UPSC NDA Math Model Paper 5

Given: Equation of line is 3x + 2y = 8
ere, we have to find the equation of the normal to the given line and passing through the midpoint of the line joining the points (6, 4) and (4, - 2).
First let's find the midpoint of he line joining the points (6, 4) and (4, - 2).
⇒(

6+4

2

,

4−2

2

)=(5,1)
We can re-write the given equation of line as: y=−

3

2

.x+4
Now by comparing the above equation with y = mx + c, we get slope of the give line i.e m1=−

3

2

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

2

.m2=−1⇒m2=

2

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)
So, the equation of the normal whose slope is m2=

2

3

and passing through the point (5,1)
⇒y−1=

2

3

.(x−5)
So, the equation of the required normal is: 2x - 3y - 7 = 0
Hence, option C is the correct answer