RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    For the different values of p and q, the line \[(p+2q)x+(p-3q)y=p-q\]passes through the fixed point

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

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

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

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

    Correct Answer: D

    Solution :

     Given, \[(p+2q)x+(p-3q)y=p-q\]       ...(i) \[\Rightarrow \] \[px+2qx+\text{ }py-3qy-p+q=0\] \[\Rightarrow \] \[(x+y-1)p+(2x-3y+1)q=0\] For different values of p and q the line (i) will pass through the point of intersection of lines \[x+y-1=0\]and\[2x-3y+1=0\]. So, Intersection point is\[\left( \frac{2}{5},\frac{3}{5} \right)\].


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