document.write( "Question 1183787: Show that in the plane R^2, the area K of the parallelogram OXZY with vertices at O(0,0), X(x1,x2), Y(y1,y2), and Z(z1,z2) is given by \r
\n" ); document.write( "\n" ); document.write( "1. K^2 = |X|^2|Y|^2 - (X*Y)^2.
\n" ); document.write( "2. K = |x1*y2 - x2*y1|
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Algebra.Com's Answer #814282 by robertb(5830) $\"\"$ $\"About$

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1. From vector calculus we know that $\"abs%28X+x+Y%29\"$ gives the area of the parallelogram spanned by the vectors $\"X\"$ and $\"Y\"$ , and $\"X+x+Y\"$ is their cross-product.\r
\n" ); document.write( "\n" ); document.write( "But $\"abs%28X+x+Y%29+=+abs%28X%29%2Aabs%28Y%29%2Asin%28theta%29\"$ , where $\"theta+\"$ is the angle between the two vectors.\r
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\r
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\n" ); document.write( "\n" ); document.write( "2. If Z= ( $\"z%5B1%5D\"$ , $\"z%5B2%5D\"$ ) is the diagonal vector of the parallelogram, then ( $\"z%5B1%5D\"$ , $\"z%5B2%5D\"$ ) = ( $\"x%5B1%5D+%2B+y%5B1%5D\"$ , $\"x%5B2%5D+%2B+y%5B2%5D\"$ ).\r
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\n" ); document.write( "\n" ); document.write( "The area of the triangle bounded by X, Y, and Z is given by

,
\n" ); document.write( "if direction of evaluation is done counter-clockwise. If the evaluation is done in clockwise manner, area is negative of the preceding value.\r
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\n" ); document.write( "\n" ); document.write( "Hence, area of triangle is given by $\"%281%2F2%29abs%28x%5B1%5Dy%5B2%5D+-+x%5B2%5Dy%5B1%5D%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "But since the triangle mentioned above is half of the parallelogram, we then have $\"K+=+abs%28x%5B1%5Dy%5B2%5D+-+x%5B2%5Dy%5B1%5D%29\"$ .
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