UPSC NDA Math Model Paper 7

Here, A (1, 2), B (4, y), C (x, 6) and D (3, 5) are the vertices of parallelogram. AB = CD and AD = BC (sides of parallelogram) AC and BD are the diagonals which bisect each other. Suppose diagonals intersect each other at (m, n) Midpoint of $AC = \left( \frac{1 + x}{2}, \frac{6 + 2}{2} \right)$ Midpoint of $BD = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right)$ (Using midpoint formula) We can see that, Midpoint of AC = midpoint of BD $\left( \frac{1 + x}{2}, \frac{6 + 2}{2} \right) = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right)$ $\therefore \frac{1 + x}{2} = \frac{7}{2} \text{ and } \frac{8}{2} = \frac{y + 5}{2}$ (Equating respective co-ordinates) $\Rightarrow 1 + x = 7 \text{ and } 8 = y + 5$ $\Rightarrow x = 6 \text{ and } y = 3$ The point of intersection of diagonals = Midpoint of AC $(m, n) = \left( \frac{1 + x}{2}, \frac{6 + 2}{2} \right)$ $= \left( \frac{1 + 6}{2}, \frac{6 + 2}{2} \right)$ $= \left( \frac{7}{2}, 4 \right)$ Hence, option 1 is correct.