General Form of a Pair of Straight Lines

Multiplying out two arbitrary (not necessarily through the origin) lines $l_1x+m_1y+n_1=0$ and $l_2x+m_2y+n_2=0$ gives the general second-degree equation

with $a=l_1l_2,\ b=m_1m_2,\ c=n_1n_2,\ 2h=l_1m_2+l_2m_1,\ 2g=l_1n_2+l_2n_1,\ 2f=m_1n_2+m_2n_1$ — a non-homogeneous equation of degree two (unlike §6.5.1's homogeneous case, since the lines need not pass through the origin now).

Condition for $(\ast)$ to represent a pair of straight lines. Not every second-degree equation factors into two linear equations — treating $(\ast)$ as a quadratic in $x$: $ax^2+2(hy+g)x+(by^2+2fy+c)=0$, so $x=\dfrac{-(hy+g)\pm\sqrt{(hy+g)^2-a(by^2+2fy+c)}}{a}$, i.e. $ax+hy+g=\pm\sqrt{(h^2-ab)y^2+2(gh-af)y+(g^2-ac)}$. For this to be a genuine straight line (linear in $x,y$), the expression under the square root must itself be a perfect square in $y$ — a discriminant-zero condition that, after simplifying and dividing by $a$, reduces to

(the determinant expansion — covered fully in the next chapter — reproduces exactly this same expression).

Results without proof, quoted for use:

1. If $(\ast)$ represents a pair of straight lines, they are parallel iff $\dfrac{a}{h}=\dfrac{h}{b}=\dfrac{g}{f}$ (equivalently $bg^2=af^2$), and the distance between the two parallel lines is then $2\sqrt{\dfrac{g^2-ac}{a(a+b)}}$ or, equivalently, $2\sqrt{\dfrac{f^2-bc}{b(a+b)}}$.
2. The homogeneous pair $ax^2+2hxy+by^2=0$ (through the origin) is parallel to $(\ast)$'s pair (same two slopes) — $(\ast)$'s slopes depend only on the coefficients of $x^2,xy,y^2$, unaffected by $g,f,c$; the origin pair meets at $(0,0)$, while $(\ast)$'s pair meets at $P\left(\dfrac{hf-bg}{ab-h^2},\ \dfrac{gh-af}{ab-h^2}\right)$.
3. The angle between $(\ast)$'s two lines equals the angle between the origin pair's two lines: $\theta=\tan^{-1}\!\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|$ — again unaffected by $g,f,c$.
4. $(\ast)$'s two lines are perpendicular iff $a+b=0$ — exactly the same condition as for the homogeneous pair, for the same reason (only $a,h,b$ govern slope; $g,f,c$ only govern position).

Separating a general pair into its two lines. First factor the second-degree terms $ax^2+2hxy+by^2$ alone into two linear factors (as in §6.5.1); then find the two constant terms by writing $ax^2+2hxy+by^2+2gx+2fy+c\equiv(l_1x+m_1y+c_1)(l_2x+m_2y+c_2)$ and comparing the coefficients of $x$ and $y$ (using $c_1c_2=c$ as a consistency check). …