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Mathematics

The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.

Straight Line Eq

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Answer

Let points be A(K, 3), B(2, -4) and C(-K + 1, -2).

Since, points are collinear.

∴ Slope of AB = Slope of BC

432K=2(4)K+1272K=2K17(K1)=2(2K)7K+7=42K7K+2K=479K=3K=39=13.\Rightarrow \dfrac{-4 - 3}{2 - K} = \dfrac{-2 - (-4)}{-K + 1 - 2} \\[1em] \Rightarrow \dfrac{-7}{2 - K} = \dfrac{2}{-K - 1} \\[1em] \Rightarrow -7(-K - 1) = 2(2 - K) \\[1em] \Rightarrow 7K + 7 = 4 - 2K \\[1em] \Rightarrow 7K + 2K = 4 - 7 \\[1em] \Rightarrow 9K = -3 \\[1em] \Rightarrow K = -\dfrac{3}{9} = -\dfrac{1}{3}.

Hence, K = 13-\dfrac{1}{3}.

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