Find the point on the $x$-axis which is equidistant from A(−2, 3) and B(6, 5).

Any point on the $x$-axis is $(x$, 0). Equate the SQUARED distances: $(x+2{)}^2+(0-3{)}^2=(x-6{)}^2+(0-5{)}^2$ → $x^2+4x+4+9=x^2-12x+36+25$ → $4x+13=-12x+61$ → $16x=48$ → $x=3$. The point is (3, 0); check that both distances equal $\sqrt{34}$. Taking the midpoint's $x$-coordinate, $\frac{-2+6}{2}=2$, gives (2, 0) and is wrong — the midpoint of AB does not sit on the $x$-axis here. (0, 3) answers the $y$-axis version of the question.