The area of triangle PQR with the vertices P( | vedclass

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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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$-1, 3$

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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 $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$,and the area is $3$ sq. units. Substituting the values: $3 = \frac{1}{2} |k(4 - 1) + 2(1 - 1) + 1(1 - 4)|$ $6 = |3k + 0 - 3|$ $6 = |3k - 3|$ This implies two cases: Case $1$: $3k - 3 = 6 \implies 3k = 9 \implies k = 3$. Case $2$: $3k - 3 = -6 \implies 3k = -3 \implies k = -1$. Thus,$k = -1$ or $k = 3$.

The area of $\triangle PQR$ with the vertices $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$ is $3$ sq. units. Then,$k = $ . . . . . . .

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