What is the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1) ? enter your answer in the box. do not round any side lengths.

Area of the rectangle is 24

Step-by-step explanation:

Given the vertices of the rectangle to be (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1)

Let the the vertices be at point ABCD respectively

so that A(-6, 3)  B(-3,6) C(1,2) D(-2 -1)

To find the area, we need to first find the length and height of the rectangle

We let AB to be the length of the rectangle and BC be the height of the triangle. so first, we will find the distance between the two points AB

A(-6, 3)  B(-3,6)

Using the distance formula

|AB| = √(  -  )²   +  ( - )²

       =√(-3+6)²    +   (6-3)²

     =√(3)²  +  (3)²

     = √(9+9)

     =√18

    =√(9 ×2)

     =√9  ×  √2

    =3√2

|AB| =  3√2

The length of the rectangle is  3√2

We will now proceed to find the height of the rectangle which is the distance between point the two point BC

B(-3,6) C(1,2)

Using the distance formula;

|BC |=  √(  -  )²   +  ( - )²

    =√(1+3)²  + (2-4)²

     =√4²  +  4²

     =  √(16 + 16)

    =√32

    =√(16 × 2)

     =√16  ×  √2

    =4√2

|BC|  = 4√2

Therefore, the height of the triangle is 4√2

Area of rectangle = l × h

                            = |AB| .  |BC|

                              = 3√2  ×  4√2

                                =12 (2)

                                =24

Therefore  area of the rectangle is 24

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5

b

step-by-step explanation: