

Answer 1

both points are equidistant from a point on x axis

let point (2,-5) be A and (-2,9) be B and point on x axis be P (x,0)

PA^2 =PB^2

(x-2)^2+ (y+5)^2 = (x+2)^2 + (y-9)^2

x^2 + 4 -4x + y^2 + 25+ 10y = x^2 + 4 +4x + y^2 + 81 -18y

8x+56-28y=0

point lies on x axis so coordinate of y axis is 0

then y value in the equation becomes zero

=> 8x+56=0

8x = -56

x = -7 required point is (-7,0)