We have an equation of line

x

3

+

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.
⇒

|4×a+3×0−12|

√42+32

= 4 ⇒

|4a−12|

√16+9

= 4
⇒

|4a−12|

√25

= 4 ⇒

|4a−12|

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).