Write down the equation of the line passing through (-3, 2) and perpendicular to the line 3y = 5 - x.

Given equation of the line is,

⇒ 3y = 5 - x

⇒ y = $-\dfrac{1}{3}x + \dfrac{5}{3}.$

Comparing with y = mx + c we get,

Slope (m₁) = $-\dfrac{1}{3}$.

Let slope of the line perpendicular to the given line be m₂.

∴ m₁ × m₂ = -1.

$\Rightarrow -\dfrac{1}{3} \times m$2 = -1 \\[1em] \Rightarrow m2 = 3.

Equation of the line having slope 3 and passing through (-3, 2) can be given by point-slope form i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 2 = 3(x - (-3)) \\[1em] \Rightarrow y - 2 = 3(x + 3) \\[1em] \Rightarrow y - 2 = 3x + 9 \\[1em] \Rightarrow 3x - y + 11 = 0.

Hence, the equation of the required line is 3x - y + 11 = 0.