Show that the triangle formed by the points A(1, 3), B(3, -1) and C(-5, -5) is a right angled triangle (by using slopes).

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by

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

Slope (m₁) of the line joining A(1, 3) and B(3, -1) is,

$= \dfrac{-1 - 3}{3 - 1} \\[1em] = \dfrac{-4}{2} \\[1em] = -2.$

Slope (m₂) of line joining B(3, -1) and C(-5, -5) is,

$= \dfrac{-5 - (-1)}{-5 - 3} \\[1em] = \dfrac{-4}{-8} \\[1em] = \dfrac{1}{2}.$

Since, m₁ × m₂ = $-2 \times \dfrac{1}{2} = -1$.

Thus AB and BC are perpendicular to each other.

Hence, △ABC is a right-angled triangle.