Question

Without using the Pythogoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.

Answer 1

Let A(4, 4), B(3, 5) and C(-1, -1) be three vertices of a $\Delta ABC$.

$\therefore$ Slope of $AB=\frac{5-4}{3-4}=\frac{1}{-1}=-1$

$\therefore$ Slope of $BC=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$

$\therefore$ Slope of $AC=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1$

Now slope of AB $\times$ slope of AC = $-1\times 1=-1$

This shows that AB $\perp$ AC. Thus $\Delta ABC$ is right angled at point A.