Question
Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.
$\begin{aligned} & \text { Slope of } A B=\frac{y_2-y_1}{x_2-x_1}=\frac{5-4}{3-4}=-1 \\ & \text { Slope of } B C=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2} \\ & \text { Slope of } A C=\frac{-1-4}{-1-4}=1\end{aligned}$
Slope of AB x slope of AC = – 1 x 1 = – 1 ∴ side AB ⊥ side AC ∴ ∆ABC is a right angled triangle right angled at A. ∴ The given points are the vertices of a right angled triangle.
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