Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.

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

The length of PQ

$= \sqrt{(6 - 0)^2 + (2 - (-4))^2}\\[1em] = \sqrt{6^2 + 6^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em]$

The length of RS

$= \sqrt{(-3 - 3)^2 + (-1 - 5)^2}\\[1em] = \sqrt{-6^2 + (-6)^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em]$

The length of QR

$= \sqrt{(3 - 6)^2 + (5 - 2)^2}\\[1em] = \sqrt{(-3)^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em]$

The length of SP

$= \sqrt{(-3 - 0)^2 + (-1 - (-4))^2}\\[1em] = \sqrt{-3^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em]$

PQ = RS
QR = SP

The length of diagonal QS =

$= \sqrt{(-3 - 6)^2 + (-1 - 2)^2}\\[1em] = \sqrt{(-9)^2 + (-3)^2}\\[1em] = \sqrt{81 + 9}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em]$

The length of diagonal PR =

$= \sqrt{(3 - 0)^2 + (5 - (-4))^2}\\[1em] = \sqrt{3^2 + 9^2}\\[1em] = \sqrt{9 + 81}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em]$

So, QS = PR

Since opposite sides are equal, and the diagonals are equal, we can conclude that the quadrilateral PQRS is a rectangle.

Hence, the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.