Reflection of the point $\left(1,1\right)$ with respect to the line $4x+3y-5=0$ is $\left(\mathrm{\alpha },\mathrm{\beta }\right)$ then $\mathrm{\alpha }+\mathrm{\beta }$

If $\left(h,k\right)$ is the image of the point $\left(x_1,y_1\right)$ 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, if  $\left(\mathrm{\alpha },\mathrm{\beta }\right)$  is the image of the point $\left(1,1\right)$ with respect to the line  $4x+3y-5=0$ then  $\frac{\mathrm{\alpha }-1}{4}=\frac{\mathrm{\beta }-1}{3}=-\frac{2\left(4\left(1\right)+3\left(1\right)-5\right)}{4^2+3^2}$
Simplify $\frac{\mathrm{\alpha }-1}{4}=\frac{\mathrm{\beta }-1}{3}=-\frac{4}{25}$
 
It implies 

$\begin{array}{c}\frac{\mathrm{\alpha }-1}{4}=-\frac{4}{25}\\ \mathrm{\alpha }-1=-\frac{16}{25}\\ \mathrm{\alpha }=\frac{9}{25}\end{array}$

and 

$\begin{array}{c}\frac{\mathrm{\beta }-1}{3}=-\frac{4}{25}\\ \mathrm{\beta }-1=-\frac{12}{25}\\ \mathrm{\beta }=\frac{13}{25}\end{array}$

Therefore, $\alpha +\beta =\frac{13}{25}+\frac{9}{25}=\boxed{\frac{22}{25}}$