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

rn

and (2, -3).

Let any point equidistant from the points (-2,5) and (2,-3) be given by P(x,y)

Then since distance between (x,y) and (-2,5) and the distance between (x,y) and (2,-3) is equal, we have by the distance formula:

(x-(-2))² + (y-5)² = (x-2)² + (y-(-3))²

On evaluating, this comes out to: x - 2y + 2 = 0

This line is the locus of all the points equidistant from (-2,5) and (2,-3) [Locus is the line or curve representing all the solutions to a given equation]

Since you want the point on the x-axis that is equidistant from these points, put y = 0 in x - 2y + 2 = 0 (because y=0 for any pt in the x-axis). You get x + 2 = 0 i.e  x = -2. Hence the point on the x-axis equidistant from the given points is (-2,0)

Note that you can also solve this problem more easily by taking y in the equation (x+2)² + (y-5)² = (x-2)² + (y+3)² to be zero, as you know that the point lies on the x-axis and y=0. Hence you can directly get x = -2. But I found the locus so as to give you a better understanding of the concept.