Question

The line $(p+2q)x+(p-3q)y=p-q$ for different values of $p$ and $q$ passes through a fixed point whose coordinates 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 $(\frac{2}{5},\frac{3}{5})$

We have,

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

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

Equation the coefficients to 0 gives us
$x+y-1=0$ and $2x-3y+1=0$

Solving the above set of linear equation gives us
$x=\frac{2}{5}$ and $y=\frac{3}{5}$.

Substitution of the above point in the original equation will give us $LHS=RHS$ independent of the values of p and q.

Hence the line $(p+2q)x+(p-3q)y=p-q$ passes through a fixed point which is

$\Rightarrow$ $(x,y)=(\frac{2}{5},\frac{3}{5})$.