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/
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