Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:

(i) perpendicular to the line 2x – y + 7 = 0

(ii) parallel to it.

Given line equation,

⇒ 2x - y + 7 = 0 ……….(1)

⇒ y = 2x + 7

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

Slope (m₁) = 2

Given line equation,

⇒ (k - 2)x + (k + 3)y - 5 = 0 ……….(2)

⇒ (k + 3)y = -(k - 2)x + 5

⇒ y = $-\dfrac{k - 2}{k + 3}x + \dfrac{5}{k + 3}$

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

Slope (m₂) = $-\dfrac{k - 2}{k + 3}$

(i) If lines (1) and (2) are perpendicular then product of their slopes = -1.

⇒ m₁ x m₂ = -1

$\Rightarrow 2 \times -\dfrac{k - 2}{k + 3} = -1 \\[1em] \Rightarrow -2(k - 2) = -1(k + 3) \\[1em] \Rightarrow -2k + 4 = -k - 3 \\[1em] \Rightarrow -k + 2k = 4 + 3 \\[1em] \Rightarrow k = 7.$

Hence, k = 7.

(ii) If lines 1 and 2 are parallel then their slopes will be equal.

⇒ $-\dfrac{k - 2}{k + 3}$ = 2

⇒ -(k - 2) = 2(k + 3)

⇒ -k + 2 = 2k + 6

⇒ 2k + k = 2 - 6

⇒ 3k = -4

⇒ k = $-\dfrac{4}{3}$.

Hence, k = $-\dfrac{4}{3}$.