Question 1183787
1.  From vector calculus we know that {{{abs(X x Y)}}} gives the area of the parallelogram spanned by the vectors {{{X}}} and {{{Y}}}, and {{{X x Y}}} is their cross-product.

But {{{abs(X x Y) = abs(X)*abs(Y)*sin(theta)}}}, where {{{theta }}} is the angle between the two vectors.

===> {{{K^2 = abs(X x Y)^2 = abs(X)^2*abs(Y)^2*sin^2(theta) = abs(X)^2*abs(Y)^2*(1-cos^2(theta)) = abs(X)^2*abs(Y)^2 -abs(X)^2*abs(Y)^2* cos^2(theta) = abs(X)^2*abs(Y)^2 - (abs(X)*abs(Y)*cos(theta))^2 = abs(X)^2*abs(Y)^2 - (X*Y)^2 }}}


2.  If Z= ({{{z[1]}}},{{{z[2]}}}) is the diagonal vector of the parallelogram, then ({{{z[1]}}},{{{z[2]}}}) = ({{{x[1] + y[1]}}}, {{{x[2] + y[2]}}}).


The area of the triangle bounded by X, Y, and Z is given by {{{(1/2)*abs(matrix(3,3,x[1],x[2],1,x[1]+y[1], x[2] + y[2], 1, y[1],y[2],1)) = (1/2)(x[1]y[2] - x[2]y[1])}}}, 
if direction of evaluation is done counter-clockwise.  If the evaluation is done in clockwise manner, area is negative of the preceding value.


Hence, area of triangle is given by {{{(1/2)abs(x[1]y[2] - x[2]y[1])}}}


But since the triangle mentioned above is half of the parallelogram, we then have {{{K = abs(x[1]y[2] - x[2]y[1])}}}.