Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.

Given equation of line,

⇒ 3x + 5y + 15 = 0

Converting it in the form y = mx + c,

⇒ 5y = -3x - 15

⇒ y = $-\dfrac{3}{5}x - 3$

Comparing with y = mx + c we get,

Slope = $-\dfrac{3}{5}$

The slope of the line parallel to the given line will also be $-\dfrac{3}{5}$.

Given, the new line has slope = $-\dfrac{3}{5}$ and passes through (0, 4).

So, equation can be given by,

⇒ y - y₁ = m(x - x₁)

$\Rightarrow y - 4 = -\dfrac{3}{5}(x - 0) \\[1em] \Rightarrow 5(y - 4) = -3x \\[1em] \Rightarrow 5y - 20 = -3x \\[1em] \Rightarrow 3x + 5y - 20 = 0,$

Hence, the equation of the line is 3x + 5y - 20 = 0.