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/
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