Two sides of a parallelogram are along the lines, x+y=3 and x−y+3=0. If its diagonals intersect at (2,4) then one of its vertex is :
A
(2,6)
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B
(2,1)
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C
(3,6)
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D
(3,5)
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Solution
The correct option is C(3,6) Let the point C is (x,y)
x+y=3⋯(1)
and x−y+3=0⋯(2)
The point of insection from the equation (1) and (2) is point A:(0,3)
So, if point C is (x,y) then, x+02=2⇒x=4 and y+32=4⇒y=5
As line BC and AD are parrallel ∴ slope of line BC=1 ∴ equation of BC is y−5=1(x−4) ⇒x−y+1=0⋯(3)
As line CD and AB are parrallel ∴ slope of line CD=−1 ∴ equation of CD is y−5=−1(x−4) ⇒x+y=9⋯(4)
Hence, from (1) and (3), point B is (1,2) and from (2) and (4), point D is (3,6).