Q.Find the combined equation of the pair of bisectors of the angles between the lines represented by $2x^2 + 5xy + 2y^2 = 0$.

Given the pair of lines $ax^2+2hxy+by^2=0$ (with slopes $m_1,m_2$ satisfying $m_1+m_2=-2h/b$ and $m_1m_2=a/b$), 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 $\dfrac{m-m_1}{1+mm_1} = -\dfrac{m-m_2}{1+mm_2}$. Cross-multiplying, expanding, and collecting terms by powers of $m$ produces

Substituting $m_1+m_2=-2h/b$ and $m_1m_2=a/b$, and simplifying (multiplying through by $-b/2$), this becomes the compact quadratic

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:

As a worked example, the pair $x^2-3xy+2y^2=0$ — i.e. $(x-y)(x-2y)=0$ — has $a=1,h=-\tfrac32,b=2$, so its bisector pair is $-\tfrac32(x^2-y^2)=(1-2)xy=-xy$, which simplifies (multiplying by $-2$) to $3x^2-2xy-3y^2=0$.

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 $h(x^2-y^2)=(a-b)xy$ 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 $hm^2+(a-b)m-h=0$ always multiply to $-h/h=-1$ (whenever $h\neq0$), 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 $h(x^2-y^2)=0$ and hence $x^2-y^2=0$ (assuming $h\neq0$) — the bisectors are then simply $x=y$ and $x=-y$, reflecting the underlying symmetry of the original pair about the $45^\circ$ line whenever the coefficients of $x^2$ and $y^2$ are equal.