If the line $4x-5y+41=0$ is the perpendicular bisector of the line segment joining the points $A\left(1,1\right)$ and B  .The image of B in  x-axis is $\left(\mathrm{a},\mathrm{b}\right)$  then the numerical value of $\mathrm{a}+\mathrm{b}$  is

The point B is the image of $A\left(1,1\right)$ with respect to the line $4\mathrm{x}-5\mathrm{y}+41=0$

If $\left(\mathrm{h},\mathrm{k}\right)$ is the image of the point $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ with respect to the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ then $\frac{\mathrm{h}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{b}}=-\frac{2\left({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Hence, 

 $\frac{\mathrm{h}-1}{4}=\frac{\mathrm{k}-1}{-5}=-\frac{2\left(4-5+41\right)}{4^2+{\left(-5\right)}^2}$           

Simplify 

 $\frac{\mathrm{h}-1}{4}=\frac{\mathrm{k}-1}{-5}=-\frac{80}{41}$           

Hence,

$\begin{array}{c}\frac{\mathrm{h}-1}{4}=-\frac{80}{41}\\ \mathrm{h}=\frac{-320}{41}+1\\ =\frac{-320+41}{41}\\ =-\frac{279}{41}\end{array}$

I$\begin{array}{c}\frac{\mathrm{k}-1}{-5}=-\frac{80}{41}\\ \mathrm{k}=\frac{400}{41}+1\\ =\frac{441}{41}\end{array}$

Hence,$\left(\mathrm{h},\mathrm{k}\right)=\left(-\frac{279}{41},\frac{441}{41}\right)$

The image of $\mathrm{B}\left(\mathrm{h},\mathrm{k}\right)$ in x-axis is  $\left(\mathrm{a},\mathrm{b}\right)=\left(\mathrm{h},-\mathrm{k}\right)$

$\begin{array}{c}a+b=h-k\\ =\frac{-441-279}{41}\\ =-\frac{720}{41}\\ =-17.56\end{array}$

Therefore,  $\mathrm{a}+\mathrm{b}=\boxed{-17.56}$