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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 $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1

Vedclass · Accepted Answer

$-6$

Vedclass · Answer

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 4$. Substituting the given vertices $(x, 0), (1, 1), (0, 2)$: $\frac{1}{2} |x(1 - 2) + 1(2 - 0) + 0(0 - 1)| = 4$ $\frac{1}{2} |-x + 2| = 4$ $|-x + 2| = 8$ This gives two cases: Case $1$: $-x + 2 = 8$ $\Rightarrow -x = 6$ $\Rightarrow x = -6$ Case $2$: $-x + 2 = -8$ $\Rightarrow -x = -10$ $\Rightarrow x = 10$ Comparing with the given options,the value $x = -6$ is present.

If the area of the triangle with vertices $(x, 0), (1, 1)$ and $(0, 2)$ is $4$ square units,then a value of $x$ is

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