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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the determinant formula:$\Delta = \frac{1}{2} \left| \begin{

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$15$ square units

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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the determinant formula:$\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} ight|$Substituting the given vertices $(-2, -3), (3, 2), (-1, -8)$:$\Delta = \frac{1}{2} \left| \begin{array}{ccc} -2 & -3 & 1 \ 3 & 2 & 1 \ -1 & -8 & 1 \end{array} ight|$Expanding along the first row:$\Delta = \frac{1}{2} [ -2(2 - (-8)) - (-3)(3 - (-1)) + 1(-24 - (-2)) ]$$\Delta = \frac{1}{2} [ -2(10) + 3(4) + 1(-22) ]$$\Delta = \frac{1}{2} [ -20 + 12 - 22 ]$$\Delta = \frac{1}{2} [ -30 ] = -15$Since the area cannot be negative,we take the absolute value:$	ext{Area} = |-15| = 15$ square units.

Find the area of the triangle with vertices at the points $(-2, -3), (3, 2), (-1, -8)$.

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