Show that the points $A (1, 2), B (5, 4), C (3, 8)$ and $D (- 1, 6)$ are the vertices of a square.

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Solution

We are going to find distance between sides
$A B = \sqrt{(5 - 1)^{2} + (4 - 2)^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}$
$B C = \sqrt{(3 - 5)^{2} + (8 - 4)^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$
$C D = \sqrt{(- 1 - 3)^{2} + (6 - 8)^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}$
$D A = \sqrt{(1 + 1)^{2} + (2 - 6)^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$
$∴$ $A B = B C = C D = D A$
Now, we are going to find distance between diagonals.
$A C = \sqrt{(3 - 1)^{2} + (6 - 2)^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$
$B D = \sqrt{(- 1 + 5)^{2} + (6 - 4)^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}$
$∴$ $A C = B D$
So, here we can see distance between all sides and diagonals are equal.
$∴$ Points $A, B, C$ and $D$ are the vertices of a square.

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