2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The distance of the point (1, 2) from the line \[x+y+5=0\]measured along the line parallel to\[3x-y=7\]is equal to

    A)  \[4\sqrt{10}\]                  

    B)  \[40\]

    C)  \[\sqrt{40}\]             

    D)                         \[10\sqrt{2}\]

    E)  \[2\sqrt{20}\]

    Correct Answer: C

    Solution :

    Let equation of line parallel to\[3x-y=7\]be\[3x-y=\lambda \]. It passes through (1, 2). \[\therefore \]  \[3-2=\lambda \Rightarrow \lambda =1\] \[\therefore \]Line is \[3x-y=1\] The point of intersection of\[x+y+5=0\]and \[3x-y-1\]is\[(-1,-4)\]. \[\therefore \]Distance between (1, 2) and\[(-1,-4)\]                 \[=\sqrt{{{(2)}^{2}}+{{(6)}^{2}}}=\sqrt{40}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner