# Perpendicular bisectors of hyperbolic lines

I want to prove the following basic property of hyperbolic lines in $$IR^{2,1}$$.

If x $$\in$$ $$H^2$$ and l is a line in $$H^2$$ then there is a unique line l' through x orthogonal to l.

I want to prove this in the hyperboloid model. Let $$\langle\,,\rangle$$ denote the Lorentz scalar product. Consider a line l. l is the intersection of $$H^2$$ with a 2-dimensional linear subspace

\begin{align*} U= & \; \{y \in IR^{2,1} \mid | \langle\,y,n \rangle=0\} \end{align*}

where $$\langle\,n,n \rangle=1$$. I have to construct a normal vector n' such that the associated plane U' contains x (that is $$\langle\,x,n'\rangle=0$$) and $$\langle\,n',n\rangle=0$$. Then the line l'=$$H^2 \cap U'$$ should intersect l orthogonally. The problem is that I don't see how to construct n' or prove the uniqueness. I cannot apply the Gram-Schmidt-Algorithm since this would also change x. Is there an elementary way to show this?

• Is the scenario relevant? If radical axis of two orthogonal circles in the Euclidean plane passes through origin, then it can be converted to a Poincare polar model. A circle cutting them orthogonality can be taken as the hyperbolic boundary.Two orthogonal (interior segments) circles are possible.A sketch needs to be added. – Narasimham Nov 14 '18 at 20:55

## 1 Answer

I would do it as follows.

Assume that $$l$$ is the X axis, i.e., the set of points with the $$y$$ coordinate equal to 0.

Let $$C=(0,0,1)$$, $$X_a = \left(\begin{array}{ccc} \cosh a&0&\sinh a\\0&1&0\\\sinh a&0&\cosh a\end{array}\right)$$ be the isometry which shifts $$C$$ $$a$$ units to the right, and $$Y_a$$ be the similar isometry which shifts $$C$$ $$a$$ units up. Hence $$l = \{X_aC: a \in \mathbb{R}\}$$.

The orthogonal line at point $$X_aC$$ is $$\{X_a Y_b C: b \in \mathbb{R}\}$$. If you multiply the matrices it is clear that $$a$$ and $$b$$ exist and are unique (if I remember correctly, for the point $$(x,y,z)$$ we have $$b = \sinh(y)$$ and $$a = \sinh(x/\cosh(b))$$).