Show that the figure ABCD whose vertices are the points (-2,2),(4,2),(-2,-1) and (4,-1) is a rectangle?

To show that the figure ABCD is a rectangle, we need to prove that the opposite sides are parallel and equal in length, and that the diagonals are equal in length and bisect each other.

1. Opposite sides are parallel:
The slope of line AB is (2-2)/(4-(-2)) = 0
The slope of line CD is (-1-(-1))/(4-(-2)) = 0
Since the slopes of AB and CD are both 0, they are parallel.

The slope of line BC is (2-(-1))/(4-(-2)) = 1
The slope of line AD is (2-(-1))/(-2-4) = 1
Since the slopes of BC and AD are both 1, they are parallel.

2. Opposite sides are equal in length:
The length of AB = √((4-(-2))^2 + (2-2)^2) = √36 = 6
The length of CD = √((4-(-2))^2 + (-1-(-1))^2) = √36 = 6
The length of BC = √((4-4)^2 + (2-(-1))^2) = √9 = 3
The length of AD = √((-2-(-2))^2 + (2-(-1))^2) = √9 = 3

3. Diagonals are equal in length and bisect each other:
The length of diagonal AC = √((-2-4)^2 + (2-(-1))^2) = √52
The length of diagonal BD = √((4-(-2))^2 + (2-(-1))^2) = √52
Since the lengths of both diagonals are equal, and they intersect at the midpoint of each other, the diagonals bisect each other.

Therefore, since all the conditions for a rectangle are satisfied, we can conclude that the figure ABCD is a rectangle.

Step 1: Calculate the Lengths of the Sides

- AD = √[(4 - (-2))^2 + (2 - 2)^2] = √[36 + 0] = 6
- BC = √[(4 - (-2))^2 + (-1 - 2)^2] = √[36 + 9] = 3√5
- AB = √[(4 - (-2))^2 + (2 - (-1))^2] = √[36 + 9] = 3√5
- CD = √[(-2 - 4)^2 + (2 - (-1))^2] = √[36 + 9] = 3√5

Step 2: Check if Opposite Sides are Equal

- AD = BC (both equal to 3√5)
- AB = CD (both equal to 3√5)

Step 3: Check if Diagonals are Equal

- AC = √[(-2 - 4)^2 + (2 - (-1))^2] = √[36 + 9] = 3√5
- BD = √[(4 - (-2))^2 + (2 - (-1))^2] = √[36 + 9] = 3√5

Conclusion:

Since opposite sides are equal and diagonals are equal, the figure ABCD is a rectangle.