The area of a triangle whose vertices are (2, | vedclass

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(A) 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: $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3

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$12, -2$

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(A) 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: $	ext{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 $(2, -6)$,$(5, 4)$,and $(k, 4)$ with $	ext{Area} = 35$. Substituting the values: $35 = \frac{1}{2} |2(4 - 4) + 5(4 - (-6)) + k(-6 - 4)|$ $70 = |2(0) + 5(10) + k(-10)|$ $70 = |50 - 10k|$ This implies two cases: $1) 50 - 10k = 70 \implies -10k = 20 \implies k = -2$ $2) 50 - 10k = -70 \implies -10k = -120 \implies k = 12$ Thus,the values of $k$ are $12$ and $-2$.

The area of a triangle whose vertices are $(2, -6)$,$(5, 4)$,and $(k, 4)$ is $35$ sq. units. Then,the value of $k$ is . . . . . . .

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