To find the centroid of triangle ABC with vertices A(4, 4), B(9, 1), and C(3, -1), we will use the formula for the centroid \( G \) of a triangle given its vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
We have:
- \( A(4, 4) \): \( (x_1, y_1) = (4, 4) \)
- \( B(9, 1) \): \( (x_2, y_2) = (9, 1) \)
- \( C(3, -1) \): \( (x_3, y_3) = (3, -1) \)
Now, we can compute the x-coordinate of the centroid \( G \):
\[ x_G = \frac{x_1 + x_2 + x_3}{3} = \frac{4 + 9 + 3}{3} = \frac{16}{3} \]
Next, we compute the y-coordinate of the centroid \( G \):
\[ y_G = \frac{y_1 + y_2 + y_3}{3} = \frac{4 + 1 - 1}{3} = \frac{4}{3} \]
Thus, the coordinates of the centroid \( G \) are:
\[ G = \left( \frac{16}{3}, \frac{4}{3} \right) \]
The centroid \( G \) is the point where the three medians intersect. Since we have found the centroid, we have effectively proven that the medians of triangle ABC meet at a single point.
The final answer, expressed in fractions, is:
The centroid \( G \) is at \( \left( \frac{16}{3}, \frac{4}{3} \right) \).