# How to prove that there is a unique geodisic segment that is pependicular to two other geodesics? [duplicate]

Here is the question

I'm not sure how to proceed with this question. An idea that I have is that I assume that geodisic from l to m is perpendicular to one of the geodisics, then I have to show that the other geodisic is parallel but not sure if that's the right way to proceed and the steps involved. I'd appreciate a hint or any help in proceeding with this.

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## marked as duplicate by Davide Giraudo, Dan Rust, Amzoti, Lord_Farin, azimutJun 29 '13 at 15:00

This question was marked as an exact duplicate of an existing question.

As far as I understand things, to talk about geodesics you need to have a metric specified. Since you've tagged this in Complex Analysis, can we assume you are working in the complex plane $\mathbb{C} \cong \mathbb{R}^2$? What sort of metric can be given on this manifold? Once you've got a metric, you'll need to solve the geodesic equation system ($i=1,2$) \begin{align} \frac{d^2 x_i}{dt^2} + \sum\limits_{j,k=1}^2 \Gamma^i \text{}_{jk} \frac{dx_j}{dt} \frac{dx_k}{dt} = 0, \end{align} where the Christoffel symbol $\Gamma^i \text{}_{jk}$ is given, for each $i,j,k=1,2$, by the equation \begin{align}\Gamma^i \text{}_{jk} = \frac{1}{2} \sum\limits_{l=1}^2 g^{il} \left[\frac{\partial g_{kl}}{\partial x_j} + \frac{\partial g_{jl}}{\partial x_k} - \frac{\partial g_{jk}}{\partial x_l}\right],\end{align} where $g_{ij}$ denotes a component of the metric and $g^{ij}$ a component of the inverse metric. This will give you a family of curves $(x_1(t),x_2(t))$, and plugging in initial and final values will give a particular solution: the geodesic between the initial and final point. Does this help?
if you want to go through it, the metric, with $y > 0,$ is $ds^2 = (dx^2 + dy^2)/y^2.$ – Will Jagy Apr 22 '13 at 18:30