The coordinates of the vertices of a rectangle are (-3,5), (0,-4), (3,7), and (6,-2). How do you find the area of this figure?

In this case it is a good idea to draw a quick sketch so that you can see what you are dealing with.
Tony B

We can use Pythagoras to solve this as appropriately projecting lines from any point will form a triangle. The vertical and horizontal lengths of which can be read of from the axis.

The distance between two points is `sqrt( (x_1-x_2)^2+(y_1-y_2)^2)`

Let the distance between two points be `d_i`

Distance between points A and B

`d_1=sqrt([3-(-3)]^2+[7-5]^2)`

`d_1=sqrt(36+4)=sqrt(2^2xx10)=2sqrt(10)`

Distance between points B and C

`d_2=sqrt([6-3]^2+[7-(-2)]^2)`

`d_2=sqrt(9+81)=sqrt(3^2xx10)=3sqrt(10)`
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Area `=d_1xxd_2 = 2sqrt10 xx 3 sqrt10 = 60`

The four vertices are `A (-3,5) B (0,-4),C (3,7),D (6,-2)`

Distance between two points `D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2`

Side `AB= sqrt ((-3-0)^2+(5+4)^2)=sqrt90 ~~ 9.49`unit

Side `BC= sqrt ((0-3)^2+(-4-7)^2)=sqrt130 ~~11.40`unit

Side `CD= sqrt ((3-6)^2+(7+2)^2)=sqrt90 = 9.49`unit

Side `AD= sqrt ((6+3)^2+(-2-5)^2)=sqrt130 = 11.40`unit

The area of the rectangle is`A_r=AB*BC`

`:.A_r==sqrt90*sqrt130=sqrt117*10 ~~108.17(2dp)`sq.unit

[Ans]