If a straight line $L$ is perpendicular to the line $5x-y=1$, the area of the triangle formed by $L$ and coordinate axis is 5, then the equation of the line is

Equation of line which is perpendicular to $ax+by+c=0$  is $bx-ay+k=0$

Equation of line K which is perpendicular to $5x-y-1=0$  is 

$\begin{array}{l}-1x-5y+k=0\\ \Rightarrow x+5y-k=0\\ \Rightarrow x+5y=k\end{array}$

Area of triangle formed by L with coordinate axes is 5

$\begin{array}{l}\frac{1}{2}\left|ab\right|=5\\ L=x+5y=k\to \frac{x}{k}+\frac{y}{\frac{k}{5}}=1\\ \Rightarrow \frac{1}{2}|k\times \frac{k}{5}|=5\\ \Rightarrow \frac{k^2}{10}=5\to k=\pm 50\\ \Rightarrow k=\pm 5\sqrt{2}\end{array}$

Required equations are $x+5y=5\sqrt{2}\text{\hspace{0.17em}and\hspace{0.17em}}x+5y=-5\sqrt{2}$.