JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    The locus of the mid-points of the perpendiculars drawn from points on the line, \[x=2y\]to the line \[x=y\] is [JEE MAIN Held on 07-01-2020 Evening]

    A) \[5x7y=0\]      

    B) \[2x3y=0\]

    C) \[3x2y=0\]      

    D) \[7x5y=0\]

    Correct Answer: A

    Solution :

    [a] Let coordinate of P is \[(2\lambda ,\lambda )\] and coordinate of mid-point M is\[({{x}_{1}},\text{ }{{y}_{1}})\].
    \[\therefore \] Coordinate of Q
    \[=(2{{x}_{1}}2\lambda ,\text{ }2{{y}_{1}}\lambda )\]
    \[\because \] Q lies on line y = x
    \[\therefore \,\,\,\lambda \,\,=2{{x}_{1}}2{{y}_{1}}\]                                 ...(i)
    (Slope of line PQ) \[\cdot \] (Slope of line y = x) = -1
    \[\therefore \,\,\,\frac{\lambda -{{y}_{1}}}{2\lambda -{{x}_{1}}}=-1\]
    \[\therefore \,\,\,\lambda =\frac{{{x}_{1}}+{{y}_{1}}}{3}\]                                 ...(ii)
    From equation (i) and (ii) : \[5{{x}_{1}}=7{{y}_{1}}\]
    \[\therefore \]   Required locus is 5x = 7y.

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