Find the value of k for which the lines kx + 2y + 3 = 0 and 8x + ky – 1 = 0 are parallel.

Since, the lines are parallel they have same gradient.

Given, kx + 2y + 3 = 0

Converting kx + 2y + 3 = 0 in the form y = mx + c, we get :

⇒ 2y = -kx - 3

⇒ y = $-\dfrac{k}{2}x - \dfrac{3}{2}$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with y = $-\dfrac{k}{2}x - \dfrac{3}{2}$, we get :

⇒ m₁ = $-\dfrac{k}{2}$

Given,

8x + ky - 1 = 0

Converting 8x + ky - 1 = 0 in the form y = mx + c we get,

⇒ ky = -8x + 1

⇒ y = $\dfrac{-8x}{k} + \dfrac{1}{k}$

Comparing y = mx + c with y = $\dfrac{-8x}{k} + \dfrac{1}{k}$, we get :

⇒ m₂ = $-\dfrac{8}{k}$

Since, lines are parallel, equating the gradients :

$\Rightarrow -\dfrac{k}{2} = -\dfrac{8}{k} \\[1em] \Rightarrow k(k) = 8(2) \\[1em] \Rightarrow k^2 = 16 \\[1em] \Rightarrow k = \sqrt{16} \\[1em] \Rightarrow k = \pm 4.$

Hence, value of k = ± 4.