Angle Between a Pair of Lines

Suppose $ax^2+2hxy+by^2=0$ represents two real lines through the origin (so $h^2>ab$, as in the previous section), with slopes $m_1$ and $m_2$. We can find $m_1+m_2$ and $m_1m_2$ directly from the coefficients, without first solving for the individual lines. Writing the equation as a quadratic in $y/x$ (dividing by $x^2$, assuming $b\neq0$):

whose roots are precisely $m_1=y_1/x_1$ and $m_2=y_2/x_2$, the two slopes. By the sum and product of roots of a quadratic,

The angle $\theta$ between two lines of slopes $m_1,m_2$ satisfies the standard formula $\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$. To express this in terms of $a,h,b$ alone, first compute $(m_1-m_2)^2$ using the identity $(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$:

Also, $1+m_1m_2 = 1+\dfrac{a}{b} = \dfrac{a+b}{b}$. Substituting both into the slope-angle formula, the factor of $b$ cancels between numerator and denominator:

This is the angle-between-the-lines formula, valid whenever $h^2>ab$ (so the square root is real) and $a+b\neq0$. When $a+b=0$ the formula's denominator vanishes, meaning $\tan\theta\to\infty$, i.e. $\theta=90^\circ$ — the two lines are perpendicular; this special case is examined fully in the next section.

As a worked example, take once more $2x^2-3xy-2y^2=0$, whose lines are $x-2y=0$ (slope $\tfrac12$) and $2x+y=0$ (slope $-2$). Here $a=2$, $h=-\tfrac32$, $b=-2$, so $a+b=0$ and the formula predicts $\theta=90^\circ$. Checking directly: the product of the slopes is $\tfrac12\times(-2)=-1$, which is exactly the condition for two lines to be perpendicular — confirming the formula.

A second worked example shows the formula handling a genuinely acute, non-special angle. Take $2x^2-7xy+3y^2=0$: here $a=2$, $h=-\tfrac72$, $b=3$, so $h^2-ab=\tfrac{49}{4}-6=\tfrac{25}{4}$ and $a+b=5$. Then

Factoring the equation directly gives $(2x-y)(x-3y)=0$ (slopes $2$ and $\tfrac13$), and indeed $\tan\theta=\left|\dfrac{2-\frac13}{1+2\times\frac13}\right|=\left|\dfrac{5/3}{5/3}\right|=1$, matching exactly. …