If (p + 1)x + y = 3 and 3y - (p - 1)x = 4 are perpendicular to each other, find the value of p.

Given 1st equation,

⇒ (p + 1)x + y = 3

⇒ y = -(p + 1)x + 3

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

⇒ Slope (m₁) = -(p + 1)

Given 2nd equation,

⇒ 3y - (p - 1)x = 4

⇒ 3y = (p - 1)x + 4

⇒ y = $\dfrac{(p - 1)x}{3} + \dfrac{4}{3}$

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

⇒ Slope (m₂) = $\dfrac{p - 1}{3}$

We know that,

Product of slopes of perpendicular lines = -1.

$\therefore -(p + 1) \times \dfrac{p - 1}{3} = -1 \\[1em] \Rightarrow \dfrac{-(p^2 - 1)}{3} = -1 \\[1em] \Rightarrow -(p^2 - 1) = -3 \\[1em] \Rightarrow p^2 - 1 = 3 \\[1em] \Rightarrow p^2 = 4 \\[1em] \Rightarrow p = \sqrt{4} = \pm 2.$

Hence, p = $\pm 2$.