Show that the points $(7, 10)$, $(-2, 5)$ and $(3, -4)$ are the vertices of an isosceles right triangle.

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Step 1: Let $A(7, 10)$, $B(-2, 5)$, $C(3, -4)$. $AB = \sqrt{(-2-7)^2 + (5-10)^2} = \sqrt{81 + 25} = \sqrt{106}$
Step 2: $BC = \sqrt{(3+2)^2 + (-4-5)^2} = \sqrt{25 + 81} = \sqrt{106}$
Step 3: $AC = \sqrt{(3-7)^2 + (-4-10)^2} = \sqrt{16 + 196} = \sqrt{212}$
Step 4: $AB = BC$ (isosceles) and $AC^2 = 212 = 106 + 106 = AB^2 + BC^2$ (right angle)
Step 5: Hence $	riangle ABC$ is an isosceles right triangle.

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