JEE Main & Advanced Sample Paper JEE Main - Mock Test - 7

  • question_answer
    The length of the perpendicular from the origin to a line is 7 and line makes an angle of \[150{}^\circ \] with the positive direction of y-axis, then the equation of the line is

    A) \[\sqrt{3}x+y=7\]                  

    B) \[\sqrt{3}x-y=14\]

    C) \[\sqrt{3}x+y+14=0\]           

    D) \[\sqrt{3}x+y-14=0\]

    Correct Answer: D

    Solution :

    Here \[p=7\]and \[\alpha =30{}^\circ \] \[\therefore \] Equation of the required line is \[x\,\cos 30{}^\circ +y\sin 30{}^\circ =7\]  or  \[x\frac{\sqrt{3}}{2}+y\times \frac{1}{2}=7\] or  \[\sqrt{3}x+y=14\]


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