The ortho centre of a triangle lies on the variable line $(1+2\lambda )x-(2+\lambda )y=4+5\lambda$  and circum centre lies on $(1+2\mu )x-(2+\mu )y=4\mu +5 \forall \lambda ,\mu \in R$, The centroid of this triangle is $\left(x_1,y_1\right)$ then $x_1+y_1$ is 

Given orthocentre lies on the line $(1+2\lambda )x-(2+\lambda )y=4+5\lambda$

$$\begin{gathered} \Rightarrow (x-2y-4)+\lambda (2x-y-5)=0 \\ P.I of x-2y-4=0 and 2x-y-5=0 is (2,-1) \end{gathered}$$

Orthocentre is H(2,-1)

similarly given circumcentre lies on the line $(1+2\mu )x-(2+\mu )y=4\mu +5$

$\Rightarrow (x-2y-5)+\mu (2x-y-4)=0$
Circumcentre is S (1,-2)

Since centriod divides orthocentre and circumcentre in the ratio 2:1

then centiod=($\frac{4}{3},\frac{-5}{3})$