Angle-Bisector Pair of a Homogeneous Pair of Lines

Given a pair of lines $ax^2+2hxy+by^2=0$ through the origin, the two lines that bisect the angles between them are also two straight lines through the origin, and they too can be written as one combined equation. To derive it, write the pair as $y=m_1x$ and $y=m_2x$ (assuming $b\neq0$), where $m_1,m_2$ are the roots of $bm^2+2hm+a=0$, so that

Let $y=mx$ be one of the two bisectors. A bisector makes an equal acute angle with each of the two lines it bisects, and since the internal and external bisectors sit on opposite sides, the two angle-equality conditions combine (with the sign convention appropriate to a bisector lying "between" the two lines) into

Cross-multiplying and expanding both sides,

Collecting all terms on one side and grouping by powers of $m$,

Now substitute $m_1+m_2=-2h/b$ and $m_1m_2=a/b$:

Multiplying throughout by $-b/2$ to clear denominators,

This quadratic in $m$ has both bisector slopes as its two roots; note that the product of its roots is $-h/h=-1$, confirming — as a built-in consistency check — that the two bisectors are always perpendicular to each other, exactly as we expect of the internal and external bisector of any angle. Writing $m=y/x$ and clearing the denominator $x^2$,

which is the combined equation of the pair of angle bisectors. As a worked example, take the pair $x^2-3xy+2y^2=0$, i.e. $(x-y)(x-2y)=0$, so $a=1$, $h=-\tfrac32$, $b=2$. The bisector pair is $-\tfrac32(x^2-y^2)=(1-2)xy=-xy$, i.e. (multiplying by $-2$) $3x^2-2xy-3y^2=0$.

It is worth noting why the "equidistant from both lines" idea and this slope-based derivation must agree: a point lies on an angle bisector precisely when its perpendicular distance to one line equals its perpendicular distance to the other, and this equal-distance condition is exactly what forces the bisector to make an equal angle with each line — the two characterisations (equal perpendicular distance, and equal angle) describe the very same set of points, just viewed through different geometric lenses. The slope-based route above is generally the faster one to execute in an examination setting, since it only ever manipulates the two symmetric quantities $m_1+m_2$ and $m_1m_2$, which come straight from $a,h,b$ without any need to solve for $m_1,m_2$ individually first. …