Note that to show that a quadrilateral is a rhombus, it is sufficient to show that
(a) ABCD is a parallelogram, i.e., AC and BD have the same midpoint.
(b) a pair of adjacent edges are equal
Let A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a rhombus.
Thus, AC and BD have the same midpoint.
Hence, ABCD is a parallelogram
Now, using Distance Formula
d(A,B)= AB = √(5 – 1)2 + (3 – 0)2
⇒ AB = √(4)2 + (3)2
⇒ AB = √16 + 9
⇒ AB = √25 = 5 units
d(B,C)= BC = √(2 – 5)2 + (7 – 3)2
⇒ BC = √(-3)2 + (4)2
⇒ BC = √9 + 16
⇒ BC = √25 = 5 units
Therefore, adjacent sides are equal.
Hence, ABCD is a rhombus.
