UPSC NDA Math Model Paper 4

Here, we have to find the the equation of the line perpendicular to the line 3x + 5y = 4 and having x-intercept - 3.
The given equation 3x+5y=4 can be re-written as: y=−

3

5

.x+

4

5

By comparing the above with y = mx + c, we get slope of the give line i.e m1=−

3

5

let slope of the normal be m2
∵ The angle between the line and the required normal is 90°
As we know that, if two lines are perpendicular then product of their slopes is – 1, i.e m1.m2=−1
⇒−

3

5

.m2=−1 ⇒m2=

5

3

As we know that, the equation of a line passing through the point (x1,y1) and having the slope ' m ' is given as: y−y1=m.(x−x1)
∵ The normal has x-intercept - 3 i.e it passes through the point (- 3, 0)
So, the equation of the normal whose slope is m2=5∕3 and passes through the point (3, 0)
⇒y−0=

5

3

.(x+3)
So, the equation of the required normal is: 5x−3y+15=0
Hence, option D is the correct answer.