Minimum curve for the distance between two points at the plane The problem is to determine the curve y=y(x) in the plane, the lenght of which is given by the functional:
\begin{equation}
I(y)=\int_{x_1}^{x_2}\sqrt{1+(y')^2}dx=\int_{x_1}^{x_2}F(x,y,y')dx
\end{equation}
which will make the distance between two points of the plane minimum. 
In other words, we need to determine the curve for which $I(y)$ is minimum.
Now I just want to ask the following perhaps naive question, but I am not sure about it:
The Euler-Lagrange equation becomes:
\begin{equation}
\frac{d}{dx}\left(\frac{\partial{F}}{\partial(y')} \right)=0
\end{equation}
since the functional $F(x,y,y')=\sqrt{1+(y')^2}$ actually depends only on $(x,y')$, which means that $\partial{F}/{\partial{y}}=0$
Also I can see that (if I am right):
\begin{equation}
\frac{\partial{F}}{\partial(y')}=F_{y'}=\frac{y'}{\sqrt{1+(y')^2}}
\end{equation}
but the book ends up with the DE: $y''=0$ and finally the curve extremizing $I(y)$ is:
\begin{equation}
y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1
\end{equation}
I cannot see why $y''=0$ from Euler Lagrange and how I end up with that solution?
Thank you!

It is a constant:
$ \frac{\partial{F}}{\partial(y')}=F_{y'}=\frac{y'}{\sqrt{1+(y')^2}} = c $
$ \dfrac{y'^2}{ {1+y'^2}} =c^2  =\dfrac{2 y'y''}{ {2 y' y''}} $ 
by Quotient rule. Cross-multiply, 
$ {2 y'y''}{y '^2 } ={2 y'y''}+  {2 y'{^3}y''} $
$  y^{'} = 0, y^{''} =0 $
$ y = c_1, y = m \, x + c_2 $ which are straight lines in x-y plane.
There is another ready integrated form when $F(y, y^{'})$ only, often very useful.
Also, if $y'$ is a constant, is n't $y'' =0$ straight forward?
