Geodesic on a plane I guess that each geodesic on a plane is a straight line. Is it right? What can I use to prove it?
I guess I have to use somehow Levi-Civita connection.
 A: This just adds a few words and references, but maybe it will help you organize your thoughts. If you use the standard coordinates for $\mathbf R^2$, then the Euclidean metric has constant matrix
\[
(g_{ij}) = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}.
\]
Therefore, the Christoffel symbols of the Levi-Civita connection are all zero. If you trace through the definition of the covariant derivative along a curve $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ applied to the tangent vector field of $\gamma$, then you end up with the geodesic equation, which in this case requires that
\[
\frac{\partial^2\gamma_1}{\partial t^2} = \frac{\partial^2\gamma_2}{\partial t^2} = 0.
\]
A: I think it is quite simple, it is a classical example of calculus of variations. You find a function that minimizes a Lagrangian that represents the function's length. You use Euler-Lagrange equation to find that the function's second derivative is zero, concluding that it must be a straight line.
For a full proof, see here.
A: Similar to what Dylan said, let $\mathbf{u}$ be a tangent vector to a path $\gamma$, then $\nabla_{u}u=0$ describes geodesic motion.  Then noting that the Christoffel symbols vanish, we have $\partial_{A}u^{B}=0$.  If we parameterize in $\tau$ we have
$$ \frac{d u^{x}(\tau)}{d\tau}=0$$
and
$$ \frac{du^y(\tau)}{d\tau}=0$$
which imples $u^x=v_{0}^x$ and $u^y=v_{0}^y$ and $x(\tau)=v_{0}^x\tau+x_0$ and $y(\tau)=v_{0}^y\tau +y_0$.  These are lines.
I hope this helps.
A: There is a much simpler answer that uses a bit of theory of ordinary differential equations: 
1) If you take a straight line in the plane, its second derivative is $0$, so it is a geodesic by definition.
2) There is a unique geodesic through any given point of the plane in any given direction. This is true because the geodesic equations have a unique solution.
3) Since there is a line through any given point in any given direction there are no other geodesics.
A: here is more basic approach:
consider two points $\mathbf p$ and $\mathbf q$ in $R^2$. we will prove that the straight line between them is the shortest curve that connects them. thus for any two point in the geodesic curve, the the curve between them is a line and therefore the the whole curve is a straight linebetween $\mathbf p$ and $\mathbf q$. 
let $\mathbf u$ be a unit vector and $\gamma$ be some curve connecting between $\mathbf p$ and $\mathbf q$.
from Cauchy–Schwarz inequality one have: $$<\mathbf u,\dot\gamma> <= \left\lVert u \right\rVert*\left\lVert \dot\gamma \right\rVert = \left\lVert \dot\gamma \right\rVert$$
since $\mathbf u$ is a unit vector.
taking the integral over the path from p to q we get:
$$\int <\mathbf u,\dot\gamma> <= \int \left\lVert \dot\gamma \right\rVert$$ 
the last integral is just the length of the curve by definition. however if we take $ \mathbf u = \frac{\mathbf p - \mathbf q}{\left\lVert \mathbf p - \mathbf q \right\rVert} $ we get that the first integral equals:
$$ \int <\mathbf u,\dot\gamma> = \frac {\int <\mathbf p - \mathbf q,\dot\gamma>}{\left\lVert \mathbf p - \mathbf q \right\rVert} = \frac { <\mathbf p - \mathbf q,\mathbf p - \mathbf q>}{\left\lVert \mathbf p - \mathbf q \right\rVert} = \left\lVert \mathbf p - \mathbf q \right\rVert $$
by definition of norms induced from the inner product.
so we get that the length of the curve is at least the length of the line passing through the points. hence the line is the shortest way from $\mathbf p$ to $\mathbf q$.
for the uniqueness note that the Cauchy–Schwarz inequality is equality iff the elements in the product are parallel. hence the tangent to the curve is always in the direction of the specific line between $\mathbf p$ and $\mathbf q$. thus the curve is the line.
