Angle Bisectors of a Pair of Lines

Combined equation of the angle bisectors of the pair ax²+2hxy+by²=0. The bisectors are always perpendicular to each other.

Let the pair $ax^2+2hxy+by^2=0$ represent lines $y=m_1x$ and $y=m_2x$, i.e. $m_1x-y=0$ and $m_2x-y=0$.

A point $P(x,y)$ lies on a bisector if it is equidistant from both lines:

$$\frac{m_1x-y}{\sqrt{m_1^2+1}} = \pm\frac{m_2x-y}{\sqrt{m_2^2+1}}$$

The combined equation of both bisectors is obtained by multiplying:

$$\left(\frac{m_1x-y}{\sqrt{m_1^2+1}}\right)^2 = \left(\frac{m_2x-y}{\sqrt{m_2^2+1}}\right)^2$$

After expanding and using $m_1+m_2 = -2h/b$ and $m_1m_2 = a/b$, simplification yields:

$$\boxed{\frac{x^2-y^2}{a-b} = \frac{xy}{h}}$$

Key properties of the bisectors

The bisectors are always perpendicular. The combined equation $h(x^2-y^2) = (a-b)xy$ is a homogeneous pair; the sum of coefficients of $x^2$ and $y^2$ is $h + (-h) = 0$, confirming perpendicularity.

If $a = b$: the bisectors are $x^2-y^2=0$, i.e. $y = \pm x$ — the coordinate diagonals.

If $h = 0$: the original pair has no $xy$ term; the coordinate axes themselves are the bisectors.