Answer
$$L=\sqrt{x^2+(\frac{2x}{x-3})^2}$$
Work Step by Step
Keep in mind that $$L=\sqrt{x^2+y^2}$$
So, thanks to equating slopes, the equality would be the following: $$\frac{y-2}{0-3}=\frac{0-2}{x-3}$$
So, with that in terms of solving for $y$, then the steps would be the following:
$$\begin{matrix}
\frac{y-2}{0-3}=\frac{0-2}{x-3}\\\\
-\frac{y-2}{3}=-\frac{2}{x-3}\\\\
-\frac{y-2}{3}(-3)=-\frac{2}{x-3}(-3)\\\\
y-2=-\frac{6}{x-3}\\\\
y-2+2=-\frac{6}{x-3}+2\\\\
y=-\frac{6}{x-3}+2\\\\
y=\frac{6}{x-3}+2(\frac{x-3}{x-3})\\\\
y=\frac{6}{x-3}+\frac{2x-6}{x-3}\\\\
y=\frac{6+2x-6}{x-3}\\\\
y=\frac{2x+6-6}{x-3}\\\\
y=\frac{2x}{x-3}
\end{matrix}$$
Since $L=\sqrt{x^2+y^2}$ and $y=\frac{2x}{x-3}$, then:
$$\begin{matrix}
L=\sqrt{x^2+y^2}\\\\
\mathbf{L=\sqrt{x^2+(\frac{2x}{x-3})^2}}
\end{matrix}$$
