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I came across the following question:

Find the equation of the bisectors of the angle between the lines represented by $$3x^2-5xy+4y^2=0$$

In the solution, they directly used the formula of combined equation of angle bisectors for a pair of straight lines passing through origin, i.e. $$h(x^2-y^2)-(a-b)xy=0 \tag{$\star$}$$ where the given lines have equation $a x^2 + 2 h x y + b y^2 = 0$.

And after using this formula we get $5x^2-2xy-5y^2=0$ as the combined equation of angle bisectors for a pair of straight lines represented by $3x^2-5xy+4y^2=0$.

Now, two questions:

  1. How is it possible that angle bisectors of two imaginary lines are real lines?

  2. For a pair of real lines passing through the origin, how is formula $(\star)$ derived?

Please guide me through this confusion.

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Sameer Nilkhan
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  • See this entry on Brilliant for a derivation of the formula. – Blue May 18 '20 at 09:09
  • Read the classic book "The Methods of Plane Projective Geometry Based on the Use of General Homogeneous Coordinates" by Maxwell, E.A to answer this sort of question. – P. Lawrence May 18 '20 at 13:48
  • Angle bisectors of imaginary lines make some sort of sense if you view them as the principal axes of a conic, in this case a degenerate one. The associated matrix to this conic is real symmetric, so has a complete set of orthogonal eigenvectors, which point in the directions of the principal axes. – amd May 18 '20 at 20:16

1 Answers1

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One way to get at the formula is to suppose that one angle bisector is inclined at some angle $\theta$ (so that the other bisector is at angle $\theta+\pi/2$), and that the two given lines are inclined by $\pm \phi$ relative to that bisector.

Then the combined equation of the lines is given by $$\left(y - x\tan(\theta+\phi)\right)\left(y-x\tan(\theta-\phi)\right) = 0 \quad\to\quad a x^2 + b y^2 + 2 h x y = 0 \tag{1}$$ where $$a = \sin(\theta+\phi)\sin(\theta-\phi) \qquad b = \cos(\theta+\phi)\cos(\theta-\phi) \qquad h = - \sin\theta\cos\theta \tag{2}$$

And the combined equation of the bisectors is given by $$(y-x\tan\theta)(y-\tan(\theta+\pi/2))=0 \quad\to\quad p (x^2-y^2) + q x y = 0 \tag{3}$$ where $$\begin{align} p &= \phantom{-}\sin\theta \cos\theta &&= -h \\ q &= -\cos 2\theta &&=\phantom{-}a-b \end{align} \tag{4}$$ Dividing $(3)$ by $-1$ gives equation $(\star)$ in the question. $\square$

Blue
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