Prove that the points $(3, 0)$, $(6, 4)$ and $(-1, 3)$ are the vertices of an isosceles triangle.

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Step 1: Let $A(3, 0)$, $B(6, 4)$, $C(-1, 3)$
Step 2: $AB = \sqrt{(3 - 6)^2 + (0 - 4)^2} = \sqrt{9 + 16} = 5$
Step 3: $BC = \sqrt{(6 + 1)^2 + (4 - 3)^2} = \sqrt{49 + 1} = \sqrt{50}$
Step 4: $CA = \sqrt{(-1 - 3)^2 + (3 - 0)^2} = \sqrt{16 + 9} = 5$. Since $AB = CA$, $	riangle ABC$ is isosceles.

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