UPSC NDA I 2023 Mathematics Solved Paper

Concept:In a parallelogram, the diagonals bisect each other. Use the intersection points of given lines to locate vertices, then find the equation of the required diagonal.Explanation:Given lines are $x=0$, $y=0$, $x+y=1$, and $6x+y=3$.Intersection of $x=0$ and $x+y=1$ gives point $D(0,1)$.Intersection of $y=0$ and $6x+y=3$ gives point $B\left(\frac{1}{2},0\right)$.Intersection of $x+y=1$ and $6x+y=3$ gives point $C\left(\frac{2}{5},\frac{3}{5}\right)$.Point $O(0,0)$ is also a vertex.These four points form a parallelogram.Diagonal BD joins $B\left(\frac{1}{2},0\right)$ and $D(0,1)$.Using the two‑point formula: $y-0 = \frac{1-0}{0-\frac{1}{2}}\left(x-\frac{1}{2}\right)$.Simplify: $y = -2\left(x-\frac{1}{2}\right)$ → $y = -2x+1$.Rearranging gives $2x+y-1=0$.Answer:$2x+y-1=0$ (Option D).