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.