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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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प्रश्न

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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उत्तर

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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अध्याय 5: Determinants - Exercise 6.3 [पृष्ठ ७१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.3 | Q 9 | पृष्ठ ७१

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