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Algebra.Com's Answer #64242 by malakumar_kos@yahoo.com(315) $\"\"$ $\"About$

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Hi how can i find the area of a triangle using a determinant and the following vertices; (1,-9), (-9, -2), (-7,6)
\n" ); document.write( "Area of the triangle whose vertices are (x1,y1)(x2,y2) &(x3,y3)is given by
\n" ); document.write( " A = x1 y1 1
\n" ); document.write( " x2 y2 1 divided by 2 [Determinant]
\n" ); document.write( " x3 y3 1\r
\n" ); document.write( "\n" ); document.write( " 1 -9 1 [Determinant]
\n" ); document.write( " -9 -2 1
\n" ); document.write( " -7 6 1 Area of triangle = 1/2.[1(-2-6)-(-9)(-9+7)+1(-54-14)]
\n" ); document.write( " = 1/2.[-8-18-68]
\n" ); document.write( " = 1/2.(-94)
\n" ); document.write( " = -47
\n" ); document.write( " Taking the absolute value of the area we get Area of the triangle = 47 sq units
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