Prove that $A (0,0), B (7,0), C (7,5)$,and $D (0,5)$ are the vertices of a rectangle.

(N/A) To prove that the points $A(0,0), B(7,0), C(7,5)$,and $D(0,5)$ form a rectangle,we must show that opposite sides are equal and the diagonals are equal.
$1$. Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(7-0)^2 + (0-0)^2} = \sqrt{49} = 7$
$BC = \sqrt{(7-7)^2 + (5-0)^2} = \sqrt{25} = 5$
$CD = \sqrt{(0-7)^2 + (5-5)^2} = \sqrt{49} = 7$
$DA = \sqrt{(0-0)^2 + (0-5)^2} = \sqrt{25} = 5$
Since $AB = CD = 7$ and $BC = DA = 5$,opposite sides are equal.
$2$. Calculate the lengths of the diagonals:
$AC = \sqrt{(7-0)^2 + (5-0)^2} = \sqrt{49 + 25} = \sqrt{74}$
$BD = \sqrt{(0-7)^2 + (5-0)^2} = \sqrt{49 + 25} = \sqrt{74}$
Since the opposite sides are equal and the diagonals are equal,the quadrilateral $ABCD$ is a rectangle.