JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • question_answer
    The figure formed by the lines \[{{x}^{2}}+4xy+{{y}^{2}}=0\] and \[x-y=4,\] is                                     [Roorkee 1980]

    A)            A right angled triangle      

    B)            An isosceles triangle

    C)            An equilateral triangle     

    D)            None of these

    Correct Answer: C

    Solution :

    \[{{S}_{1}}=\frac{1}{-2+\sqrt{4-1}}=\frac{1}{-2+\sqrt{3}}=-(\sqrt{3}+2)\]            \[{{S}_{2}}=\frac{1}{-2-\sqrt{4-1}}=\frac{1}{-2-\sqrt{3}}=(\sqrt{3}-2)\] and \[{{S}_{3}}=1\].            \[{{\theta }_{13}}={{\tan }^{-1}}\left| \frac{-(\sqrt{3}+2)-1}{1-(\sqrt{3}+2)} \right|={{\tan }^{-1}}\left| \frac{-(\sqrt{3}+3)}{-(\sqrt{3}+1)} \right|\]                  \[={{\tan }^{-1}}(\sqrt{3})=60{}^\circ \].            \[{{\theta }_{23}}={{\tan }^{-1}}\left| \frac{\sqrt{3}-2-1}{1+\sqrt{3}-2} \right|={{\tan }^{-1}}\left| \frac{\sqrt{3}-3}{\sqrt{3}-1} \right|\]             \[={{\tan }^{-1}}(\sqrt{3})=60{}^\circ \].

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