Given equation of line

⇒ 4x + 3y = 12

⇒ 4x + 3y – 12 = 0

Comparing above equation with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

⇒ |4a – 12| = 4 × 5

⇒ (4a – 12) = 20

⇒ 4a – 12 = 20 or - (4a – 12) = 20

⇒ 4a = 20 + 12 or 4a = -20 + 12

⇒ a = 32/4 or a = -8/4

⇒ a = 8 or a = -2

∴ The required points on the x – axis are (-2, 0) and (8, 0)