Given line is $\frac{x}{3}+\frac{y}{4}$ = 1 ⇒ 4x + 3y – 12 = 0 ... (i) Let a point on y-axis be (0, k) If its distance from line (i) is 4 units, then $\left|\frac{4\times0+3k-12}{\sqrt{4^{2}+3^{2}}}\right|$ = 4 ⇒ |3k – 12| = 4 × 5 ⇒ 3k – 12 = ± 20 ⇒ 3k = 12 ± 20 ⇒ 3k = 12 + 20 or, 3k = 12 – 20 ⇒ 3k = 32 or, 3k = – 8 ⇒ k = $\frac{32}{8}$ or $\frac{-8}{3}$ ∴ Required points are $\left(0-\frac{8}{3}\right),\left(0,\frac{32}{3}\right)$