UPSC NDA Math Model Paper 5

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.