Let AB = 4 x + 5y = 0 => y = -4/5 x

let AD = 7x + 2y = 0 => y = -7/2 x

so the point *A = (0, 0) * as these two intersect at origin.

let BD = 11 x + 7y = 9 we know it does not pass through origin. So C cannot be on that line.

Intersection of AB and BD : 11 x - 28/5 x = 9 => x = 5/3 and so y = -4/3

so* B (5/3, -4/3)*

intersection of AD and BD : 11 x - 49/2 x = 9 => x = -2/3 and so y = 7/3

so *D(-2/3, 7/3)*

Now midpoint of BD = O = ((5-2)/3/2 , (7-4)/3/2 ) = (1/2, 1/2)

Line OA is the other diagonal AC, so its equation is : y = x as its slope is 1/2 / 1/2 = 1 and it passes throug h origin.

O is the midpoint of AC. Hence* C = (1, 1) * that is simple to find.

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equation of BC: parallel to AD 7x + 2y = K

(1, 1) lies on it => * 7x+2y =9 as K = 9*

equation of CD : it is parallel to AB. hence it is 4 x + 5 y = K

(1, 1) is on it... hence , K = 9

* so CD: 4x + 5y = 9*