Find the area of the triangle whose vertices | vedclass

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(C) Given the vertices of the triangle are $(x_1, y_1) = (-8, 4)$,$(x_2, y_2) = (-6, 6)$,and $(x_3, y_3) = (-3, 9)$.The formula for the area of a tria

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$0$

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(C) Given the vertices of the triangle are $(x_1, y_1) = (-8, 4)$,$(x_2, y_2) = (-6, 6)$,and $(x_3, y_3) = (-3, 9)$.The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Substituting the given values into the formula:$	ext{Area} = \frac{1}{2} |-8(6 - 9) + (-6)(9 - 4) + (-3)(4 - 6)|$$= \frac{1}{2} |-8(-3) - 6(5) - 3(-2)|$$= \frac{1}{2} |24 - 30 + 6|$$= \frac{1}{2} |30 - 30| = \frac{1}{2}(0) = 0$Since the area is $0$,the points are collinear,and the area of the triangle formed by them is $0$.

Find the area of the triangle whose vertices are $(-8, 4), (-6, 6)$ and $(-3, 9).$

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