Answer
An arc length parametrization:
${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {1 + {m^2}} }},\frac{{ms}}{{\sqrt {1 + {m^2}} }}} \right)$
Work Step by Step
We have $y=m x$. Setting $t=x$, the line can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,mt} \right)$.
The derivative is ${\bf{r}}'\left( t \right) = \left( {1,m} \right)$. The speed is
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,m} \right)\cdot\left( {1,m} \right)} = \sqrt {1 + {m^2}} $
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
Since $m$ is a constant, we get
$s\left( t \right) = \sqrt {1 + {m^2}} \mathop \smallint \limits_0^t {\rm{d}}u = t\sqrt {1 + {m^2}} $
Thus, the inverse of $s\left( t \right)$ is $t = \frac{s}{{\sqrt {1 + {m^2}} }}$.
Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {t,mt} \right)$ gives the arc length parametrization:
${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {1 + {m^2}} }},\frac{{ms}}{{\sqrt {1 + {m^2}} }}} \right)$
