4B. The point $T(1,2)$ is the centroid of the triangle $\triangle A B C$, the point $D(3,4)$ is the midpoint of side $B C$, and $S\left(\frac{14}{9}, \frac{10}{9}\right)$ is the circumcenter of the triangle $A B C$. Find the coordinates of the vertices of the triangle $A B C$.
Solution. One property of centroids gives us $\overline{A T}: \overline{T D}=2: 1$. According to this, $A(-3,-2)$. Therefore, the square of the radius of the circumscribed circle of triangle $\triangle A B C$ is $r^{2}=\overline{A S}^{2}=\frac{2465}{61}$, and the equation of the circle is

$$
(9 x-14)^{2}+(9 y-10)^{2}=2465
$$

The slope of the perpendicular bisector $D S$ of side $B C$ is $k=2$. Therefore, the slope of side $B C$ is

$$
k^{\prime}=\frac{1}{k}=-\frac{1}{2},
$$

and its equation is

$$
x+2 y-11=0
$$

By solving the system of equations

(1) and (2) we get

$$
B(7,2) \text { and } C(-1,6)
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_56aeac1763a3f1e8ce2ag-10.jpg?height=592&width=648&top_left_y=1045&top_left_x=820)