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(C) 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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$0, 8$

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(C) 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 $(-2, 0)$,$(0, 4)$,and $(0, k)$,and the area is $4$ sq. units.Substituting the values into the formula:$4 = \frac{1}{2} |-2(4 - k) + 0(k - 0) + 0(0 - 4)|$$4 = \frac{1}{2} |-8 + 2k|$$8 = |-8 + 2k|$This implies two cases:Case $1$: $-8 + 2k = 8 \implies 2k = 16 \implies k = 8$Case $2$: $-8 + 2k = -8 \implies 2k = 0 \implies k = 0$Thus,the possible values of $k$ are $0$ and $8$.

The area of a triangle is $4$ sq. units,and its vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$. Find the value of $k$.

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