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

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Solution

Given: $k x + 3 k y + 1 = 0$ ....... $(1)$
Slope of line $(1) = - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f y} = - \frac{k}{3 k} = \frac{- 1}{3}$
Given: $x - 5 k y - 2 = 0$ ....... $(2)$
Slope of line $(2) = - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f y} = - \frac{1}{- 5 k} = \frac{1}{5 k}$
As the lines $(1)$ and $(2)$ are perpendicular, the product of their slopes $= - 1$
$\Rightarrow \frac{- 1}{3} \times \frac{1}{5 k} = - 1$
$\Rightarrow 15 k = 1$
$∴ k = \frac{1}{15}$

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