UPSC NDA I 2015 Mathematics Paper

Concept:The circumcenter is the intersection point of the perpendicular bisectors of the sides of the triangle.Explanation:Coordinates: $A(-2,3)$, $B(2,1)$, $C(1,2)$.Midpoint of $BC$: $\left(\frac{2+1}{2},\frac{1+2}{2}\right) = \left(\frac{3}{2},\frac{3}{2}\right)$.Slope of $BC$: $\frac{2-1}{1-2} = -1$.Slope of perpendicular bisector $DE$: $1$ (since product $=-1$).Equation of $DE$: $y - \frac{3}{2} = 1\left(x - \frac{3}{2}\right) \Rightarrow y = x$ …(1).Midpoint of $AC$: $\left(\frac{-2+1}{2},\frac{3+2}{2}\right) = \left(-\frac{1}{2},\frac{5}{2}\right)$.Slope of $AC$: $\frac{3-2}{-2-1} = -\frac{1}{3}$.Slope of perpendicular bisector $EF$: $3$.Equation of $EF$: $y - \frac{5}{2} = 3\left(x + \frac{1}{2}\right) \Rightarrow 2y - 6x = 8$ …(2).Solve (1) and (2): Substituting $y=x$ in (2): $2x-6x=8 \Rightarrow -4x=8 \Rightarrow x=-2$.From (1): $y=x=-2$.Thus circumcenter is $(-2,-2)$.Answer:$(-2,-2)$