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

Note that to show that a quadrilateral is a square, it is sufficient to show that

(a) ABCD is a parallelogram, i.e., AC and BD bisect each other

(b) a pair of adjacent edges are equal

(c) the diagonal AC and BD are equal.

Let the vertices of a quadrilateral are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).

Coordinates of the midpoint of AC are

Coordinates of the midpoint of BD are

Thus, AC and BD have the same midpoint.

Hence, ABCD is a parallelogram

Now, Using Distance Formula, we get

AB = √(x₂ – x₁)² + (y₂ – y₁)²

= √[(6 – 4)² + (4 - 3)²]

= √(2)² + (1)²

= √(4 + 1)

= √5 units

BC = √[(5 – 6)² + (6 - 4)²]

= √(-1)² + (2)²

= √(1 + 4)

= √5 units

Therefore, AB = BC = √5 units

Now, check for the diagonals

AC = √(5 – 4)² + (6 – 3)²

= √(1)² + (3)²

= √1 + 9

= √10 units

and

BD = √(3 - 6)² + (5 – 4)²

⇒ BD = √(-3)² + (1)²

⇒ BD = √9 + 1

⇒ BD = √10 units

∴ AC = BD

Hence, ABCD is a square.