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)

Let coordinates be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).

By formula,

Distance between two points (D) = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Calculating sides, we get :

$AB = \sqrt{[3 - (-3)]^2 + [1 - 5]^2} \\[1em] = \sqrt{[3 + 3]^2 + [-4]^2} \\[1em] = \sqrt{6^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units} \\[1em] BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-3)^2 + 2^2} \\[1em] = \sqrt{9 + 4} \\[1em] = \sqrt{13} \text{ units} \\[1em] CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} \\[1em] = \sqrt{(-1)^2 + (-7)^2} \\[1em] = \sqrt{1 + 49} \\[1em] = \sqrt{50} \text{ units} \\[1em] AD = \sqrt{[-1 - (-3)]^2 + [-4 - 5]^2} \\[1em] = \sqrt{[-1 + 3]^2 + [-9]^2} \\[1em] = \sqrt{[2]^2 + [-9]^2} \\[1em] = \sqrt{4 + 81} \\[1em] = \sqrt{85}.$

Since, AB ≠ BC ≠ CD ≠ AD,

Hence, no special quadrilateral can be formed from the given vertices.