Angle Between a Pair of Lines

Acute angle between the two lines represented by ax²+2hxy+by²=0.

For the pair $ax^2+2hxy+by^2=0$ with slopes $m_1, m_2$, use the angle formula:

$$\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

From Vieta's: $m_1+m_2 = -2h/b$ and $m_1m_2 = a/b$.

Compute $(m_1-m_2)^2$:

$$(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2 = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4(h^2-ab)}{b^2}$$

And:

$$1 + m_1m_2 = 1 + \frac{a}{b} = \frac{a+b}{b}$$

Substituting:

$$\tan\theta = \frac{\frac{2\sqrt{h^2-ab}}{b}}{\frac{a+b}{b}} = \frac{2\sqrt{h^2-ab}}{a+b}$$ $$\boxed{\tan\theta = \frac{2\sqrt{h^2-ab}}{a+b}}$$

The formula requires $h^2 \geq ab$ for real lines and $a+b \neq 0$ (otherwise the lines are perpendicular and $\tan\theta$ is undefined).