Show that the points A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now

$A B=\sqrt{(3-6)^{2}+(0-4)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}$

$=\sqrt{9+16}=\sqrt{25}=5$

$B C=\sqrt{(6+1)^{2}+(4-3)^{2}}=\sqrt{(7)^{2}+(1)^{2}}$

$=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2}$

$A C=\sqrt{(3+1)^{2}+(0-3)^{2}}=\sqrt{(4)^{2}+(-3)^{2}}$

$=\sqrt{16+9}=\sqrt{25}=5$

$\because A B=A C$ and $A B^{2}+A C^{2}=B C^{2}$

Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.