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 - y_3) + x_2(y_3

Vedclass · Accepted Answer

$0, 8$

Vedclass · Answer

(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 $(k, 0), (4, 0), (0, 2)$ into the formula:$4 = \frac{1}{2} |k(0 - 2) + 4(2 - 0) + 0(0 - 0)|$$4 = \frac{1}{2} |-2k + 8|$$8 = |-2k + 8|$This implies $-2k + 8 = 8$ or $-2k + 8 = -8$.Case $1$: $-2k + 8 = 8 \implies -2k = 0 \implies k = 0$.Case $2$: $-2k + 8 = -8 \implies -2k = -16 \implies k = 8$.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 $(k, 0), (4, 0), (0, 2)$.

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