For what value of k are the points (k, 2 - 2k | vedclass

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(C) Three points are collinear if the area of the triangle formed by them is $0$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(

Vedclass · Accepted Answer

$-1, 1/2$

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(C) Three points are collinear if the area of the triangle formed by them is $0$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.Substituting the given points:$\frac{1}{2} |k(2k - (6 - 2k)) + (1 - k)((6 - 2k) - (2 - 2k)) + (-4 - k)((2 - 2k) - 2k)| = 0$Simplifying the terms inside:$k(4k - 6) + (1 - k)(4) + (-4 - k)(2 - 4k) = 0$$4k^2 - 6k + 4 - 4k - 8 + 16k - 2k + 4k^2 = 0$$8k^2 + 4k - 4 = 0$$2k^2 + k - 1 = 0$Factoring the quadratic equation:$(2k - 1)(k + 1) = 0$Thus,$k = 1/2$ or $k = -1$.

For what value of $k$ are the points $(k, 2 - 2k)$,$(1 - k, 2k)$,and $(-4 - k, 6 - 2k)$ collinear?

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