If y + (2p + 1)x + 3 = 0 and 8y - (2p - 1)x = 5 are mutually perpendicular, find the value of p.

Given,

⇒ y + (2p + 1)x + 3 = 0

⇒ y = -(2p + 1)x - 3

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

⇒ Slope (m₁) = -(2p + 1)

Given,

⇒ 8y - (2p - 1)x = 5

⇒ 8y = (2p - 1)x + 5

⇒ y = $\dfrac{2p - 1}{8} + \dfrac{5}{8}$

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

⇒ Slope (m₂) = $\dfrac{2p - 1}{8}$.

We know that,

Product of slopes of perpendicular lines = -1.

$\therefore -(2p + 1) \times \dfrac{2p - 1}{8} = -1 \\[1em] \Rightarrow \dfrac{-(4p^2 - 1)}{8} = -1 \\[1em] \Rightarrow -(4p^2 - 1) = -8 \\[1em] \Rightarrow 4p^2 - 1 = 8 \\[1em] \Rightarrow 4p^2 = 9 \\[1em] \Rightarrow p^2 = \dfrac{9}{4} \\[1em] \Rightarrow p = \sqrt{\dfrac{9}{4}} = \pm \dfrac{3}{2} = \pm 1\dfrac{1}{2}.$

Hence, p = $\pm 1\dfrac{1}{2}$.