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Question
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
\[∆ = \begin{vmatrix}k & 2 - 2k & 1 \\ - k + 1 & 2k & 1 \\ - 4 - k & 6 - 2k & 1\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}k & 2 - 2k & 1 \\ - 2k + 1 & 4k - 2 & 0 \\ - 4 - k & 6 - 2k & 1\end{vmatrix} = 0 \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ \Rightarrow \begin{vmatrix}k & 2 - 2k & 1 \\ - 2k + 1 & 4k - 2 & 0 \\ - 4 - 2k & 4 & 0\end{vmatrix} = 0 \left[\text{ Applying } R_3 \to R_3 - R_1 \right]\]
\[ \Rightarrow \begin{vmatrix}- 2k + 1 & 4k - 2 \\ - 4 - 2k & 4\end{vmatrix} = 0\]
\[ \Rightarrow - 8k + 4 + 16k - 8 + 8 k^2 - 4k = 0 \]
\[ \Rightarrow 8 k^2 + 4k - 4 = 0\]
\[ \Rightarrow \left( 8k - 4 \right)\left( k + 1 \right) = 0\]
\[ \Rightarrow k = - 1\text{ or }k = \frac{1}{2}\]
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