1. **State the problem:** We have a square with side length 8 units. One vertex E is at (1, 3), and vertex F lies on the same horizontal line (same y-coordinate) as E but 8 units to the right since the square's sides are parallel to the axes. 2. **Find coordinates of F:** Since E is at (1, 3) and the side length is 8, moving 8 units right along the x-axis gives $$F = (1 + 8, 3) = (9, 3).$$ 3. **Determine the other vertices:** The square has four vertices. We know E and F are on the top side (horizontal). The square extends 8 units down (since the square is aligned with axes and E is not at the bottom). 4. **Find coordinates of G:** G is the vertex directly below F, so it has the same x-coordinate as F and y-coordinate 8 units less than F's y-coordinate: $$G = (9, 3 - 8) = (9, -5).$$ 5. **Verify:** The side FG is vertical and length 8, consistent with the square's properties. **Final answer:** The coordinates of point G are **$(9, -5)$**.