Pair of Lines Passing Through the Origin

The simplest case is a pair of lines both through the origin: $y-m_1x=0$ and $y-m_2x=0$. Their combined equation is

which suggests the general form: any homogeneous second-degree equation

(every term of degree exactly $2$; 'homogeneous' means the degree is constant across all terms) represents a pair of straight lines through the origin, with slopes $m_1,m_2$ recoverable from the coefficients. Being homogeneous of degree $2$ is precisely what guarantees the represented lines pass through the origin — the origin trivially satisfies any such equation.

Separating the two lines. Divide the equation by $x^2$ and substitute $m=y/x$ (the slope for a line through the origin): $ax^2+2hxy+by^2=0$ becomes $bm^2+2hm+a=0$, a quadratic in $m$ whose two roots $m_1,m_2$ are the slopes of the two component lines — either factor the original quadratic in $x,y$ directly, or solve this quadratic in $m$ and write $y=m_1x,\ y=m_2x$. …