Show that $(0,6), (-5,3)$ and $(3,1)$ are the vertices of an isosceles right triangle.

(N/A) Let the vertices be $A(0,6), B(-5,3)$,and $C(3,1)$.
Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:
$AB^2 = (-5 - 0)^2 + (3 - 6)^2 = (-5)^2 + (-3)^2 = 25 + 9 = 34$
$BC^2 = (3 - (-5))^2 + (1 - 3)^2 = (8)^2 + (-2)^2 = 64 + 4 = 68$
$AC^2 = (3 - 0)^2 + (1 - 6)^2 = (3)^2 + (-5)^2 = 9 + 25 = 34$
Since $AB^2 + AC^2 = 34 + 34 = 68 = BC^2$,the triangle satisfies the Pythagorean theorem,so it is a right-angled triangle with $\angle BAC = 90^\circ$.
Also,since $AB^2 = AC^2 = 34$,we have $AB = AC$.
Therefore,the triangle is an isosceles right triangle.