Distance between parallel homogeneous lines

Question

I came across a question online, although we haven't been taught this at school. However, I'm curious as to how it could be solved. Please do explain the formula/steps as well. Walk me through the process. Thanks!

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

Answer 1

Hint: your equation is equivalent to $$(2+3x-y)(4+3x-y)=0$$ The you can convert one equation into the Hessian Normalform: $$\frac{-3x+y-4}{\pm\sqrt{10}}$$ and plug ing one Point from the other equation Write your equation in the form $$(3x-y)^2+6(3x-y)+8=0$$

Answer 2

Hint: Use Dr. Graubner's hint about the factorization, which will give you two line equations of the form $$ y = a x + b $$

Hint: A line can be also described as \begin{align} y &= a x + b \iff \\ b &= -a x + y = (-a, 1) \cdot (x, y)^\top = n \cdot u \iff \\ \left( n / \lVert n \rVert \right) \cdot u &= b / \lVert n \rVert = d \end{align} where $d$ is the signed distance of the line to the coordinate system origin $(0,0)^\top$ and $\lVert n \rVert = \lVert (n_x, n_y)^\top \rVert = \sqrt{n_x^2 + n_y^2}$ is the length of a normal vector $n$ of the line. So $$ d = \frac{b}{\sqrt{a^2 + 1}} $$

Hint: Use the difference in distances

Answer 3

Factor $$9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$$

$$ (3x-y+4)(3x-y+2)=0$$

The lines are $ 3x-y+4=0$ and $3x-y+2=0$

The distance from a point $(x_0, y_0)=(0,4)$ on the first line to the second line is found to be

$$ \frac {|3x_0-y_0+2|}{\sqrt {9+1}}=\frac {2}{\sqrt {10}} $$