Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.

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Solution

If the points (k, 2 − 2 k), (− k + 1, 2k) and (− 4 − k, 6 − 2k) are collinear, then

$∆ = |\begin{matrix} k & 2 - 2 k & 1 \\ - k + 1 & 2 k & 1 \\ - 4 - k & 6 - 2 k & 1 \end{matrix}| = 0 \Rightarrow |\begin{matrix} k & 2 - 2 k & 1 \\ - 2 k + 1 & 4 k - 2 & 0 \\ - 4 - k & 6 - 2 k & 1 \end{matrix}| = 0 [Applying R_{2} \to R_{2} - R_{1}] \Rightarrow |\begin{matrix} k & 2 - 2 k & 1 \\ - 2 k + 1 & 4 k - 2 & 0 \\ - 4 - 2 k & 4 & 0 \end{matrix}| = 0 [Applying R_{3} \to R_{3} - R_{1}] \Rightarrow |\begin{matrix} - 2 k + 1 & 4 k - 2 \\ - 4 - 2 k & 4 \end{matrix}| = 0 \Rightarrow - 8 k + 4 + 16 k - 8 + 8 k^{2} - 4 k = 0 \Rightarrow 8 k^{2} + 4 k - 4 = 0 \Rightarrow (8 k - 4) (k + 1) = 0 \Rightarrow k = - 1 or k = \frac{1}{2}$

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Similar questions

For what value of k are the points $(k, 2 - 2 k), (- k + 1, 2 k), (- 4 - k, 6 - 2 k)$ collinear?

(k, 2 - 2 k)

(- k + 1, 2 k)

(- 4 - k, 6 - 2 k)

(k, 2 - 2 k), (- k + 1, 2 k), (- 4 - k, 6 - 2 k)

k =

k

(k, 2 - 2 k)

(1 - k, 2 k)

(- 4 - k, 6 - 2 k)

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