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Question
Find the area of the triangle with vertice at the point:
(3, 8), (−4, 2) and (5, −1)
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Solution
\[∆ = \frac{1}{2}\begin{vmatrix}3 & 8 & 1 \\ - 4 & 2 & 1 \\ 5 & - 1 & 1\end{vmatrix} \]
\[ ∆ = \frac{1}{2}\begin{vmatrix}3 & 8 & 1 \\ - 7 & - 6 & 0 \\ 5 & - 1 & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ ∆ = \frac{1}{2}\begin{vmatrix}3 & 8 & 1 \\ - 7 & - 6 & 0 \\ 2 & - 9 & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\]
\[ ∆ = \frac{1}{2}\begin{vmatrix}- 7 & - 6 \\ 2 & - 9\end{vmatrix}\]
\[ ∆ = \frac{1}{2}\left( 63 + 12 \right)\]
\[ ∆ = \frac{1}{2}\left( 75 \right) = \frac{75}{2}\text{square units }\]
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