What point on the x-axis is equidistant from the points (7, 6) and (-3, 4) ?

Let the required point on the x-axis be (x, 0).

Given (x, 0) is equidistant from (7, 6) and (-3, 4).

i.e. distance between (x, 0) and (7, 6) = distance between (x, 0) and (-3, 4)

$\sqrt{(7 - x)^2 + (6 - 0)^2} = \sqrt{((-3) - x)^2 + (4 - 0)^2}\\[1em] ⇒ \sqrt{(7 - x)^2 + (6)^2} = \sqrt{(-3 - x)^2 + (4)^2}\\[1em] ⇒ (7 - x)^2 + (6)^2 = (-3 - x)^2 + (4)^2\\[1em] ⇒ 49 + x^2 - 14x + 36 = 9 + x^2 + 6x + 16\\[1em] ⇒ x^2 - 14x + 85 = x^2 + 6x + 25\\[1em] ⇒ - 14x + 85 = 6x + 25\\[1em] ⇒ - 14x - 6x = 25 - 85\\[1em] ⇒ - 20x = -60\\[1em] ⇒ x = \dfrac{60}{20}\\[1em] ⇒ x = 3$

Hence, the point on the x-axis which is equidistant from the points (7, 6) and (-3, 4) is (3, 0).