Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4.

Given,

⇒ x - 3y = 4

⇒ 3y = x - 4

⇒ y = $\dfrac{1}{3}x - \dfrac{4}{3}$

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

Slope = $\dfrac{1}{3}$

Since, parallel lines have equal slope.

∴ Slope of line parallel to x - 3y = 4 is $\dfrac{1}{3}$.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = $\dfrac{1}{3}(x - 5)$

⇒ 3(y + 3) = x - 5

⇒ 3y + 9 = x - 5

⇒ x - 3y - 5 - 9 = 0

⇒ x - 3y - 14 = 0.

Hence, equation of the line passing through (5, -3) and parallel to x - 3y = 4 is x - 3y - 14 = 0.