Prove that the points $A(1, -2)$,$B(2, 3)$,$C(-3, 2)$,and $D(-4, -3)$ are the vertices of a rhombus.

(N/A) Let the vertices be $A(1, -2)$,$B(2, 3)$,$C(-3, 2)$,and $D(-4, -3)$.
To prove that $ABCD$ is a rhombus,we must show that all four sides are equal in length,but the diagonals are not equal.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(2 - 1)^2 + (3 - (-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$
$BC = \sqrt{(-3 - 2)^2 + (2 - 3)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$
$CD = \sqrt{(-4 - (-3))^2 + (-3 - 2)^2} = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$
$DA = \sqrt{(1 - (-4))^2 + (-2 - (-3))^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$
Since $AB = BC = CD = DA = \sqrt{26}$,all sides are equal.
Now,calculate the lengths of the diagonals $AC$ and $BD$:
$AC = \sqrt{(-3 - 1)^2 + (2 - (-2))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$BD = \sqrt{(-4 - 2)^2 + (-3 - 3)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$
Since all sides are equal $(AB=BC=CD=DA)$ and the diagonals are not equal $(AC \neq BD)$,the quadrilateral $ABCD$ is a rhombus.