UPSC NDA I 2025 Mathematics Paper

Concept:The perpendicular bisector of a line segment passes through its midpoint and has slope equal to the negative reciprocal of the segment's slope. The distance formula is used to find points at a given distance.Explanation:First, find the midpoint P of AB: $P = \left(\frac{3+1}{2}, \frac{-1+1}{2}\right) = (2,0)$.Slope of AB: $m_{AB} = \frac{1-(-1)}{1-3} = \frac{2}{-2} = -1$.Slope of perpendicular bisector = 1 (negative reciprocal).Equation of perpendicular bisector through P: $y - 0 = 1(x - 2)$ → $y = x - 2$.Any point Q on this line is $(x, x-2)$.Distance from P(2,0) to Q must be $\sqrt{2}$.By distance formula: $\sqrt{(x-2)^2 + ((x-2)-0)^2} = \sqrt{2}$.Square both sides: $(x-2)^2 + (x-2)^2 = 2$ → $2(x-2)^2 = 2$ → $(x-2)^2 = 1$.So $x-2 = \pm 1$ → $x = 3$ or $x = 1$.If $x = 3$, then $y = 3-2 = 1$ → Q = (3,1).If $x = 1$, then $y = 1-2 = -1$ → Q = (1,-1), which is not in the options.Thus the only possible coordinate among the given options is (3,1).Answer:Option B: (3,1).