Let 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)
Thus, it’s clearly seen that the mid-point of AC and BD are same.
So, ABCD is a parallelogram.
Now,
AC = √[(6 – 4)2 + (3 – 5)2]
= √[(2)2 + (-2)2]
= √[4 + 4]
= √8 units
And,
BD = √[(7 – 3)2 + (6 – 2)2]
= √[(4)2 + (4)2]
= √[16 + 16]
= √32 units
Since, AC ≠ BD
Therefore, ABCD is not a rectangle.