Plot the points A(3, 2), B(-2, 2), C(-2, -3) and D(-3, 3). Proof that ABCD is a rectangle.

To plot the points $A(3, 2)$, $B(-2, 2)$, $C(-2, -3)$, and $D(3, -3)$ on a graph paper:

• For $A(3, 2)$, start at the origin $(0,0)$, move 3 units to the right along the x-axis, then 2 units up along the y-axis.
• For $B(-2, 2)$, start at the origin, move 2 units to the left along the x-axis, then 2 units up along the y-axis.
• For $C(-2, -3)$, start at the origin, move 2 units to the left along the x-axis, then 3 units down along the y-axis.
• For $D(3, -3)$, start at the origin, move 3 units to the right along the x-axis, then 3 units down along the y-axis. Connect the points in order $A \to B \to C \to D \to A$ to form the quadrilateral.

Length of $AB$:

$$AB = \sqrt{(-2 - 3)^2 + (2 - 2)^2} = \sqrt{(-5)^2 + (0)^2} = \sqrt{25} = 5 \text{ units}$$

Length of $BC$:

$$BC = \sqrt{(-2 - (-2))^2 + (-3 - 2)^2} = \sqrt{(0)^2 + (-5)^2} = \sqrt{25} = 5 \text{ units}$$

Length of $CD$:

$$CD = \sqrt{(3 - (-2))^2 + (-3 - (-3))^2} = \sqrt{(5)^2 + (0)^2} = \sqrt{25} = 5 \text{ units}$$

Length of $DA$:

$$DA = \sqrt{(3 - 3)^2 + (2 - (-3))^2} = \sqrt{(0)^2 + (5)^2} = \sqrt{25} = 5 \text{ units}$$

Since $AB = CD = 5$ units and $BC = DA = 5$ units, the opposite sides are equal. In fact, all four sides are equal.

Length of $AC$:

$$AC = \sqrt{(-2 - 3)^2 + (-3 - 2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} \text{ units}$$

Length of $BD$:

$$BD = \sqrt{(3 - (-2))^2 + (-3 - 2)^2} = \sqrt{(5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} \text{ units}$$

Since $AC = BD = \sqrt{50}$ units, the diagonals are equal.

A quadrilateral is a rectangle if its opposite sides are equal and its diagonals are equal.
From Step 2, we found that $AB = CD$ and $BC = DA$.
From Step 3, we found that $AC = BD$.
Therefore, $ABCD$ is a rectangle. (Specifically, since all four sides are equal and the diagonals are equal, $ABCD$ is a square, which is a special type of rectangle).

The proof that $ABCD$ is a rectangle is complete.