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

Let the points be A(3, 0), B(6, 4) and C(-1, 3).

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{(6 - 3)^2 + (4 - 0)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] \therefore BC = \sqrt{(-1 - 6)^2 + (3 - 4)^2} \\[1em] = \sqrt{(-7)^2 + (-1)^2} \\[1em] = \sqrt{49 + 1} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.\\[1em] \therefore AC = \sqrt{(-1 - 3)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5.

∴ AB = AC = 5

∴ ΔABC is an isosceles triangle

AB² + AC² = 5² + 5²

= 25 + 25

= 50.

BC² = $(5\sqrt{2})^2$ = 50.

Since, AB² + AC² = BC².

Hence, proved that (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.