6.1. On the parabola $y=x^{2}-4 x-1$, three different points $A\left(x_{a}, y_{a}\right), B\left(x_{b}, y_{b}\right), C\left(x_{c}, y_{c}\right)$ are taken. It is known that $x_{c}=5$ and $y_{a}=y_{b}$. Find the abscissa of the point of intersection of the medians of triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e273449a9e482933432fg-2.jpg?height=463&width=323&top_left_y=331&top_left_x=204)
# Answer: 3

Solution. Since $y_{a}=y_{b}$, points $A$ and $B$ are symmetric with respect to the axis of the parabola, which means the midpoint $K$ of segment $A B$ has an abscissa equal to the abscissa of the vertex of the parabola $x_{k}=2$. Next, by the property of the median, the point of intersection of the medians $M$ divides the median $C K$ in the ratio $2: 1$, counting from the vertex, so the abscissa $x_{m}$ of point $M$ is found from the equality $x_{c}-x_{k}=3\left(x_{m}-x_{k}\right)$. Finally: $x_{m}=\frac{x_{c}+2 x_{k}}{3}=3$.