Linear Algebra Solutions 39

# Linear Algebra Solutions 39 - O ¨¨ P 2 ¡ ¨ ¡ ¨ ¨¨...

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Unformatted text preview: O ¨¨ P 2 ¡ ¨ ¡ ¨ ¨¨ ¡ ¨ ¡ P3 ¨ ¡d ¡ ¡ ¡ d ¡ d ¡ ¡ d¡P ¨1 ¡ ¨¨ ¡ ¨¨ ¡¨¨ ¡ ¨ Section 4.1 1. We ﬁrst prove that the area of a triangle P1 P2 P3 , where the points are in anti–clockwise orientation, is given by the formula 1 2 x1 x2 y1 y2 + x2 x3 y2 y3 + x3 x1 y3 y1 . Referring to the above diagram, we have Area P1 P2 P3 = Area OP1 P2 + Area OP2 P3 − Area OP1 P3 1 x2 x3 1 x1 x3 1 x1 x2 + − , = 2 y1 y2 2 y2 y3 2 y1 y3 which gives the desired formula. We now turn to the area of a quadrilateral. One possible conﬁguration occurs when the quadrilateral is convex as in ﬁgure (a) below. The interior diagonal breaks the quadrilateral into two triangles P1 P2 P3 and P1 P3 P4 . Then Area P1 P2 P3 P4 = Area P1 P2 P3 + Area P1 P3 P4 = 1 2 x1 x2 y1 y2 + x2 x3 y2 y3 42 + x3 x1 y3 y1 ...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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