UPSC NDA I 2025 Mathematics Paper

Given points A(3,−1) and B(1,1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √‌2 units from P.
Compute the midpoint P of AB :
P=(‌

3+1

2

,‌

−1+1

2

)=(2,0)
Compute the slope of AB :
mAB=‌

1−(−1)

1−3

=‌

2

−2

=−1
Therefore, the equation of line AB is:
y−1=−1(x−1)⟹x+y−2=0
The perpendicular bisector of AB must pass through P(2,0) and have slope perpendicular to -1 (i.e. slope +1 ):
y−0=1(x−2)⟹y=x−2
Any point Q on this bisector satisfies y=x−2. Write Q=(x,x−2).
We require the distance PQ=√‌2. Since P(2,0),
‌‌ Distance ‌2=(x−2)2+((x−2)−0)2=2
‌⇒(x−2)2+(x−2)2=2⇒2(x−2)2=2⇒(x−2)2=1
Thus,

⇒x−2=±1⇒x=3‌ or ‌x=1
If x=3, then y=3−2=1. So one solution is Q(3,1).
If x=1, then y=1−2=−1. So the other solution is Q(1,−1).