Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square.

Solution:

Let A (4, 3); B (6, 4); C (5, 6) and D (3, 5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a square.

So we should find the lengths of sides of quadrilateral ABCD.

$\mathrm{AB}=\sqrt{(6-4)^{2}+(4-3)^{2}}$

$=\sqrt{4+1}$

$=\sqrt{5}$

$B C=\sqrt{(6-5)^{2}+(4-6)^{2}}$

$=\sqrt{1+4}$

$=\sqrt{5}$

$\mathrm{CD}=\sqrt{(3-5)^{2}+(5-6)^{2}}$

$=\sqrt{4+1}$

$=\sqrt{5}$

$\mathrm{AD}=\sqrt{(3-4)^{2}+(5-3)^{2}}$

$=\sqrt{1+4}$

$=\sqrt{5}$

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

$\mathrm{AC}=\sqrt{(5-4)^{2}+(6-3)^{2}}$

$=\sqrt{1+9}$

$=\sqrt{10}$

$\mathrm{BD}=\sqrt{(6-3)^{2}+(4-5)^{2}}$

$=\sqrt{9+1}$

$=\sqrt{10}$

All the sides as well as the diagonals are equal. Hence ABCD is a square.