Question

Distance between parallel lines $9x^2-6xy+y^2+18x-6y+8=0$ is $\frac{m}{\sqrt{n}}$.Find $m+n$

Answer 1

$9x^2-6xy+y^2+18x-6y+8=0(3x-y+2)(3x-y+4)=0\therefore Parallellinesare3x-y+2=03x-y+4=0$

Distance between lines =$\frac{4-2}{\sqrt{3^2+1^2}}$

=$\frac{2}{\sqrt{1}0}$

$\therefore \frac{m}{\sqrt{n}}=\frac{2}{\sqrt{10}}m+n=12$