2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The lines\[(a+2b)x+(a-3b)y=a-b\]for different values of a and b pass through the fixed point whose coordinates are

    A)  \[\left( \frac{2}{5},\frac{2}{5} \right)\]                  

    B)  \[\left( \frac{3}{5},\frac{3}{5} \right)\]

    C)  \[\left( \frac{1}{5},\frac{1}{5} \right)\]  

    D)         \[\left( \frac{3}{5},\frac{2}{5} \right)\]

    E)  \[\left( \frac{2}{5},\frac{3}{5} \right)\]

    Correct Answer: E

    Solution :

    Given equation is \[(a+2b)\text{ }x+(a-3b)\text{ }y=a-b\] It can be rewritten as \[a(x+y-1)+b(2x-3y+1)=0\] This is the form of intersection of two lines. \[\therefore \]  \[x+y-1=0\] and        \[2x-3y+1=0\] On solving, we get \[x=\frac{2}{5}\]and\[y=\frac{3}{5}\]


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