Find the equation of the line through the point (-1, -2) and perpendicular to the line 3x + 4y = 5.

Given,

Equation : 3x + 4y = 5

⇒ 4y = -3x + 5

⇒ y = $-\dfrac{3}{4}x + \dfrac{5}{4}$.

Comparing above equation with y = mx + c, we get :

(Slope) m = $-\dfrac{3}{4}$

Let slope of line perpendicular to above line be m₁.

We know that,

Product of slope of perpendicular lines is -1.

$\therefore m$1 \times -\dfrac{3}{4} = -1 \\[1em] \Rightarrow m1 = \dfrac{-4}{-3} = \dfrac{4}{3}.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get :

Equation of line passing through (-1, -2) and slope $\dfrac{4}{3}$ is

$\Rightarrow y - (-2) = \dfrac{4}{3}[x - (-1)]$

⇒ 3(y + 2) = 4(x + 1)

⇒ 3y + 6 = 4x + 4

⇒ 3y = 4x + 4 - 6

⇒ 3y = 4x - 2.

Hence, equation of the line is 3y = 4x - 2.