To find the centroid of triangle \( ABC \) with vertices \( A(1,3) \), \( B(6,6) \), and \( C(3,1) \), we use the formula for the centroid \( G \) of a triangle given by the coordinates:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
where \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are the coordinates of the vertices \( A \), \( B \), and \( C \).
Plugging in the coordinates:
- \( A(1, 3) \): \( x_1 = 1, y_1 = 3 \)
- \( B(6, 6) \): \( x_2 = 6, y_2 = 6 \)
- \( C(3, 1) \): \( x_3 = 3, y_3 = 1 \)
Now, we calculate the \( x \) and \( y \) coordinates of the centroid:
\[ G_x = \frac{1 + 6 + 3}{3} = \frac{10}{3} \]
\[ G_y = \frac{3 + 6 + 1}{3} = \frac{10}{3} \]
Therefore, the coordinates of the centroid \( G \) are:
\[ G\left(\frac{10}{3}, \frac{10}{3}\right) \]
Thus, the centroid is at:
\[ \text{centroid} = \left(\frac{10}{3}, \frac{10}{3}\right) \]
This point confirms that the medians of the triangle intersect at a single point (the centroid).