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.

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

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