CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The distance between the pair of parallel lines \[{{x}^{2}}+4xy+4{{y}^{2}}+3x+6y-4=0\]is:

    A)  \[\sqrt{5}\]                       

    B)         \[\frac{2}{\sqrt{5}}\]

    C)  \[\frac{1}{\sqrt{5}}\]                    

    D)         \[\frac{\sqrt{5}}{2}\]

    E)  \[\sqrt{\frac{5}{2}}\]

    Correct Answer: A

    Solution :

    \[\because \]The given equation of pair of parallel lines is \[{{x}^{2}}+4xy+4{{y}^{2}}+3x+6y-4=0\] Here, \[a=1,b=4g=\frac{3}{2}\]and \[c=-4.\] Then, required distance\[=2\sqrt{\frac{{{g}^{2}}-ac}{a(a+b)}}\]                 \[=2\sqrt{\frac{\frac{9}{4}+4}{5}}=\frac{2.\sqrt{16+9}}{2.\sqrt{5}}\]                 \[=\frac{2\sqrt{25}}{2\sqrt{5}}=\frac{5}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}\]                 \[=\sqrt{5}\]


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