The sides of a rhombus ABCD are parallel to the lines,x−y+2=0 and7x−y+3=0. If the diagonals of the rhombus intersect at P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is

The sides of a rhombus ABCD are parallel to the lines,xy+2=0 and7xy+3=0. If the diagonals of the rhombus intersect at P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is

  1. A

    2

  2. B

    5/2

  3. C

    7/4

  4. D

    7/2

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    Solution:

    Let coordinates of A be (0, a)

    The diagonals intersect at P(1, 2)

    We know that the diagonals will be parallel to the

    angle bisectors of the two sides y=x+2 and y=7x+3

    i.e. xy+22=±7xy+352

    5x5y+10=±(7xy+3)

    2x+4y7=0 and 12x6y+13=0

    m1=1/2 and m2=2

    (where m1 and m2 are the slopes of the given two lines) Let one diagonal be parallel to 2x + 4y- 7 = 0 and other be parallel to 12x6y+13=0

    The vertex A could be on any of the two diagonals.

    Hence, slope of AP is either1/2 or 2.

    2a10=2 or2a10=12a=0 ora=52

    But a0                                                                     Thus, ordinate of A is 5/2.

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