Without using the distance formula, show that the points A(4, -2), B(-4, 4) and C(10, 6) are the vertices of a right-angled triangle.

By formula,

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

$\text{Slope of AB }(m$1) = \dfrac{4 - (-2)}{-4 - 4} \\[1em] = \dfrac{4 + 2}{-8} \\[1em] = -\dfrac{6}{8}. \\[1em] \text{Slope of AC }(m2) = \dfrac{6 - (-2)}{10 - 4} \\[1em] = \dfrac{6 + 2}{6} \\[1em] = \dfrac{8}{6}. \\[1em] m1 \times m2 = -\dfrac{6}{8} \times \dfrac{8}{6} = -1.

Since, m₁.m₂ = -1.

∴ AB ⊥ AC.

Hence, proved that ABC is a right-angled triangle at A.