Question

The distance between the parallel lines $9x^2-6xy+y^2+18x-6y+8=0$ is

• A. $\frac{2}{\sqrt{10}}$
• B. $\frac{1}{\sqrt{10}}$
• C. $\frac{4}{\sqrt{10}}$
• D. None of these

Answer 1

Answer: (A)

$9x^2-6xy+y^2+18x-6y+8=0$
$\Rightarrow \left(\left(3x^2\right)-2\times \left(3x\right)\times y+y^2\right)+6\left(3x-y\right)+8=0$
$\Rightarrow {\left(3x-y\right)}^2+6\left(3x-y\right)+8=0$
Let $3x-y=z$
$\therefore z^2+6z+8=0$
$\Rightarrow z^2+4z+2z+8=0$
$\Rightarrow z\left(z+4\right)+2\left(z+4\right)=0$
$\Rightarrow \left(z+2\right)\left(z+4\right)=0$
$\Rightarrow z=-2,z=-4$
$3x-y+2=0...\left(i\right)$ or $3x-y+4=0$
If $P_1$ be the distance of line $\left(i\right)$ from the origin, then
$P_1=\frac{2}{\sqrt{9+1}}=\frac{2}{\sqrt{10}}$
Also, if $P_2$ be the distance of line \left(ii\right) from the origin,then
$P_2=\frac{4}{\sqrt{10}}$
So, distance between lines
$P=P_2-P_1=\frac{4}{\sqrt{10}}-\frac{2}{\sqrt{10}}=\frac{2}{\sqrt{10}}$