A value of k such that the straight lines $y - 3 k x + 4 = 0$ and $(2 k - 1) x - (8 k - 1) y - 6 = 0$ are perpendicular is

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$\frac{1}{6}$

$- \frac{1}{6}$

answer is A.

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Detailed Solution

Given lines $3 k x - y - 4 = 0$

Let $a_{1} = 3 k, b_{1} = - 1$

And $(2 k - 1) x - (8 k - 1) y - 6 = 0$

$a_{2} = 2 k - 1, b_{2} = - (8 k - 1)$

The lines are perpendicular if $a_{1} a_{2} + b_{1} b_{2} = 0$

$\begin{array}{l} \Rightarrow 3 k (2 k - 1) + 1 (8 k - 1) = 0 \\ \Rightarrow 6 k^{2} - 3 k + 8 k - 1 = 0 \\ \Rightarrow 6 k^{2} + 5 k - 1 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 6 k^{2} + 6 k - k - 1 = 0 \\ \Rightarrow 6 k (k + 1) - 1 (k + 1) = 0 \\ \Rightarrow (k + 1) (6 k - 1) = 0 \\ ∴ k = - 1, \frac{1}{6} \end{array}$