Angle Between the Pair of Lines

Acute angle θ between the two lines of ax² + 2hxy + by² = 0 (or the general pair). Uses the identity tan θ = (m₁−m₂)/(1+m₁m₂) with m₁−m₂ expressed via the discriminant.

The angle $\theta$ between lines with slopes $m_1$ and $m_2$:

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

Now:

$$(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2 = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4(h^2-ab)}{b^2}$$ $$m_1-m_2 = \frac{2\sqrt{h^2-ab}}{b} \quad (h^2 > ab)$$ $$1+m_1m_2 = 1 + \frac{a}{b} = \frac{a+b}{b}$$

Therefore:

$$\tan\theta = \left|\frac{2\sqrt{h^2-ab}/b}{(a+b)/b}\right| = \frac{2\sqrt{h^2-ab}}{|a+b|}$$ $$\tan\theta = \frac{2\sqrt{h^2-ab}}{a+b}$$

(taking the acute angle, so $|a+b|$ in the denominator — sign of $a+b$ does not matter for $\tan\theta > 0$).

Special cases:

• $a+b = 0$: $\tan\theta \to \infty \Rightarrow \theta = 90°$ — perpendicular lines
• $h^2 = ab$: $\tan\theta = 0 \Rightarrow \theta = 0°$ — coincident lines
• This formula applies equally to the general pair $ax^2+2hxy+by^2+2gx+2fy+c=0$ — the angle between the lines depends only on $a$, $h$, $b$.