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What is the area of a rectangle with vertices at (-4,0),(-3,1),(0,-2), and (-1,-3)?

User Xdeleon
by
6.2k points

1 Answer

4 votes

Answer:

The area of Rectangle with given vertices is 6 unit ²

Explanation:

Given points of vertices of rectangle as :

A = ( - 4, 0)

B = ( - 3 , 1)

C = ( 0 , - 2)

D = ( - 1 , - 3)

Now the measure of side AB =
\sqrt{(x_2 - x_1)^(2) + (y_2 - y_1)^(2)}

So, AB =
\sqrt{( - 3 + 4)^(2) + (1 - 0)^(2)}

AB =
√(2) unit

The measure of side BC =
\sqrt{(x_2 - x_1)^(2) + (y_2 - y_1)^(2)}

BC =
\sqrt{(0 +3)^(2) + (- 2 - 1)^(2)}

BC = 3
√(2) unit

The measure of side CD =
\sqrt{(x_2 - x_1)^(2) + (y_2 - y_1)^(2)}

CD =
\sqrt{( - 1 - 0)^(2) + ( - 3 + 2)^(2)}

CD =
√(2) unit

The measure of side DA =
\sqrt{(x_2 - x_1)^(2) + (y_2 - y_1)^(2)}

DA =
\sqrt{(- 4 + 1)^(2) + (0 + 3)^(2)}

DA = 3
√(2) unit

So, the measure of side AB = The measure of side CD =
√(2) unit

And The measure of side BC = The measure of side DA = 3
√(2) unit

So, Let Length = AB = CD

And Width = BC = DA

∴ The area of Rectangle = Length × Width unit²

Or, The area of Rectangle = AB × BC

So, The area of Rectangle =
√(2) unit × 3
√(2) unit

∴ The area of Rectangle = 6 unit ²

Hence The area of Rectangle with given vertices is 6 unit ² Answer

User Madhav Jha
by
5.3k points
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