1. Stating the problem: We have points $A(-4,6)$ and $B(-1,-3)$, and we want to find point $P$ on the $x$-axis such that $P$ is equidistant from $A$ and $B$. Since $P$ lies on the $x$-axis, its coordinates are $P(x,0)$. 2. Use distance formula: The distance from $P$ to $A$ is $$d(P,A) = \sqrt{(x - (-4))^2 + (0 - 6)^2} = \sqrt{(x + 4)^2 + 36}.$$ The distance from $P$ to $B$ is $$d(P,B) = \sqrt{(x - (-1))^2 + (0 - (-3))^2} = \sqrt{(x + 1)^2 + 9}.$$ 3. Set the distances equal because $P$ is equidistant from $A$ and $B$: $$\sqrt{(x + 4)^2 + 36} = \sqrt{(x + 1)^2 + 9}.$$ 4. Square both sides to eliminate the square roots: $$(x + 4)^2 + 36 = (x + 1)^2 + 9.$$ 5. Expand squares: $$(x^2 + 8x + 16) + 36 = (x^2 + 2x + 1) + 9.$$ Simplify: $$x^2 + 8x + 52 = x^2 + 2x + 10.$$ 6. Subtract $x^2$ from both sides: $$8x + 52 = 2x + 10.$$ 7. Rearrange terms: $$8x - 2x = 10 - 52,$$ $$6x = -42.$$ 8. Solve for $x$: $$x = \frac{-42}{6} = -7.$$ 9. The coordinate of $P$ on the x-axis is therefore $$P(-7, 0).$$