Find the value of $k$ so that the line $2 k x + 5 y - 6 = 0$ may be parallel to $x + 3 k y - 1 = 0$ .

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Solution

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

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