2. The point $T(1,2)$ is the centroid of triangle $ABC$; the point $D(3,4)$ is the midpoint of side $BC$, and the point $S\left(\frac{14}{9}, \frac{10}{9}\right)$ is the circumcenter of triangle $ABC$. Find the vertices of triangle $ABC$.
Solution. The centroid $T$ divides the median $A D$ in the ratio $1: 2$, so for $A\left(a_{1}, a_{2}\right)$ we have

$$
1=\frac{a_{1}+2 \cdot 3}{3}, \quad 2=\frac{a_{2}+2 \cdot 4}{3},
$$

from which we get $a_{1}=-3, a_{2}=-2$, i.e., $A(-3,-2)$.

Let $B\left(b_{1}, b_{2}\right), C\left(c_{1}, c_{2}\right)$. The point $D(3,4)$ is the midpoint of $B C$, so we have

$$
b_{1}+c_{1}=6, \quad b_{2}+c_{2}=8
$$

The point $S$ is the center of the circumscribed circle, so the radius $r$ of the circle will be

$$
r^{2}=\overline{S A}^{2}=\left(-3-\frac{14}{9}\right)^{2}+\left(-2-\frac{10}{9}\right)^{2}=\frac{2465}{81}
$$

From $\overline{S B}^{2}=r^{2}$ and $\overline{S C}^{2}=r^{2}$, we get

$$
\begin{aligned}
& \left(9 b_{1}-14\right)^{2}+\left(9 b_{2}-10\right)^{2}=2465 \\
& \left(9 c_{1}-14\right)^{2}+\left(9 c_{2}-10\right)^{2}=2465
\end{aligned}
$$

From (1), (2), and (3) we get

$$
b_{1}=7, b_{2}=2, c_{1}=-1, c_{2}=6 \text { or } b_{1}=-1, b_{2}=6, c_{1}=7, c_{2}=2 \text {. }
$$

Thus, $B(7,2), C(-1,6)$ or $B(-1,6), C(7,2)$.