Distance Between Parallel Lines

Distance between the two parallel lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0, when h² = ab. Equivalently written as 2√((f²−bc)/(b(a+b))) using the y-coefficient form.

When $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents two parallel lines ($h^2=ab$, $af^2=bg^2$), they have the form:

$$(\sqrt{a}x+\sqrt{b}y+p)(\sqrt{a}x+\sqrt{b}y+q) = 0$$

where $p+q = 2g/\sqrt{a} = 2f/\sqrt{b}$ and $pq = c$.

The two lines are $\sqrt{a}x+\sqrt{b}y+p = 0$ and $\sqrt{a}x+\sqrt{b}y+q = 0$.

Distance between two parallel lines $lx+my+p=0$ and $lx+my+q=0$:

$$d = \frac{|p-q|}{\sqrt{l^2+m^2}} = \frac{|p-q|}{\sqrt{a+b}}$$

Now:

$$(p-q)^2 = (p+q)^2-4pq = \frac{4g^2}{a} - 4c = \frac{4(g^2-ac)}{a}$$ $$|p-q| = \frac{2\sqrt{g^2-ac}}{\sqrt{a}}$$

Therefore:

$$d = \frac{2\sqrt{g^2-ac}/\sqrt{a}}{\sqrt{a+b}} = 2\sqrt{\frac{g^2-ac}{a(a+b)}}$$

Alternative form using $f$: Since $af^2=bg^2$, an equivalent expression is $2\sqrt{(f^2-bc)/(b(a+b))}$.

Check: For $d > 0$, we need $g^2 > ac$ (for distinct parallel lines; $g^2 = ac$ gives coincident lines).