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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-II

  • question_answer
    The sides of a rhombus \[ABCD\] are parallel to the lines, \[x-y+2=0\] and \[7x-y+3=0\]. 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 is                            [JEE Online 15-04-2018 (II)]

    A)         2                             

    B)          \[\frac{7}{4}\]   

    C) \[\frac{7}{2}\]                   

    D)          \[\frac{5}{2}\]

    Correct Answer: D

    Solution :

    Let the coordinate A be \[(0,c)\] Equations of the parallel lines are given: \[x-y+2=0\] and \[7x-y+3=0\] We know that the diagonals will be parallel to the angle bisectors of the two sides \[y=x+2\] and \[y=7x+3\] \[\frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{5\sqrt{2}}\] \[5x-5y+10=\pm (7x-y+3)\] Parallel equations of the diagonals are \[2x+4y-7=0\] and \[12x-6y+13=0\] Slopes of diagonals \[m=\frac{-1}{2}\]and \[2\] We know that the slope of diagonal from \[A(0,c)\] and passing through \[P(1,2)\] is \[(2-c)\] therefore \[2-c=2\Rightarrow c=0\], but it is given that \[A\] is not origin, so \[2-c=\frac{-1}{2}\Rightarrow c=\frac{5}{2}\] \[\therefore coordinate\,\,of\,\,A\,\,is(0,5/2)\]


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