To solve this problem, let's analyze the given information and use the section formula.

Given:

1. $C$ is the midpoint of line segment $AB$.
2. The coordinates of $A$ are $(12, 3)$, and the coordinates of $B$ are $(20, 13)$.
3. The coordinates of $E$ are $(28, 2)$.
4. The ratio of the lengths $DC : CE = 2 : 3$.
5. We need to find the coordinates of point $D$.

Steps to Solve:

1. Find the Coordinates of $C$: Since $C$ is the midpoint of $AB$, the coordinates of $C$ can be calculated as: $C = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right)$ Substituting the coordinates of $A$ and $B$: $C = \left( \frac{12 + 20}{2}, \frac{3 + 13}{2} \right) = \left( \frac{32}{2}, \frac{16}{2} \right) = (16, 8)$

2. Use the Section Formula to Find $D$: Since $D$ divides $CE$ in the ratio $2 : 3$, we can use the section formula, which states: $D = \left( \frac{m \cdot x_2 + n \cdot x_1}{m + n}, \frac{m \cdot y_2 + n \cdot y_1}{m + n} \right)$ where $C = (x_1, y_1)$ and $E = (x_2, y_2)$, and the ratio $m : n = 2 : 3$.

   Substituting the values: $D = \left( \frac{2 \cdot 28 + 3 \cdot 16}{2 + 3}, \frac{2 \cdot 2 + 3 \cdot 8}{2 + 3} \right)$ Calculating each coordinate: $x_D = \frac{56 + 48}{5} = \frac{104}{5} = 20.8$ $y_D = \frac{4 + 24}{5} = \frac{28}{5} = 5.6$

Conclusion:

The coordinates of point $D$ are $(20.8, 5.6)$.

Would you like more details or have any questions about this solution?

Here are 5 related questions to expand understanding:

1. How would the solution change if the ratio $DC : CE$ were different?
2. What if $C$ was not the midpoint of $AB$; how would that affect the coordinates of $C$?
3. Can you determine the length of line segments $DC$ and $CE$ with these coordinates?
4. What if point $D$ were closer to $C$ than $E$; how would that affect the coordinates?
5. How could this problem be solved if only the graphical representation were given?

Tip: The section formula is a useful tool for dividing a line segment into any given ratio, and it can be applied in various geometry problems involving coordinates.