The point P (a,b) lies on the straight line 3x-2y=13 and the point Q (b, a) lies on the staright line 4x-y=5 then the equation of line PQ is

$\begin{array}{l}P\left(a,b\right)lieson3x–2y–13=0\Rightarrow 3a-2b-13=0--\left(1\right)\\ Q\left(b,a\right)lieson4x-y-5=0\Rightarrow 4b-a-5=0\Rightarrow a-4b+5=0--\left(2\right)\\ Solving\left(1\right)\&\left(2\right)\\ ab1\\ \begin{array}{cccc}-2& -13& 3& -2\\ -4& 5& 1& -4\end{array}\\ \Rightarrow \frac{a}{-10-52}=\frac{b}{-13-15}=\frac{1}{-12+2}\\ \Rightarrow \frac{a}{-62}=\frac{b}{-28}=\frac{1}{-10}\\ \therefore a=\frac{-62}{-10}=\frac{31}{5},b=\frac{-28}{-10}=\frac{14}{5}\\ \therefore P\left(\frac{31}{5},\frac{14}{5}\right)andQ\left(\frac{14}{5},\frac{31}{5}\right)\end{array}$

$\begin{array}{l}SlopeofPQ=\frac{\frac{31}{5}-\frac{14}{5}}{\frac{14}{5}-\frac{31}{5}}=\frac{17}{-17}=-1\\ Equationof\overline{PQ}y=\frac{14}{5}=-1\left(x-\frac{31}{5}\right)\\ \Rightarrow 5y-14=-5x+31\\ \Rightarrow 5x+5y-45=0\\ \Rightarrow x+y=9\end{array}$