Question

ABCD is a parallelogram equation of AB and AD 4 x + 5y=0 and 7 x + 2 Y = 0 and the equation of diagonal BD is 11x + 7y + 9=0 the equation of AC is??
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Answer 1

Answer:

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.

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  4x + 5 y=K

(1, 1) is on it... hence ,      K = 9

so  CD:  4x + 5y = 9

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