JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    If the two lines \[x+\left( a1 \right)y=1\]and \[2x+{{a}^{2}}y=1\]\[(a\in R-\{0,1\})\]are perpendicular, then the distance of their point of intersection from the origin is :-        [JEE Main 9-4-2019 Afternoon]

    A) \[\frac{2}{5}\]                        

    B) \[\frac{2}{\sqrt{5}}\]

    C) \[\frac{\sqrt{2}}{5}\]                         

    D) \[\sqrt{\frac{2}{5}}\]

    Correct Answer: D

    Solution :

    \[\left( \frac{-1}{a-1} \right)\left( \frac{-2}{{{a}^{2}}} \right)=-1\]           \[2=-({{a}^{2}})(a-1)\]           \[{{a}^{3}}-{{a}^{2}}+2=0\]           \[(a+1)({{a}^{2}}-2a+2)=0\]           \[\therefore \]\[a=-1\]           \[\left. \begin{align}   & {{L}_{1}}:x-2y+1=0 \\  & {{L}_{2}}:2x+y-1=0 \\ \end{align} \right\}\]           \[0(0,0)\,\,\,\,\,\,\,\,\,\,P\left( \frac{1}{5},\frac{3}{5} \right)\]           \[OP=\sqrt{\frac{1}{25}+\frac{9}{25}}=\sqrt{\frac{10}{25}}=\sqrt{\frac{2}{5}}\]


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