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

#### Solution

Let A (4, 5); B (7, 6); C (6, 3) and D (3, 2) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point P(x,y) of two points `A(x_1, y_1)` and `B(x_2,y_2)` we use section formula as,

`P(x,y) = ((x_1 + x_2)/2, (y_1 + y_2)/2)`

So the mid-point of the diagonal AC is,

`Q(x,y) = ((4 + 6)/2, (5 + 3)/2)`

= (5,4)

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

Now to check if ABCD is a rectangle, we should check the diagonal length.

`AC = sqrt((6 - 4)^2 + (3 - 5)^2)`

`= sqrt(4 + )4`

`= 2sqrt2`

Similarly,

`BD = sqrt((7 - 3)^2 + (6 - 2)^2)`

`= sqrt(16 + 16)`

`= 4sqrt2`

Diagonals are of different lengths.

Hence ABCD is not a rectangle.