Find the value of k if the points A(2, 3),B(4 | vedclass

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(B) Since the given points $A(2, 3)$,$B(4, k)$,and $C(6, -3)$ are collinear,the area of the triangle formed by these points must be $0$.The formula fo

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$0$

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(B) Since the given points $A(2, 3)$,$B(4, k)$,and $C(6, -3)$ are collinear,the area of the triangle formed by these points must be $0$.The formula for the area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$Substituting the given coordinates:$\frac{1}{2} |2(k - (-3)) + 4(-3 - 3) + 6(3 - k)| = 0$$\frac{1}{2} |2(k + 3) + 4(-6) + 6(3 - k)| = 0$$\frac{1}{2} |2k + 6 - 24 + 18 - 6k| = 0$$\frac{1}{2} |-4k| = 0$$-2k = 0$Therefore,$k = 0$.

Find the value of $k$ if the points $A(2, 3)$,$B(4, k)$,and $C(6, -3)$ are collinear.

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