If 2x – 3y – 5= 0 is the perpendicular bisector $\text{ of the line segment joining }(3,-4)\text{ and }(\mathrm{\alpha },\mathrm{\beta })\text{. }$$\text{ then find }\mathrm{\alpha }+\mathrm{\beta }\text{. }$

$\text{ If }(\mathrm{\alpha },\mathrm{\beta })\text{ is the image of the point }(3,-4)\text{ w.r.t }$st.line 2x – 3y – 5 = 0 then
$\frac{\mathrm{h}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{-2\left({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}$
$\begin{array}{l}\frac{\mathrm{\alpha }-3}{2}=\frac{\mathrm{\beta }+4}{-3}=\frac{-2(2.3-3(-4)-5)}{(2{)}^2+(3{)}^2}\\ \Rightarrow \frac{\mathrm{\alpha }-3}{2}=\frac{\mathrm{\beta }+4}{-3}=\frac{-2(6+12-5)}{13}\\ \therefore \mathrm{\alpha }=-1,\mathrm{\beta }=2\\ \therefore \mathrm{\alpha }+\mathrm{\beta }=1\end{array}$