Find the distance!

Given that $x^2 - 4xy + 4y^2 + x - 2y - 6 = 0$ represents a pair of parallel lines, we can find the \two $y$ - and $x$ -intercepts by putting $x=0$ and $y=0$ respectively, we have:

$\begin{cases} x=0 & \implies 4y^2 - 2y - 6 = 0 & \implies 2(y+1)(2y-3) = 0 & \implies y = - 1, \dfrac 32 \\ y=0 & \implies x^2 +x - 6 = 0 & \implies (x+3)(x-2) = 0 & \implies x = - 3, 2 \end{cases}$

From the four intercepts we note that the two parallel lines are $\begin{cases} 2y = x + 3 \\ 2y = x - 2 \end{cases}$ as shown in the figure. Also note that $(x+3-2y)(x-2-2y) = x^2 - 4xy + 4y^2 +x - 2y - 6$ .

Let us drop a perpendicular from $y$ -intercept $A\left(0, \frac 32\right)$ to the line $2y=x-1$ at point $C$ . Then $AC$ is the distance between the two parallel lines. Since the gradient of the parallel lines is $\frac 12$ , then $\angle BAC = \tan^{-1} \frac 12$ , where $B(0,-1)$ is the other $y$ -intercept. We note that $AC = AB \cos \angle BAC = \frac 52 \times \frac 2{\sqrt 5} = \boxed{\sqrt 5}$ .

$x^{2}-4xy+4y^{2} +x-2y-6=0$ represents 2 parallel lines. So we can write the equation as:

$(x-2y + c_{1})(x-2y+c_{2})$ because they are parallel the coeffient of $x$ and $y$ are the same and when we simplify we get:

$x^{2}-4xy+4y^{2} +(c_{1}+c_{2})x-2(c_{1}+c_{2})y-c_{1}\cdot c_{2}=0$

$\therefore c_{1} + c_{2} = 1$ and $c_{1}\cdot c_{2} = -6$

$\Rightarrow c_{1} = 3$ and $c_{2} =-2$ OR $\Rightarrow c_{2} = 3$ and $c_{1} =-2$

$\therefore$ distance $= \left | \frac{c_{1}-c_{2}}{\sqrt{(1)^{2} + (-2)^{2}}} \right | = \boxed{\sqrt{5}}$ ,where $1$ and $-2$ are coeffients of $x$ and $y$ .