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Calculate the area of triangle ABC, if the vertices are A(2 ; 3), B(-3 ; -1),  and C(6 ; -2)
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Workings:

We shall use the determinant.

$A(2_{x_1},3_{y_1}), B(-3_{x_2},-1_{y_2}), C(6_{x_3},-2_{y_3})$

Area = $\frac{1}{2}\begin{vmatrix} (x_1 - x_2) & (x_1 - x_3) \\ (y_1 - y_2) & (y_1 - y_3) \end{vmatrix} = \begin{vmatrix} (2- (-3)) & (2- 6) \\ (3 - (-1)) & (3 - (-2)) \end{vmatrix} = \begin{vmatrix} 5 & -4 \\ 4 & 5 \end{vmatrix}$

The determinant of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\det(A) = ad -bc$

$\therefore \det\left(\begin{vmatrix} 5 & -4 \\ 4 & 5 \end{vmatrix}\right) = 5(5) - (-4\times4) =25 + 16 = 41$

$\therefore Area = \frac{1}{2} \times 41 = 20,5$ square units
by Wooden (166 points)
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