$\left.\quad \begin{array}{ll}\text { Coordinate method on the plane }\end{array}\right]$

Given points $A(5 ; 5), B(8 ;-3)$ and $C(-4 ; 1)$. Find the coordinates of the intersection point of the medians of triangle $A B C$.
## First method.

It is known that the medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex. Therefore, if $D\left(x_{1} ; y_{1}\right)$ is the midpoint of segment $B C$, then $A M: M D=2: 1$. It is also known that the coordinates of the midpoint of a segment are the arithmetic means of the corresponding coordinates of its endpoints. Thus,

$$
x_{1}=\frac{8-4}{2}=2, y_{1}=\frac{-3+1}{2}=-1
$$

Since the point $M\left(x_{0} ; y_{0}\right)$ divides the segment $A D$ in the ratio $2: 1$, counting from point $A$, then by the theorem of proportional segments, the projection of point $M$ on the $O X$ axis divides the projection of segment $A D$ on this axis in the same ratio, i.e.,

$$
\frac{x_{0}-5}{2-x_{0}}=2
$$

From this, we find that $x_{0}=3$. Similarly, we find that $y_{0}=1$.

## Second method.

Let $M\left(x_{0} ; y_{0}\right)$ be the point of intersection of the medians of triangle $A B C$. Since the coordinates of the point of intersection of the medians of a triangle are the arithmetic means of the corresponding coordinates of the vertices of the triangle, then

$$
x_{0}=\frac{5+8-4}{3}=3, y_{0}=\frac{5-3+1}{3}=1
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_89dcdfe8766a8b2a52b5g-33.jpg?height=555&width=838&top_left_y=2098&top_left_x=610)

## Answer

$(3 ; 1)$.