A(8, 2),B(2, 4),and C(6, 7) are the vertices | vedclass

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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 formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(

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$13$

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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 formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $A(8, 2)$,$B(2, 4)$,and $C(6, 7)$.Here,$x_1 = 8, y_1 = 2, x_2 = 2, y_2 = 4, x_3 = 6, y_3 = 7$.Substituting these values into the formula:Area $= \frac{1}{2} |8(4 - 7) + 2(7 - 2) + 6(2 - 4)|$Area $= \frac{1}{2} |8(-3) + 2(5) + 6(-2)|$Area $= \frac{1}{2} |-24 + 10 - 12|$Area $= \frac{1}{2} |-26|$Area $= \frac{26}{2} = 13$.Therefore,the area of $\Delta ABC$ is $13$ square units.

$A(8, 2)$,$B(2, 4)$,and $C(6, 7)$ are the vertices of $\Delta ABC$. Find the area of $\Delta ABC$.

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