Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

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

The length of AB

$= \sqrt{(1 - 5)^2 + (5 - 6)^2}\\[1em] = \sqrt{(-4)^2 + (-1)^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17}$

The length of BC

$= \sqrt{(2 - 1)^2 + (1 - 5)^2}\\[1em] = \sqrt{1^2 + (-4)^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17}$

The length of CD

$= \sqrt{(6 - 2)^2 + (2 - 1)^2}\\[1em] = \sqrt{4^2 + 1^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17}$

The length of DA

$= \sqrt{(6 - 5)^2 + (2 - 6)^2}\\[1em] = \sqrt{1^2 + 4^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17}$

AB = BC = CD = DA = $\sqrt{17}$

The length of diagonal AC =

$= \sqrt{(2 - 5)^2 + (1 - 6)^2}\\[1em] = \sqrt{(-3)^2 + (-5)^2}\\[1em] = \sqrt{9 + 25}\\[1em] = \sqrt{34}$

The length of diagonal BD =

$= \sqrt{(6 - 1)^2 + (2 - 5)^2}\\[1em] = \sqrt{(-5)^2 + (-3)^2}\\[1em] = \sqrt{25 + 9}\\[1em] = \sqrt{34}$

So, AC = BD

Since all sides and diagonals are equal, the points form a square.

Hence, the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.