Given the pair of lines ax2+2hxy+by2=0 (with slopes m1,m2 satisfying m1+m2=−2h/b and m1m2=a/b), the two lines that bisect the angles between them are themselves a pair of straight lines through the origin, and — just as with the original pair — they can be captured in one combined equation rather than derived one bisector at a time.
If y=mx is one of the bisectors, it makes an equal angle with each of the two original lines, which (using the tangent-of-angle-between-two-lines formula on both sides, with the sign convention that places the bisector "between" the two lines) leads to 1+mm1m−m1=−1+mm2m−m2. Cross-multiplying, expanding, and collecting terms by powers of m produces
m2(m1+m2)−2m(m1m2−1)−(m1+m2)=0.
Substituting m1+m2=−2h/b and m1m2=a/b, and simplifying (multiplying through by −b/2), this becomes the compact quadratic
hm2+(a−b)m−h=0,
whose two roots are exactly the two bisector slopes. A nice consistency check falls straight out of this: the product of the two roots is −h/h=−1, confirming that the two angle bisectors are always themselves perpendicular to one another — as any pair of internal/external bisectors of an angle must be. Writing m=y/x and clearing the denominator gives the combined equation of the bisector pair:
h(x2−y2)=(a−b)xy.
As a worked example, the pair x2−3xy+2y2=0 — i.e. (x−y)(x−2y)=0 — has a=1,h=−23,b=2, so its bisector pair is −23(x2−y2)=(1−2)xy=−xy, which simplifies (multiplying by −2) to 3x2−2xy−3y2=0.
This formula is significant because it entirely bypasses the longer route of finding both individual lines first, then constructing each bisector separately via equal-perpendicular-distance reasoning — a single substitution into h(x2−y2)=(a−b)xy gives the whole bisector pair directly from the original coefficients. It is also worth noting a built-in structural guarantee: because the two roots of the bisector-slope quadratic hm2+(a−b)m−h=0 always multiply to −h/h=−1 (whenever h=0), the two bisectors it produces are always perpendicular to one another, no matter what a,h,b happen to be — which matches the elementary geometric fact that the internal and external bisector of any angle are themselves at right angles. This is a genuinely useful self-check: after computing a bisector pair, confirming that its own a′+b′=0 (perpendicularity condition, from the earlier section) holds is an easy way to catch an arithmetic slip in the substitution.
A further special case is worth flagging: if a=b, the right-hand side (a−b)xy vanishes, forcing h(x2−y2)=0 and hence x2−y2=0 (assuming h=0) — the bisectors are then simply x=y and x=−y, reflecting the underlying symmetry of the original pair about the 45∘ line whenever the coefficients of x2 and y2 are equal.