Determinants and the Area of a Triangle

Date: 12/14/98 at 13:50:41
From: Frank Chiaravalli
Subject: Matrices and determinants

The area of a triangle having vertices (A,B), (C,D), and (E,F) is the
absolute value of the determinant of M, where:

           | A B 1 |
   M = 1/2 | C D 1 |
           | E F 1 |

How did the textbook arrive at this formula? 

Many thanks for any help you can give us. Quite a few students have 
been asking about this.

Date: 12/14/98 at 14:41:59
From: Doctor Anthony
Subject: Re: Matrices and determinants

Draw a figure with vertices (A,B), (C,D), (E,F) in the first quadrant.  
For the sake of argument let (A,B) be nearest the y axis, (C,D) 
farthest from the y axis and (E,F) between the other two vertices and 
lower than either so that it is closest to the x axis.

Now draw verticals from the vertices to the x axis:

   

The area of the triangle is found by finding the area of the largest 
trapezium (that with one boundary: the line joining (A,B) to (C,D)) and 
then subtracting two smaller trapezia, those with other two sides of 
the triangle as boundaries.

The large trapezium has area (1/2)(B+D)(C-A), and the smaller trapezia 
have areas (1/2)(B+F)(E-A) and (1/2)(F+D)(C-E). For more information 
on the area of trapezia, see:

   http://mathforum.org/dr.math/faq/formulas/faq.quad.html   

So the area of the triangle is:

   (1/2)[(B+D)(C-A) - (B+F)(E-A) - (F+D)(C-E)]

   (1/2)[BC-BA+DC-DA - BE+BA-FE+FA - FC+FE-DC+DE]

   (1/2)[BC - DA - BE + FA - FC + DE]

   (1/2)[-AD - BE - CF + ED + FA + BC]

compared with:

        |A  B  1|      
   (1/2)|C  D  1|  = (1/2)[AD + BE + CF - ED - FA - BC]
        |E  F  1|   

and apart from being opposite in sign the two expressions are the same. 
So the determinant gives twice the area of the triangle.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/