Thank you for reporting, we will resolve it shortly
Q.
For the different values of p and q, the line $ (p+2q)x+(p-3q)y=p-q $ passes through the fixed point
Rajasthan PETRajasthan PET 2006
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) $ .