The vertices of triangle $A B C$ are $A (5, 3), B (- 4, - 2)$ and $C (4, 4)$ . Find the length of the altitude from $A$ to $B C$ .

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Solution

$\begin{matrix} B C = \sqrt{{(- 4 - 4)}^{2} + {(- 2 - 4)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 ∴ a r (Δ A B C) = \frac{1}{2} (B C) (A D) = \frac{1}{2} (10) (A D) = 5 (A D) . . . . . . . . . . . (1) a r (Δ A B C) u sin g t h e t h r e e v e r t i c e s = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 5 & 3 & 1 - 4 & - 2 & 1 4 & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{2} [5 (- 2 - 4) + 4 (3 - 4) + 4 (3 + 2)] = \frac{1}{2} [- 30 - 4 + 20] = \frac{1}{2} [- 34 + 20] = 7........... (2) e q u a t i n g t h e a r e a s f r o m (1) & (2) 5 (A D) = 7 A D = \frac{7}{5} u n i t s = 1.4 u n i t s \end{matrix}$

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