Finding Area Using Determinants

Date: 07/27/99 at 23:24:38
From: victoria
Subject: Linear algebra - determinants and area

When using the determinant to find the area of a region, 1's are put 
in the last column. For example: find area of region with vertices 
(1,.7), (1.5,2.5) and (3.2,1)

     Area = +/- 1/2 * |1     .7  1 |
                      |1.5  2.5  1 | = 1.905 km ^2
                      |3.2  1    1 |

that is, plus or minus 1/2 times the determinant of (x-coordinates in 
1st column, y-coordinates in 2nd column and 1's in the last column.) I 
have been searching for a proof of why this works with 1's in the last 
column. I know how to use it, but don't understand it. Can you please 
help? 

Thank you very much.

Date: 07/28/99 at 03:38:02
From: Doctor Floor
Subject: Re: Linear algebra - determinants and area

Dear Victoria,

Thanks for your question.

One explanation is given in the Dr. Math archives:

Determinants and the Area of a Triangle
http://mathforum.org/dr.math/problems/chiaravalli12.14.98.html   

Here is another:

Let the vertices of triangle ABC have coordinates A(xA,yA), B(xB,yB), 
and C(xC,yC). Define two of the vectors v from A to B as 
v = (xB-xA,yB-yA) and u from A to C as u = (xC-xA,yC-yA).

The area of the parallelogram created by these two vectors is given by 
the determinant:

     |xB-xA xC-xA|
     |yB-yA yC-yA|   ......(1)

which equals (xB-xA)(yC-yA)-(xC-xA)(yB-yA).

See these answers from the Dr. Math archives:

Determinant of a Matrix
http://mathforum.org/dr.math/problems/geraghty11.5.97.html   
http://mathforum.org/dr.math/problems/carlino11.16.97.html   

This can be rewritten as the determinant:

     |xA yA 1|
     |xB yB 1|   ......(2)
     |xC yC 1|

This can be seen by subtracting the first row from the second and 
third, so that (2) stays equal to:

     |xA    yA    1|
     |xB-xA xC-xA 0|
     |yB-yA yC-yA 0|

and this determinant is clearly equal to (1), as desired.

If you have more questions, write us back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/