Find the value of 'p' if the lines 5x - 3y + 2 = 0 and 6x - py + 7 = 0 are perpendicular to each other. Hence find the equation of a line passing through (-2, -1) and parallel to 6x - py + 7 = 0.

Given lines,

⇒ 5x - 3y + 2 = 0 and 6x - py + 7 = 0

⇒ 3y = 5x + 2 and py = 6x + 7

⇒ y = $\dfrac{5}{3}x + \dfrac{2}{3} \text{ and } y = \dfrac{6}{p}x + \dfrac{7}{p}$

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

Slope of 1st line = $\dfrac{5}{3}$

Slope of 2nd line = $\dfrac{6}{p}$

Since, product of slopes of perpendicular lines = -1.

$\therefore \dfrac{5}{3} \times \dfrac{6}{p} = -1 \\[1em] \Rightarrow 5 \times \dfrac{2}{p} = -1 \\[1em] \Rightarrow \dfrac{10}{p} = -1 \\[1em] \Rightarrow p = -10 \\[1em]$

Given,

⇒ 6x - py + 7 = 0

⇒ 6x - (-10)y + 7 = 0

⇒ 6x + 10y + 7 = 0

⇒ 10y = -6x - 7

⇒ y = $-\dfrac{6}{10}x - \dfrac{7}{10}$

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

Slope = $-\dfrac{6}{10}$

Since, parallel lines have equal slope.

∴ Slope of line parallel to line 6x + 10y + 7 = 0 is $-\dfrac{6}{10}$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

Equation of line passing through (-2, -1) and slope $-\dfrac{6}{10}$ is

$\Rightarrow y - (-1) = -\dfrac{6}{10}[x - (-2)] \\[1em] \Rightarrow 10(y + 1) = -6(x + 2) \\[1em] \Rightarrow 10y + 10 = -6x - 12 \\[1em] \Rightarrow 10y + 10 + 6x + 12 = 0 \\[1em] \Rightarrow 6x + 10y + 22 = 0 \\[1em] \Rightarrow 2(3x + 5y + 11) = 0 \\[1em] \Rightarrow 3x + 5y + 11 = 0$

Hence, p = -10 and the equation of line is 3x + 5y + 11 = 0.