Distance Between Parallel Lines and Point of Intersection

Once we know that $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a genuine pair of straight lines (via the condition of the previous section), two further practical questions arise: if the two lines are parallel, how far apart are they; and if they intersect, where exactly do they meet?

Distance between parallel lines. Two lines are parallel exactly when they share the same homogeneous (degree-2) part, i.e. when $h^2=ab$, so we may write them as $lx+my+c_1=0$ and $lx+my+c_2=0$ for the same $l,m$ but different constants $c_1,c_2$. Their product is

so matching coefficients, $a=l^2$, $b=m^2$, $2g=l(c_1+c_2)$, $c=c_1c_2$. The distance between the two parallel lines is the standard formula $\dfrac{|c_1-c_2|}{\sqrt{l^2+m^2}}$. To express $|c_1-c_2|$ via $a,g,c$, use $(c_1-c_2)^2=(c_1+c_2)^2-4c_1c_2 = \dfrac{4g^2}{l^2} - 4c = \dfrac{4(g^2-ac)}{a}$ (since $l^2=a$), so $|c_1-c_2| = 2\sqrt{\dfrac{g^2-ac}{a}}$. Since $l^2+m^2=a+b$, the distance becomes

As a worked check, $4x^2+12xy+9y^2-10x-15y+4=0$ has $a=4,h=6,b=9$ (indeed $h^2=36=ab$, confirming parallel lines), $g=-5,c=4$, giving $d=2\sqrt{\dfrac{25-16}{4\times13}}=2\sqrt{\dfrac{9}{52}}=\dfrac{3\sqrt{13}}{13}$ — matching the direct computation from the factored lines $2x+3y-1=0$ and $2x+3y-4=0$.

Point of intersection. When the lines are not parallel ($h^2\neq ab$), their point of intersection is found using elementary calculus: the point where two linear factors meet is where both partial derivatives of the quadratic expression vanish simultaneously. Differentiating $ax^2+2hxy+by^2+2gx+2fy+c$ with respect to $x$ and to $y$ and setting each to zero gives

a pair of simultaneous linear equations solved by Cramer's rule:

For example, $x^2-y^2-x+5y-6=0$ has $a=1,h=0,b=-1,g=-\tfrac12,f=\tfrac52$; solving $x-\tfrac12=0$ and $-y+\tfrac52=0$ gives the intersection point $\left(\tfrac12,\tfrac52\right)$, which indeed satisfies both factored lines $x-y+2=0$ and $x+y-3=0$. …