A (h, k), B (1, 1) and C (2, 1) are the verti | 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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$k = 3$ or $k = -1$

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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 $A(h, k), B(1, 1),$ and $C(2, 1)$.Area $= \frac{1}{2} |h(1 - 1) + 1(1 - k) + 2(k - 1)| = 1$$\frac{1}{2} |0 + 1 - k + 2k - 2| = 1$$\frac{1}{2} |k - 1| = 1$$|k - 1| = 2$This implies $k - 1 = 2$ or $k - 1 = -2$.If $k - 1 = 2$,then $k = 3$.If $k - 1 = -2$,then $k = -1$.Therefore,the possible values of $k$ are $3$ or $-1$.

$A (h, k), B (1, 1)$ and $C (2, 1)$ are the vertices of $\Delta ABC$. If the area of $\Delta ABC$ is $1$,then find the possible values of $k$.

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