Find the values of k if the points A(k+1, 2k) | vedclass

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(A) We know that if three points are collinear,the area of the triangle formed by these points is zero.Since the points $A(k+1, 2k)$,$B(3k, 2k+3)$,and

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$2, \frac{1}{2}$

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(A) We know that if three points are collinear,the area of the triangle formed by these points is zero.Since the points $A(k+1, 2k)$,$B(3k, 2k+3)$,and $C(5k-1, 5k)$ are collinear,the area of $	riangle ABC = 0$.The formula for the area of a triangle is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.Here,$x_1 = k+1, x_2 = 3k, x_3 = 5k-1$ and $y_1 = 2k, y_2 = 2k+3, y_3 = 5k$.Substituting these values:$\frac{1}{2} [(k+1)(2k+3 - 5k) + 3k(5k - 2k) + (5k-1)(2k - (2k+3))] = 0$$\frac{1}{2} [(k+1)(-3k+3) + 3k(3k) + (5k-1)(-3)] = 0$$(-3k^2 + 3k - 3k + 3) + 9k^2 - 15k + 3 = 0$$6k^2 - 15k + 6 = 0$Dividing by $3$,we get $2k^2 - 5k + 2 = 0$.Factoring the quadratic equation: $2k^2 - 4k - k + 2 = 0 \Rightarrow 2k(k-2) - 1(k-2) = 0$.$(k-2)(2k-1) = 0$.Thus,$k = 2$ or $k = \frac{1}{2}$.

Find the values of $k$ if the points $A(k+1, 2k)$,$B(3k, 2k+3)$ and $C(5k-1, 5k)$ are collinear.

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