To find the centroid of triangle ABC with vertices \( A(-2, -3) \), \( B(4, 0) \), and \( C(-2, 2) \), we use the formula for the centroid, which is given by:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Here, \( (x_1, y_1) = A(-2, -3) \), \( (x_2, y_2) = B(4, 0) \), and \( (x_3, y_3) = C(-2, 2) \).
First, we calculate the x-coordinates of the centroid:
\[ x_G = \frac{-2 + 4 - 2}{3} = \frac{0}{3} = 0 \]
Next, we calculate the y-coordinates of the centroid:
\[ y_G = \frac{-3 + 0 + 2}{3} = \frac{-1}{3} \]
Thus, the coordinates of the centroid \( G \) are:
\[ G\left( 0, -\frac{1}{3} \right) \]
The centroid \( G(0, -\frac{1}{3}) \) is indeed a single point where the three medians of the triangle intersect.
Comparing with the given options:
- (−1, \( \frac{5}{3} \))
- (0, \( \frac{5}{3} \))
- (0, −13)
- (0, −\( \frac{1}{3} \))
The correct answer is:
\[ \boxed{(0, -\frac{1}{3})} \]