The problem asks us to find the coordinates of point $D$ such that points $A(-2, 3)$, $B(2, 8)$, $C(5, 2)$, and $D$ form a parallelogram.

1. Problem Description

The problem asks us to find the coordinates of point $D$ such that points $A(-2, 3)$, $B(2, 8)$, $C(5, 2)$, and $D$ form a parallelogram.

2. Solution Steps

In a parallelogram $ABCD$, the midpoints of the diagonals $AC$ and $BD$ coincide. Let $D = (x, y)$.

The midpoint of $AC$ is given by $M_{AC} = (\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}) = (\frac{-2 + 5}{2}, \frac{3 + 2}{2}) = (\frac{3}{2}, \frac{5}{2})$.

The midpoint of $BD$ is given by $M_{BD} = (\frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}) = (\frac{2 + x}{2}, \frac{8 + y}{2})$.

Since $ABCD$ is a parallelogram, $M_{AC} = M_{BD}$. Therefore,

$\frac{3}{2} = \frac{2 + x}{2}$ and $\frac{5}{2} = \frac{8 + y}{2}$.

Solving for $x$:

$3 = 2 + x$, which gives $x = 3 - 2 = 1$.

Solving for $y$:

$5 = 8 + y$, which gives $y = 5 - 8 = -3$.

Thus, the coordinates of point $D$ are $(1, -3)$.

3. Final Answer

(1, -3)