Let’s A$(4,5)$, B$(7,6)$,C$(6,3)$ and D$(3,2)$ be the given points.

And, P be the point of intersection of AC and BD.

Coordinates of the mid-point of AC are

$(4+6/2,5+3/2)=(5,4)$

Coordinates of the mid-point of BD are

$(7+3/2,6+2/2)=(5,4)$

Therefore, it’s clearly seen that the mid-point of AC and BD are same.

thus, ABCD is a parallelogram.

Now we have,

$AC=\sqrt{\left[ {{\left( 6-4 \right)}^{2}}+{{(3-5)}^{2}} \right]}$

$=\sqrt{\left[ {{\left( 2 \right)}^{2}}+{{\left( -2 \right)}^{2}} \right]}$

$=\sqrt{\left[ 4+4 \right]}$

$=\sqrt{8}units$

And,

$BD=\sqrt{\left[ {{\left( 7-3 \right)}^{2}}+{{\left( 6-2 \right)}^{2}} \right]}$

$=\sqrt{\left[ {{\left( 4 \right)}^{2}}+{{\left( 4 \right)}^{2}} \right]}$

$=\sqrt{\left[ 16+16 \right]}$

$=\sqrt{32}Units$

Since,AC ≠ BD

Hence, ABCD is not a rectangle.