If the lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

Given,

Lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other. Then the product of their slopes is -1.

Converting 3x + by + 5 = 0 in the form y = mx + c.

⇒ by = -3x - 5

⇒ y = $-\dfrac{3}{\text{b}}\text{x} - \dfrac{5}{\text{b}}$.

Comparing with y = mx + c we get,

Slope of first line = m₁ = $-\dfrac{3}{\text{b}}$.

Converting ax - 5y + 7 = 0 in the form y = mx + c.

⇒ 5y = ax + 7

⇒ y = $\dfrac{\text{a}}{5}\text{x} + \dfrac{7}{5}$.

Comparing with y = mx + c we get,

Slope of second line = m₂ = $\dfrac{\text{a}}{5}$.

For perpendicular lines, product of their slopes is -1.

∴ m₁.m₂ = -1.

$\Rightarrow -\dfrac{3}{b} \times \dfrac{a}{5} = -1 \\[1em] \Rightarrow -\dfrac{3a}{5b} = -1 \\[1em] \Rightarrow -3a = -5b \\[1em] \Rightarrow 3a = 5b.$

Hence, the relation between a and b is given by 3a = 5b.