Let equation of line $L$, perpendicular to $5x-y=1$ be $x+5y=c$

Given that are of $\Delta AOB$  is $5$.

We know

$\left\{ {are,A = \frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_2}} \right) + {x_2}\left( {{y_3} – {y_1}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right]} \right\}$

$ \Rightarrow 5 = \frac{1}{2}\left[ {c\left( {\frac{c}{5}} \right)} \right]$

$\because \left( \begin{array}{l}
\left( {{x_1},{y_1}} \right) = \left( {10,0} \right)\left( {{x_3},{y_3}} \right) = \left( {0,\frac{c}{5}} \right)\\
\left( {{x_2},{y_2}} \right) = \left( {c,0} \right)
\end{array} \right)$

$ \Rightarrow c =  \pm \sqrt {50} $

$\therefore $ Equation of line $L$ is $x + 5y =  \pm \sqrt {50} $

Distance betwween $L$ and line $x+5y=0$ is 

$d = \left. {\frac{{ \pm \sqrt {50}  – 0}}{{\sqrt {{1^2} + {5^2}} }}} \right| = \frac{{\sqrt {50} }}{{\sqrt {26} }} = \frac{5}{{\sqrt {13} }}$