Name the type of quadrilateral formed,if any,by the following points,and give reasons for your answer: $(-1,-2), (1,0), (-1,2), (-3,0)$.

(D) Let the points $A(-1,-2), B(1,0), C(-1,2),$ and $D(-3,0)$ be the vertices of the quadrilateral.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
Diagonal $AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$
Diagonal $BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$
Since all sides are equal $(AB = BC = CD = DA = 2\sqrt{2})$ and both diagonals are equal $(AC = BD = 4)$,the quadrilateral is a square.