Find the equation of the line that has x-intercept –3 and is perpendicular to the line 3x + 5y = 1.

Let point where line touches x-axis be A. So, A = (-3, 0)

Given equation of line,

⇒ 3x + 5y = 1

⇒ 5y = -3x + 1

⇒ y = $-\dfrac{3}{5}x + \dfrac{1}{5}$

Comparing above equation with y = mx + c we get, m₁ = $-\dfrac{3}{5}$

Let slope of line perpendicular to 3x + 5y = 1 be m₂

⇒ m₁ × m₂ = -1

⇒ $-\dfrac{3}{5} \times m_2 = -1$

⇒ $m_2 = \dfrac{5}{3}$

By point-slope form,

Equation of line with slope = $\dfrac{5}{3}$and passing through (-3, 0) is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{5}{3}$[x - (-3)]

⇒ 3y = 5[x + 3]

⇒ 3y = 5x + 15

⇒ 5x - 3y + 15 = 0.

Hence, equation of required line is 5x - 3y + 15 = 0.