Two sides of a parallelogram are along the lines,x+y=3 and x−y+3=0.If its diagonals intersectat (2, 4), then one of its vertex is 

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

 

  1. A

    (3, 6)

  2. B

    (2, 6)

  3. C

    (2, 1)

  4. D

    (3, 5)

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    Solution:

    Let, equation of AB is x+y=3                                                 ..(i)

    and, equation of AD is x−y=−3                                            ..(ii)

    Solving (i) and (ii), we get x=0,y=3

    ∴ Coordinates of A are (0, 3).

    Now,  x1+02=2