Area of Polygon in 2-dimensional Coordinate System

 

Motivation:

 

                     Let’s start with the following problem.   We have two vectors  and .

                     What is the area enclosed by ,  ,  and ?

                                           

 

                     Let the angle between  and  be denoted as .  From Cosine law, we have

                           cos  =  =

 

                       (because   = (  -  )  (  -  ) =   - 2  )

 

                  So,

                            OAB = ½  sin

                                       = ½

                                       = ½

                     

 

 

                    The result here is independent from the coordinate system we choose. Let’s just put it into a 2-dimensional coordinate system by using O as the origin and label A and B as A(a₁,a₂) and B(b₁, b₂) .

 

 

 

                                               

 

                 |  |² |  |² – (  )² = (a₁² + a₂²)( b₁² + b₂²) – (a₁b₁+a₂b₂)²

                                                           = (a₁b₂ – a₂b₁)²

 

 

                    Thus,   OAB = ½ | (a₁b₂-a₂b₁)|

 

                   But let’s take a close look at the diagram:

                   

 

                   If the positions of A and B are as shown,  then we have

                                                   b₁a₂ > a₁b₂

 

                   So,

                                     OAB = ½ ( b₁a₂ – a₁b₂ )

 

                   We will use it as a basic tool to find the area of a polygon in 2-dimensional coordinate system.  First, we write down the result above as a theorem.

 

Theorem  Let A(a₁,a₂) and B(b₁, b₂) be two points in the 2-dimensional coordinate system.    Then

                                                     OAB = ½ | (a₁b₂-a₂b₁)|

 

 

 

               Furthermore, we consider the scenario that is a little more complicated:

 

                                             

 

                        ABC =  OAB +  OBC -  OAC

                                   = ½ (  | a₁b₂ – a₂b₁|  + | b₁c₂ – b₂c₁| - | a₁c₂ –a₂c₁| )

 

                So, we can use the theorem above to find the area of   ABC .  However, if we observe more,   we can find some rule behind them.  The following diagram is summarizing the rule associated with the area:

 

                                

                                 “blue” means positive;  “red” means negative.  And we only need to arrange the coordinates of all the points counterclockwise and multiply the result by ½  .

                                   

                               area =  ½ |  (a₁b₂ – a₂b₁ ) + ( b₁c₂ – b₂c₁ )+ ( c₁a₂ – c₂a₁ ) |

 

 

                        Similarly, the same approach can be applied to find the area of a polygon.