Question

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. $\left(\frac{2}{5},\frac{3}{5}\right)$

Answer 1

Answer: (D)

Given, $(p+2q)x+(p-3q)y=p-q$ ...(i)
$\Rightarrow$ $px+2qx+py-3qy-p+q=0$
$\Rightarrow$ $(x+y-1)p+(2x-3y+1)q=0$
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
$\left(\frac{2}{5},\frac{3}{5}\right)$ .