Question

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

Answer 1

Any point A on the x-axis is of the form (a,0)

If this is equidistant from P= (2,-5) and Q=(-2,9) then AP^2 = AQ^2

So by distance formula

(a-2)^2 +5^2 = (a+2)^2 +9^2

Solving we get a=-7

Hence the point on the x-axis which is equidistant from (2,-5) and (-2,9) is (-7,0)