Without using Pythagoras Theorem, prove that the points A(1, 3), B(3, –1) and C(–5, –5) are the vertices of a right-angled triangle.

By using slope formula,

m = $\dfrac{y$2 - y1}{x2 - x1}

Given, points A(1, 3), B(3, –1)

Substituting values we get,

$m_{AB} = \dfrac{-1 - 3}{3 - 1} = -\dfrac{4}{2} = -2$

Given, points B(3, –1) and C(–5, –5)

Substituting values we get,

$m_{BC} = \dfrac{-5 - (-1)}{-5 - 3} = \dfrac{-5 + 1}{-8} = \dfrac{-4}{-8} = \dfrac{1}{2}$

Check for perpendicularity,

$\Rightarrow m${AB} \times m{BC} = -2 \times \dfrac{1}{2} = -1.

Since the product of the gradients of AB and BC is -1, the side AB is perpendicular to the side BC.

∠ABC = 90°.

Hence, proved the points A(1, 3), B(3, –1) and C(–5, –5) are the vertices of a right-angled triangle.