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

Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.

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 = \(\sqrt{(7-5)^2 + (-2 + 2)^2} \)
\( = \sqrt{2^2 - 0} \)
\( = \sqrt{4} \)
\( = 2\)
Since AB = BC.

Therefore, A, B and C are vertices of an isosceles triangle.