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

Vertices of the triangle are A(-3,0), B(1,-3), C(4,1).
Distance between two points = √(x2-x1)²+(y2-y1)²
AB = √(1+3)²+(-3-0)² = √16+9 = 5
BC = √(4-1)²+(1+3)² = √9+16 = 5
AC = √(4+3)²+(1-0)² = √49+1 = √50 = 5√2
AB = BC
Therefore, ΔABC is an isosceles triangle
(AB)²+(BC)² = 5²+5² = 50
and (AC)² = (5√2)² = 50
∴ (AB)²+(BC)² = (AC)²
So, the triangle satisfies the Pythagoras theorem and hence it is a right-angled triangle.