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If $(k, 2 - 2 k), (- k + 1, 2 k), (- 4 - k, 6 - 2 k)$ are collinear, then $k =$

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\frac{1}{2}

answer is C.

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Detailed Solution

A (k, 2 - 2 k), B (- k + 1, 2 k), C (- 4 - k, 6 - 2 k)

Slope of AB $=$ Slope of BC

$\frac{2 k - 2 + 2 k}{- k + 1 - k} = \frac{6 - 2 k - 2 k}{- 4 - k + k - 1}$

$\frac{4 k - 2}{- 2 k + 1} = \frac{6 - 4 k}{- 5}$

$2 k^{2} + k - 1 = 0$

$2 k (k + 1) - 1 (k + 1) = 0$

$(k + 1) (2 k - 1) = 0$

$k = - 1 (o r) k = \frac{1}{2}$

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