Using the distance formula,show that the points $A(7,3)$,$B(3,0)$,$C(0,-4)$,and $D(4,-1)$ are the vertices of a rhombus.

(N/A) Let the vertices be $A(7,3)$,$B(3,0)$,$C(0,-4)$,and $D(4,-1)$.
To show that $ABCD$ is a rhombus,we must prove that all four sides are equal in length $(AB = BC = CD = DA)$ and the diagonals are not equal $(AC \neq BD)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(3 - 7)^2 + (0 - 3)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$BC = \sqrt{(0 - 3)^2 + (-4 - 0)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$CD = \sqrt{(4 - 0)^2 + (-1 - (-4))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$DA = \sqrt{(7 - 4)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Since $AB = BC = CD = DA = 5$,all sides are equal.
Now,calculate the diagonals:
$AC = \sqrt{(0 - 7)^2 + (-4 - 3)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}$.
$BD = \sqrt{(4 - 3)^2 + (-1 - 0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Since $AC \neq BD$,the diagonals are not equal.
Therefore,$ABCD$ is a rhombus.