If the points (k, 2 - 2k),(1 - k, 2k) and (-k | vedclass

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(A) The points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is zero.The area of the triangle is

Vedclass · Accepted Answer

$1/2, -1$

Vedclass · Answer

(A) The points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is zero.The area of the triangle 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:$\frac{1}{2} |k(2k - (6 - 2k)) + (1 - k)((6 - 2k) - (2 - 2k)) + (-k - 4)((2 - 2k) - 2k)| = 0$Simplifying the terms inside the brackets:$k(4k - 6) + (1 - k)(4) + (-k - 4)(2 - 4k) = 0$$4k^2 - 6k + 4 - 4k + (-2k + 4k^2 - 8 + 16k) = 0$$4k^2 - 10k + 4 + 4k^2 + 14k - 8 = 0$$8k^2 + 4k - 4 = 0$Dividing by $4$:$2k^2 + k - 1 = 0$$(2k - 1)(k + 1) = 0$Thus,$k = 1/2$ or $k = -1$.

If the points $(k, 2 - 2k)$,$(1 - k, 2k)$ and $(-k - 4, 6 - 2k)$ are collinear,then the possible values of $k$ are

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