Show that $(-2,-3), (6,3), (3,7),$ and $(-5,1)$ are the vertices of a rectangle.

(A) Let the vertices be $A(-2,-3), B(6,3), C(3,7),$ and $D(-5,1).$
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB^2 = (-2-6)^2 + (-3-3)^2 = (-8)^2 + (-6)^2 = 64 + 36 = 100 \implies AB = 10$
$BC^2 = (6-3)^2 + (3-7)^2 = (3)^2 + (-4)^2 = 9 + 16 = 25 \implies BC = 5$
$CD^2 = (3 - (-5))^2 + (7-1)^2 = (8)^2 + (6)^2 = 64 + 36 = 100 \implies CD = 10$
$DA^2 = (-5 - (-2))^2 + (1 - (-3))^2 = (-3)^2 + (4)^2 = 9 + 16 = 25 \implies DA = 5$
Since opposite sides are equal ($AB=CD=10$ and $BC=DA=5$),the quadrilateral is a parallelogram.
Next,we check the diagonal $AC$:
$AC^2 = (-2-3)^2 + (-3-7)^2 = (-5)^2 + (-10)^2 = 25 + 100 = 125$
In $\triangle ABC$,$AB^2 + BC^2 = 100 + 25 = 125 = AC^2$.
By the converse of the Pythagorean theorem,$\angle B = 90^\circ$.
Since the parallelogram has one right angle,it is a rectangle.