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Let A(2, 1), B(5, 2), C(6, 4) and D(3, 3) be the vertices of a parallelogram ABCD. Since, the diagonals of a parallelogram bisect each other.
AC 2 = (6 - 2) 2 + (4 - 1) 2 = (4) 2 + (3) 2 = 16 + 9 = 25
BC 2 = (6 - 5) 2 + (4 - 2) 2 = (1) 2 + (2) 2 = 1 + 4 = 5
AB 2 = (5 - 2) 2 + (2 - 1) 2 = (3) 2 + (1) 2 = 9 + 1 = 10
DC 2 = (6 - 3) 2 + (4 - 3) 2 = (3) 2 + (1) 2 = 9 + 1 = 10
AD 2 = (3 - 2) 2 + (3 - 1) 2 = (1) 2 + (2) 2 = 1 + 4 = 5
Since BC = AD and DC = AB, ABCD is a parallelogram.
AB 2 + BC 2 = 10 + 5 = 15
AB 2 + BC 2 ≠ AC 2
ΔABC is not right angled. Therefore parallelogram ABCD is not a rectangle.

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