Question

The line $\left(p+2q\right)x+\left(p-3q\right)y=p-q$ for different values of $p$ and $q$, passes through the fixed point

• A. $\left(\frac{3}{2},\frac{5}{2}\right)$
• B. $\left(\frac{2}{5},\frac{2}{5}\right)$
• C. $\left(\frac{3}{2},\frac{3}{5}\right)$
• D. $\left(\frac{2}{5},\frac{3}{5}\right)$

Answer 1

The correct option is D $\left(\frac{2}{5},\frac{3}{5}\right)$

$(p+2q)x+(p-3q)y=p-q$

$p(x-1+y)+q(2x-3y+1)=0$

intersection point of $(x-1+y=0),(2x-3y+1=0)$

$x+y=1\times 2$

$2x+2y=2$

$2x-3y=-1\_\_\_\_\_\_\_5y=3$

$y=\frac{3}{5}$ put in $x+y=1\Rightarrow x=1-\frac{3}{5}=\frac{2}{5}$

$y=\frac{3}{5}$ ; $x=\frac{2}{5}$

$\therefore$ the fixed point is $(\frac{2}{5},\frac{3}{5})$