Angle between lines represented by $ax^2 + 2hxy + by^2 = 0$

This section asks: given the homogeneous pair $ax^2+2hxy+by^2=0$, what is the angle between its two lines?

If we know the slope of a line we can always find the angle it makes with the coordinate axes, so the natural approach is to work through the slopes $m_1, m_2$ of the pair (assuming $b \neq 0$, so both slopes are defined; if $b=0$ one line is the $Y$-axis and the angle can be found directly from the other line's slope).

Since $m_1m_2 = \dfrac{a}{b}$, and two lines with slopes $m_1,m_2$ are perpendicular exactly when $m_1m_2=-1$, substituting gives $\dfrac{a}{b}=-1$, i.e. $a=-b$, i.e. $a+b=0$. So: the lines represented by $ax^2+2hxy+by^2=0$ are perpendicular to each other if and only if $a+b=0$.

When the lines are not perpendicular, the acute angle between them is given by the following theorem.

Theorem 4.4. The acute angle $\theta$ between the lines represented by $ax^2+2hxy+by^2=0$ is given by $\tan\theta = \dfrac{2\sqrt{h^2-ab}}{a+b}$.

Proof. Let $m_1,m_2$ be the slopes of the two lines, so $m_1+m_2=\dfrac{-2h}{b}$ and $m_1m_2=\dfrac{a}{b}$. Using the algebraic identity $(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$, substitute: $(m_1-m_2)^2 = \dfrac{4h^2}{b^2} - \dfrac{4a}{b} = \dfrac{4h^2-4ab}{b^2} = \dfrac{4(h^2-ab)}{b^2}$. Taking the square root, $m_1-m_2 = \pm\dfrac{2\sqrt{h^2-ab}}{b}$.

Since $\theta$ is the acute angle between the two lines, $\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right| = \dfrac{\left|\pm\dfrac{2\sqrt{h^2-ab}}{b}\right|}{\left|1+\dfrac{a}{b}\right|} = \dfrac{2\sqrt{h^2-ab}}{a+b}$ (taking the sign of $a+b$ so the value comes out as the acute-angle tangent, i.e. non-negative). $\blacksquare$

Remark (coincidence). The lines represented by $ax^2+2hxy+by^2=0$ are coincident exactly when $m_1=m_2$, i.e. $m_1-m_2=0$, i.e. $\dfrac{2\sqrt{h^2-ab}}{b}=0$, i.e. $h^2-ab=0$, i.e. $h^2=ab$. This matches the coincidence remark already noted in section 4.2.

Worked Examples

Example 1. Show that lines represented by $3x^2-4xy-3y^2=0$ are perpendicular.

Here $a=3,h=-2,b=-3$, so $a+b=3+(-3)=0$, and by the perpendicularity condition the lines are perpendicular.

Example 2. Show that lines represented by $x^2+4xy+4y^2=0$ are coincident.

Here $a=1,h=2,b=4$, so $h^2-ab=4-4=0$; the lines are coincident.

Example 3. Find the acute angle between the lines represented by (i) $x^2+xy=0$, (ii) $x^2-4xy+y^2=0$, (iii) $3x^2+2xy-y^2=0$, (iv) $2x^2-6xy+y^2=0$, (v) $xy+y^2=0$.

1. $a=1,h=\dfrac12,b=0$: $\tan\theta=\dfrac{2\sqrt{1/4-0}}{1+0}=\dfrac{2(1/2)}{1}=1$, so $\theta=45^\circ=\dfrac{\pi}{4}$.
2. $a=1,h=-2,b=1$: $\tan\theta=\dfrac{2\sqrt{4-1}}{1+1}=\dfrac{2\sqrt3}{2}=\sqrt3$, so $\theta=60^\circ=\dfrac{\pi}{3}$.
3. $a=3,h=1,b=-1$: $\tan\theta=\dfrac{2\sqrt{1+3}}{3-1}=\dfrac{2(2)}{2}=2$, so $\theta=\tan^{-1}(2)$.
4. $a=2,h=-3,b=1$: $\tan\theta=\dfrac{2\sqrt{9-2}}{2+1}=\dfrac{2\sqrt7}{3}$, so $\theta=\tan^{-1}\left(\dfrac{2\sqrt7}{3}\right)$.
5. $a=0,h=\dfrac12,b=1$: $\tan\theta=\dfrac{2\sqrt{1/4-0}}{0+1}=1$, so $\theta=45^\circ=\dfrac{\pi}{4}$. Example 4. Find the combined equation of lines through the origin making angle $\dfrac{\pi}{6}$ with the line $3x+y-6=0$. …