Check whether $(5,-2), (6,4),$ and $(7,-2)$ are the vertices of an isosceles triangle.

(A) Let the points $A(5,-2), B(6,4),$ and $C(7,-2)$ represent the vertices of the triangle.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(6-5)^2 + (4-(-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$
$BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$
$CA = \sqrt{(5-7)^2 + (-2-(-2))^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2$
Since $AB = BC = \sqrt{37}$,two sides of the triangle are equal in length.
Therefore,the given points are the vertices of an isosceles triangle.