Find the value of $k$ so that the line $k x + y - 3 = 0$ may be parallel to $2 x - 3 y + 4 = 0$

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Solution

Given: $k x + y - 3 = 0$ ....... $(1)$
Slope of line $(1) = - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f y} = - \frac{k}{1} = - k$
Given: $2 x - 3 y + 4 = 0$ ....... $(2)$
Slope of line $(2) = - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f y} = - \frac{2}{- 3} = \frac{2}{3}$
As the lines $(1)$ and $(2)$ are parallel, $k = \frac{2}{3}$
$∴ k = \frac{2}{3}$

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