Name the type of quadrilateral formed,if any,by the following points,and give reasons for your answer: $(-3, 5), (3, 1), (0, 3), (-1, -4)$.

(NONE) Let the points $A(-3, 5), B(3, 1), C(0, 3),$ and $D(-1, -4)$ be the vertices of the quadrilateral.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$
$BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$
$CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$
$DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$
Since all four sides $AB, BC, CD,$ and $DA$ have different lengths,the points do not form any special type of quadrilateral (like a square,rectangle,or parallelogram). It is a general quadrilateral.