Show that the points (2, 1), (5, 2), (6, 4) and (3, 3) are the angular pointsof a parallelogram. Is the figure a rectangle?

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.