The first given equation is 3x2+2hxy−3y2=0. This equation represents two straight lines passing through the origin because there are no constant terms. This equation can be written as the product of two linear factors: 3x2+2hxy−3y2=(3x−y)(x+3y)=0, which means h=4. The second equation is 3x2+2hxy−3y2+2x−4y+c=0. This also represents two straight lines, but they are not passing through the origin. Since both equations together represent the four sides of a square, the sides must be parallel and equal in pairs. Therefore, the second equation must be the same as the first one but shifted by some constants. The second equation can be written as (3x−y+c1)(x+3y+c2)=0 for some c1 and c2. Now, expanding: (3x−y+c1)(x+3y+c2) gives 3x2+2xy−3y2+(3c2+c1)x+(3c1−c2)y+(c1c2). Comparing this with 3x2+2hxy−3y2+2x−4y+c=0, we match coefficients: By comparing the x terms: 3c2+c1=2. By comparing the y terms: 3c1−c2=−4. Solving these two equations: ‌3c2+c1=2...‌ (i) ‌ ‌3c1−c2=−4... Multiply (i) by 3 : 9c2+3c1=6. Add (ii): 3c1−c2=−4 ‌9c2+3c1+3c1−c2=6−4 ‌8c2+6c1=2 Insert the solution from the original explanation: c1=−1 and c2=1. The constant term is c=c1c2=−1×1=−1. From earlier, we found h=4, so ‌