Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram.

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).

$A B=\sqrt{(5-2)^{2}+(2-1)^{2}}=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10}$ units

$B C=\sqrt{(6-5)^{2}+(4-2)^{2}}=\sqrt{(1)^{2}+(2)^{2}}=\sqrt{1+4}=\sqrt{5}$ units

$C D=\sqrt{(3-6)^{2}+(3-4)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9+1}=\sqrt{10}$ units

$A D=\sqrt{(3-2)^{2}+(3-1)^{2}}=\sqrt{(1)^{2}+(2)^{2}}=\sqrt{1+4}=\sqrt{5}$ units

Thus, $A B=C D=\sqrt{10}$ units and $B C=A D=\sqrt{5}$ units

So, quadrilateral $A B C D$ is a parallelogram

Also, $A C=\sqrt{(6-2)^{2}+(4-1)^{2}}=\sqrt{(4)^{2}+(3)^{2}}=\sqrt{16+9}=\sqrt{25}=5$ units

$B D=\sqrt{(3-5)^{2}+(3-2)^{2}}=\sqrt{(-2)^{2}+(1)^{2}}=\sqrt{4+1}=\sqrt{5}$ units

But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.