If in triangle $ABC,$$A=\left(1,10\right)$, circum centre$\left(-\frac{1}{3},\frac{2}{3}\right)$  and ortho centre $=\left(\frac{11}{3},\frac{4}{3}\right)$ , then the coordinates of the midpoint of the side opposite to  A are

Orthocentre$=O=\left(\frac{11}{3},\frac{4}{3}\right)$ Circum centre$=S=\left(-\frac{1}{3},\frac{2}{3}\right)$

 Centroid G divides $\overline{OS}$ in the ratio $\overline{OS}$   2:1
$\therefore$ Centriod $G=\left(1,\frac{8}{9}\right)$

Let D be the mid point of BC, then G divides $\overline{AD}$ in the ratio $2:1$

$D=\frac{3G-A}{2}=\left(1,\frac{-11}{3}\right)$