Find the value of k for which the points (8,1 | vedclass

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(C) For points to be collinear,the area of the triangle formed by them must be $0$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For points to be collinear,the area of the triangle formed by them must be $0$.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 $(8,1), (k,-4),$ and $(2,-5)$ into the formula:$\frac{1}{2} |8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4))| = 0$$\frac{1}{2} |8(-4 + 5) + k(-6) + 2(1 + 4)| = 0$$\frac{1}{2} |8(1) - 6k + 2(5)| = 0$$|8 - 6k + 10| = 0$$18 - 6k = 0$$6k = 18$$k = 3$

Find the value of $k$ for which the points $(8,1), (k,-4),$ and $(2,-5)$ are collinear.

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