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Prove that the points (4, 5), (7, 6), (6, 3), (3, 2) are the vertices of a parallelogram. Is it a rectangle?

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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.

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