The perpendicular distance from $\left(5,4\right)$ to the line passing through the point  $\left(1,1\right)$and having the slope $5$ is

$\text{ The equation of the line passing through the point }(1,1)\text{. and having slope }5\text{ is }$

$\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-1=5\left(x-1\right)\\ y-1=5x-5\\ 5x-y-4=0\end{array}$

$\text{ The perpendicular distance from a point }P\left(x_1,y_1\right)\text{ to the line }cx+by+c=0\text{ is }\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$\text{ Hence the perpendicular distance from a point }(5,4)\text{ to the line }5x-y-4=0\text{ is }$

$\left|\frac{5\left(5\right)-4-4}{\sqrt{25+1}}\right|=\boxed{\frac{17}{\sqrt{26}}}$