If the points (k, 3), (2, k), and (-k, 3) are | vedclass

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(D) For three points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero.The area of the t

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$0, 3$

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(D) For three points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be 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 $(k, 3), (2, k),$ and $(-k, 3)$:$\frac{1}{2} |k(k - 3) + 2(3 - 3) + (-k)(3 - k)| = 0$$|k(k - 3) + 0 - k(3 - k)| = 0$$|k^2 - 3k - 3k + k^2| = 0$$|2k^2 - 6k| = 0$$2k(k - 3) = 0$Thus,$k = 0$ or $k = 3$.

If the points $(k, 3), (2, k),$ and $(-k, 3)$ are collinear,then the values of $k$ are

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