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

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$12, -2$

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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:$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $(2, -6), (5, 4)$ and $(k, 4)$,and the area is $35$ $sq$ $units$.Substituting the values:$35 = \frac{1}{2} |2(4 - 4) + 5(4 - (-6)) + k(-6 - 4)|$$35 = \frac{1}{2} |2(0) + 5(10) + k(-10)|$$35 = \frac{1}{2} |50 - 10k|$$70 = |50 - 10k|$This implies:$50 - 10k = 70$ or $50 - 10k = -70$Case $1$: $50 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2$Case $2$: $50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12$Thus,the possible values for $k$ are $12$ and $-2$.

If the area of a triangle is $35$ $sq$ $units$ with vertices $(2,-6), (5,4)$ and $(k, 4)$,then $k$ is:

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