State whether the following statement is true or false. Justify your answer.
The points $A(-1, -2)$,$B(4, 3)$,$C(2, 5)$,and $D(-3, 0)$ in that order form a rectangle.

(A) True.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Length of sides:
$AB = \sqrt{(4 - (-1))^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
$BC = \sqrt{(2 - 4)^2 + (5 - 3)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$CD = \sqrt{(-3 - 2)^2 + (0 - 5)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
Since opposite sides are equal ($AB = CD$ and $BC = DA$),it is a parallelogram.
$2$. Length of diagonals:
$AC = \sqrt{(2 - (-1))^2 + (5 - (-2))^2} = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}$
$BD = \sqrt{(-3 - 4)^2 + (0 - 3)^2} = \sqrt{(-7)^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58}$
Since the diagonals are equal $(AC = BD)$,the parallelogram is a rectangle.