If the area of a triangle is 35 sq. units wit | 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: $	ext{Area} = \frac{1}{2} |x_1(y_2-y_3

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$12$,-$2$

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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: $	ext{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 35$.
Substituting the given vertices $(2, -6)$, $(5, 4)$, and $(k, 4)$ into the formula:
$\frac{1}{2} |2(4-4) + 5(4 - (-6)) + k(-6-4)| = 35$
$\frac{1}{2} |2(0) + 5(10) + k(-10)| = 35$
$|50 - 10k| = 70$
This gives two possible cases:
Case $1$: $50 - 10k = 70 \implies -10k = 20 \implies k = -2$.
Case $2$: $50 - 10k = -70 \implies -10k = -120 \implies k = 12$.
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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