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Question
Find the area of the triangle with vertice at the point:
(0, 0), (6, 0) and (4, 3)
Solution
\[∆ = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{vmatrix} \]
\[ ∆ = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ ∆ = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\]
\[ ∆ = \frac{1}{2}\begin{vmatrix}6 & 0 \\ 4 & 3\end{vmatrix}\]
\[ ∆ = \frac{1}{2}\left( 18 - 0 \right)\]
\[ ∆ = \frac{1}{2}\left( 18 \right) = 9\text{ square units }\]
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