Find the point on the x-axis that is equidistant from the points (-4,-3) and (-1,5)

Question

Hello. I need some help with the equation in the title. I am very stuck.

Answer 1

A point on the x-axis, can be expressed as (x,0)

So, we are trying to find x so that the distance from (x,0) to (-4,-3) is the same as the distance from (x,0) to (-1,5).

So, √[(x-(-4))² + (0-(-3))²] = √[(x-(-1))² + (0-5)²

x² + 8x + 16 + 9 = x² + 2x + 1 + 25

8x + 25 = 2x + 26

6x = 1

x = 1/6

The required point is (1/6, 0).