Here, we have to find the points which are on the y-axis such that the perpendicular distance between the points and the line
x
3
−
y
4
=1 is 3 units. The given equation of line can be re-written as: 4x−3y−12=0 Let P = (0, y) Here a = 4, b = - 3 and d = 3 Now substitute x1=0 and y1=y in the equation 4x−3y−12=0 ⇒|4.x1−3.y1−12|=|0−3y−12| ⇒√a2+b2=√42+(−3)2=5 As we know that, the perpendicular distance d from P(x1,y1) to the line ax+by+c=0 is given by d=|
ax1+by1+c
√a2+b2
| ⇒d=|
4.x1−3.y1−12
√42+(−3)2
|=3 ⇒|−3y−12|=15 ⇒ y = 1 or - 9 So, the points are: (0, 1) and (0, - 9) Hence, option D is the correct answer.