UPSC NDA I and II 2020 Math Solved Paper

We know that diagonal of a square bisect each other perpendicularly.
Equation of a diagonal : 3x+2y=5 (given).
Now, equation of other diagonal that is perpendicular to the given diagonal =2x−3y=K.
As vertex point (1,-1) does not lies on 3x+2y=5
{∵3(1)+2(−1)≠5}
Then, point (1,−1), must be on the diagonal 2x−3y=K
Then, 2(1)−3(−1)=K
∴K=5
Hence, equation of other diagonal : 2x−3y=5.