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

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$\frac{47}{2}$ square units

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(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ 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 $(2,7), (1,1),$ and $(10,8)$:$\Delta = \frac{1}{2} \left| \begin{array}{ccc} 2 & 7 & 1 \ 1 & 1 & 1 \ 10 & 8 & 1 \end{array} ight|$Expanding the determinant along the first row:$\Delta = \frac{1}{2} [2(1 	imes 1 - 8 	imes 1) - 7(1 	imes 1 - 10 	imes 1) + 1(1 	imes 8 - 10 	imes 1)]$$\Delta = \frac{1}{2} [2(1 - 8) - 7(1 - 10) + 1(8 - 10)]$$\Delta = \frac{1}{2} [2(-7) - 7(-9) + 1(-2)]$$\Delta = \frac{1}{2} [-14 + 63 - 2]$$\Delta = \frac{1}{2} [47] = \frac{47}{2}$ square units.

Find the area of the triangle with vertices at the points $(2,7), (1,1), (10,8)$.

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