Find the values of k if the area of the trian | vedclass

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(A) The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is given by the formula:$\Delta = \frac{1}{2} |x_{1}(y_{2} -

Vedclass · Accepted Answer

$0, 8$

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(A) The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ 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 the area is $4$ square units,we have $\Delta = \pm 4$.Substituting the vertices $(-2, 0), (0, 4), (0, k)$ into the formula:$4 = \frac{1}{2} |-2(4 - k) + 0(k - 0) + 0(0 - 4)|$$4 = \frac{1}{2} |-8 + 2k|$$8 = |-8 + 2k|$This implies two cases:Case $1$: $-8 + 2k = 8 \implies 2k = 16 \implies k = 8$Case $2$: $-8 + 2k = -8 \implies 2k = 0 \implies k = 0$Thus,the values of $k$ are $0$ and $8$.

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

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