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

Solution:

Let the points be A(5, –2), B(6, 4) and C(7, –2).

$\therefore \quad \mathrm{AB}=\sqrt{(\mathbf{6}-\mathbf{5})^{2}+\mathbf{4}-(\mathbf{- 2})^{2}}$

$=\sqrt{(1)^{2}+(B)^{2}}=\sqrt{1+36}=\sqrt{37}$

$B C=\sqrt{(7-6)^{2}+(-2-4)^{2}}$

$=\sqrt{(1)^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37}$

$A C=\sqrt{(7-5)^{2}+(-2-(-2))^{2}}$

$=\sqrt{(+2)^{2}+(0)^{2}}=\sqrt{4+0}=2$

We have $\mathrm{AB}=\mathrm{BC} \neq \mathrm{AC}$.

$\therefore \quad \triangle \mathrm{ABC}$ is an isosceles triangle.