JEE Main & Advanced Mathematics Straight Line Question Bank Distance between two lines, Perpendicular distance of the line from a point Position of point w.r.t. line

  • question_answer
    The points on the x-axis whose perpendicular distance from the line \[\frac{x}{a}+\frac{y}{b}=1\] is a, are            [RPET 2001; MP PET 2003]

    A)            \[\left[ \frac{a}{b}(b\pm \sqrt{{{a}^{2}}+{{b}^{2}}}),\,0 \right]\]               

    B)            \[\left[ \frac{b}{a}(b\pm \sqrt{{{a}^{2}}+{{b}^{2}}}),\,0 \right]\]

    C)            \[\left[ \frac{a}{b}(a\pm \sqrt{{{a}^{2}}+{{b}^{2}}}),\,0 \right]\]               

    D)            None of these

    Correct Answer: A

    Solution :

               Let the point be \[(h,0)\], then \[a=\pm \frac{bh+0-ab}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]                    Þ \[bh=\pm a\sqrt{{{a}^{2}}+{{b}^{2}}}+ab\Rightarrow h=\frac{a}{b}(b\pm \sqrt{{{a}^{2}}+{{b}^{2}}})\]                    Hence the point is \[\left\{ \frac{a}{b}(b\pm \sqrt{{{a}^{2}}+{{b}^{2}}}),0 \right\}\].


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