The area of a triangle with vertices A(k, 1), | 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: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2

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$5$ and $-3$

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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: 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 $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$,and the area is $6$ sq. units. Substituting the values: $6 = \frac{1}{2} |k(4 - 1) + 2(1 - 1) + 1(1 - 4)|$. $12 = |3k + 0 - 3|$. $12 = |3k - 3|$. This gives two cases: Case $1$: $3k - 3 = 12 \implies 3k = 15 \implies k = 5$. Case $2$: $3k - 3 = -12 \implies 3k = -9 \implies k = -3$. Thus,the values of $k$ are $5$ and $-3$.

The area of a triangle with vertices $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$ is $6$ sq. units. Find the value of $k$.

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