Prove that the lines 2x + 3y + 8 = 0 and 27x – 18y + 10 = 0 are perpendicular to each other.

Converting 2x + 3y + 8 = 0 in the form y = mx + c we get,

⇒ 3y = -2x - 8

⇒ y = $-\dfrac{2}{3}x - \dfrac{8}{3}$

Comparing, we get slope of this line : m₁ = $-\dfrac{2}{3}$

Converting 27x – 18y + 10 = 0 in the form y = mx + c we get,

⇒ -18y = -27x - 10

⇒ y = $\dfrac{-27x}{-18} - \dfrac{10}{-18}$

⇒ y = $\dfrac{3x}{2} + \dfrac{5}{9}$

Comparing, we get slope of this line : m₂ = $\dfrac{3}{2}$

Product of Gradients

m₁ × m₂ = $-\dfrac{2}{3} \times \dfrac{3}{2}$

= $-\dfrac{6}{6}$

= -1

Since the product of the gradients is -1. The lines are perpendicular to each other.

Hence, proved that lines 2x + 3y + 8 = 0 and 27x – 18y + 10 = 0 are perpendicular to each other.