Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (-1, -2).

Given equation of line,

⇒ 3x + 8y = 12,

Converting it in the form y = mx + c,

⇒ 8y = -3x + 12

⇒ y = $-\dfrac{3}{8}x + \dfrac{12}{8}$

Comparing with y = mx + c we get,

Slope (m₁) = $-\dfrac{3}{8}$.

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

∴ m₁ × m₂ = -1.

$\Rightarrow -\dfrac{3}{8} \times m$2 = -1 \\[1em] \Rightarrow m2 = \dfrac{8}{3}.

The equation of the line having slope = $\dfrac{8}{3}$ and passing through the point (-1, -2) will be

y - y₁ = m(x - x₁)

$\Rightarrow y - (-2) = \dfrac{8}{3}(x - (-1)) \\[1em] \Rightarrow 3(y + 2) = 8(x + 1) \\[1em] \Rightarrow 3y + 6 = 8x + 8 \\[1em] \Rightarrow 8x - 3y + 8 - 6 = 0 \\[1em] \Rightarrow 8x - 3y + 2 = 0.$

Hence, the equation of the line is 8x - 3y + 2 = 0.