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JEE Main & Advanced Mathematics Straight Line Question Bank Slope of line, Equation of line in different forms

  • question_answer
    The vertices of a triangle OBC are \[(0,\ 0),\ (-3,\ -1)\] and \[(-1,\ -3)\ \]respectively. Then the equation of line parallel to BC which is at \[\frac{1}{2}\]unit distant from origin and cuts OB and OC, is                                       [IIT 1976]

    A)            \[2x+2y+\sqrt{2}=0\]           

    B)            \[2x+2y-\sqrt{2}=0\]

    C)            \[2x-2y+\sqrt{2}=0\]             

    D)            None of these

    Correct Answer: A

    Solution :

               Gradient of \[BC=-1\] and its equation is \[x+y+4=0\]. Therefore the equation of line parallel to \[BC\]is \[x+y+\lambda =0\]. Also it is \[\frac{1}{2}\]unit distant from origin. Thus \[\frac{\lambda }{\sqrt{2}}=\frac{1}{2}\Rightarrow \lambda =\frac{\sqrt{2}}{2}.\]                    Hence the required equation of line is \[2x+2y+\sqrt{2}=0\].


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