Question

The line $(p+2q)x+(p-3q)y=p-q$ for different values of $p$ and $q$ passes through a fixed point whose co-ordinates are

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

Answer 1

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

The given equation of line is

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

$px+2qx+py-3qy-p+q=0$

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

For different value of p the given equation can be 0 only if

$x+y-1=0$ and $2x-3y+1=0$

i.e $x+y=\mathrm{1.......}\left(1\right)$

$2x-3y=-\mathrm{1.......}\left(2\right)$

Solving (1) and (2), we get

$2(1-y)-3y=-1$

$2-2y-3y=-1$

$-5y=-3$

$y=\frac{3}{5}$

Putting in (1)

$x=1-\frac{3}{5}$

$x=\frac{2}{5}$

Hence the given line will pass through $(x,y)=(\frac{2}{5},\frac{3}{5})$