Given line is

x

3

+

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
|

4×0+3×k−12

√42+32

| = 4
⇒ |3k – 12| = 4 × 5 ⇒ 3k – 12 = ± 20 ⇒ 3k = 12 ± 20
⇒ 3k = 12 + 20 or, 3k = 12 – 20 ⇒ 3k = 32 or, 3k = – 8
⇒ k =

32

8

or

−8

3

∴ Required points are (0−

8

3

),(0,

32

3

)