Solving the equations 4x + 5y = 0 and 11x + 7y − 9 = 0, we have x = 5/3, y = −4/3. Let A = (5/3, −4/3). Solving the equations 7x + 2y = 0 and 11x + 7y− 9 = 0, we have x = −2/3, y = 7/3. Let C = (−2/3, 7/3). Therefore,
C(-2/3,2/7),O(0,0) and A(5/3,4/3)
are the three consecutive vertices of the parallelogram. So its fourth vertex B (see Fig.) is
(-2/3 + 5/3, 7/3 - 4/3) = (1,1)
Therefore, the vertices of the parallelogram OABC are O(0, 0), A(5/3, −4/3), B(1, 1) and C(−2/3, 7/3). Since the side BC is parallel to OA and passes through (1, 1), its equation is
y - 1 = -4/5(x - 1) or 4x + 5y - 9 = 0
Also, the equation of the side AB is
y - 1 = -7/2(x - 1) or 7x + 2y - 9 = 0.
and the second diagonal is x−y = 0.