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  4. If the points (k, 2-2 k),(1-k, 2 k) and (-k-4,6-2 k) are collinear, then possible values of k are

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Q. If the points $(k, 2-2 k),(1-k, 2 k)$ and $(-k-4,6-2 k)$ are collinear, then possible values of $k$ are

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Solution:

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Since A, B, C are collinear
Slope of $A B=$ Slope of $B C$
$\frac{2-2 k -2 k }{ k -1+ k }=\frac{2 k -6+2 k }{1- k + k +4} $
$\Rightarrow \frac{2-4 k }{2 k -1}=\frac{4 k -6}{5}$
$\Rightarrow 10-20 k =(4 k -6)(2 k -1) $
$\Rightarrow (4 k -6)(2 k -1)+10(2 k -1)=0 $
$\Rightarrow k =1 / 2,-1$