Scalar product invariance
Let's play it according to the rules of tensor calculus.
Introduce polar coordinates and calculate the local base
vectors of that system:
$$
\left. \begin{matrix} x = r\cos(\phi) \\
y = r\sin(\phi) \end{matrix} \right\} \quad \Longrightarrow \quad
\left\{ \begin{matrix} \vec{c}_r =
\left[ \begin{matrix} \partial x/\partial r\\
\partial y/\partial r\end{matrix} \right]
= \left[\begin{matrix}\cos(\phi)\\ \sin(\phi)\end{matrix}\right] \\
\vec{c}_\phi =
\left[ \begin{matrix} \partial x/\partial \phi\\
\partial y/\partial \phi\end{matrix} \right]
= r \left[\begin{matrix}-\sin(\phi)\\ \cos(\phi)\end{matrix}\right]
\end{matrix} \right.
$$
Next calculate the metric tensor:
$$
\left[ \begin{matrix} g_{rr} & g_{r\phi} \\ g_{\phi r} & g_{\phi\phi}
\end{matrix} \right] =
\left[ \begin{matrix} \left(\vec{c}_r\cdot\vec{c}_r\right) &
\left(\vec{c}_r\cdot\vec{c}_\phi\right)\\
\left(\vec{c}_r\cdot\vec{c}_\phi\right) &
\left(\vec{c}_\phi\cdot\vec{c}_\phi\right) \end{matrix}\right] =
\left[ \begin{matrix} 1 & 0 \\ 0 & r^2 \end{matrix}\right]
$$
Then all we can say is something about the infinitesimal length:
$$
ds^2 = \left[ \begin{matrix} dr & d\phi \end{matrix} \right]
\left[ \begin{matrix} 1 & 0 \\ 0 & r^2 \end{matrix}\right]
\left[ \begin{matrix} dr \\ d\phi \end{matrix} \right]
\quad \Longrightarrow \quad
ds^2 = dx^2 + dy^2 = dr^2 + r^2 d\phi^2
$$
It is shown in the answer by Upax how to integrate this to obtain
a finite length. Substitute $r=t\sqrt{2}$ and $\phi=\pi/4$ into:
$$
s = \int_0^1 \sqrt{\left(\frac{dr}{dt}\right)^2+r^2\left(\frac{d\phi}{dt}\right)^2} dt =
\int_0^1 \sqrt{\left(\sqrt{2}\right)^2+\left(t\sqrt{2}\right)^2\cdot 0}\; dt = \sqrt{2}
$$