We have an equation of line $\frac{x}{3}+\frac{y}{4}$ = 1, which can be written as 4x + 3y – 12 = 0 ... (i) Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units. ⇒ $\frac{\left|4\times a+3\times 0-12\right|}{\sqrt{4^{2}+3^{2}}}$ = 4 ⇒ $\frac{\left|4a-12\right|}{\sqrt{16+9}}$ = 4 ⇒ $\frac{\left|4a-12\right|}{\sqrt{25}}$ = 4 ⇒ $\frac{\left|4a-12\right|}{5}$ = 4 ⇒ |4a – 12| = 20 ⇒ 4a – 12 = ± 20 ⇒ 4a = 12 ± 20 ⇒ a = 3 ± 5 i.e. a = 3 + 5 or a = 3 – 5 ⇒ a = 8 or a = – 2 Hence, the required points on the x-axis are (8, 0) and (–2, 0).