Find the value of $k$ so that the line $k x - 6 y - 3 = 0$ may be parallel to $x - k y + 5 = 0$ .

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Solution

Given: $k x - 6 y - 3 = 0$ ....... $(1)$
Slope of line $(1) = - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f y} = - \frac{k}{6}$
Given: $x - k y + 5 = 0$ ....... $(2)$
Slope of line $(2) = - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f y} = - \frac{1}{- k}$
As the lines $(1)$ and $(2)$ are parallel, $- \frac{k}{6} = - \frac{1}{- k}$
$\Rightarrow k^{2} = 6$
$∴ k = \pm \sqrt{6}$

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