If the straight lines 3x – 5y = 7 and 4x + ay + 9 = 0 are perpendicular to each other, find the value of a.

Converting 3x - 5y + 7 = 0 in the form y = mx + c we get,

⇒ 3x - 5y + 7 = 0

⇒ 5y = 3x + 7

⇒ y = $\dfrac{3}{5}x + \dfrac{7}{5}$

Comparing, we get slope of first line = m₁ = $\dfrac{3}{5}$

Converting 4x + ay + 9 = 0 in the form y = mx + c we get,

⇒ 4x + ay + 9 = 0

⇒ ay = -4x - 9

⇒ y = $-\dfrac{4}{a}x - \dfrac{9}{a}$

Comparing, we get slope of second line = m₂ = $-\dfrac{4}{a}$

Given, two lines are perpendicular so product of their slopes will be equal to -1,

$\Rightarrow m$1 \cdot m2 = -1 \\[1em] \Rightarrow \dfrac{3}{5} \times -\dfrac{4}{a} = -1 \\[1em] \Rightarrow -\dfrac{12}{5a} = -1 \\[1em] \Rightarrow a = \dfrac{12}{5}.

Hence,the value of a = $\dfrac{12}{5}$.