Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Given,

⇒ 4x + 5y = 6

⇒ 5y = -4x + 6

⇒ y = $-\dfrac{4}{5}x + \dfrac{6}{5}$

Comparing above equations with y = mx + c we get,

Slope (m₁) = $-\dfrac{4}{5}$

Let slope of line perpendicular to 4x + 5y = 6 be m₂.

Since, product of slopes of perpendicular lines be -1,

$\Rightarrow m$1 \times m2 = -1 \\[1em] \Rightarrow -\dfrac{4}{5} \times m2 = -1 \\[1em] \Rightarrow m2 = -1 \times -\dfrac{5}{4} = \dfrac{5}{4}. \\[1em]

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $\dfrac{5}{4}$[x - (-2)]

⇒ 4(y - 1) = 5(x + 2)

⇒ 4y - 4 = 5x + 10

⇒ 5x - 4y + 10 + 4 = 0

⇒ 5x - 4y + 14 = 0.

Hence, equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6 is 5x - 4y + 14 = 0.