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Q.
The distance between the parallel lines $9x^2 - 6xy + y^2 +18x - 6y + 8 = 0$ is
Straight Lines
Solution:
$9x^{2}-6xy+y^{2}+18x-6y+8=0$
$\Rightarrow \left(\left(3x^{2}\right)-2\times \left(3x\right)\times y+y^{2}\right)+6\left(3x-y\right)+8=0$
$\Rightarrow \left(3x-y\right)^{2}+6\left(3x-y\right)+8=0$
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
$P=P_{2}-P_{1}=\frac{4}{\sqrt{10}}-\frac{2}{\sqrt{10}}=\frac{2}{\sqrt{10}}$