## 1. Triangle Area in various coordinate systems

Using the subdivision of the triangle shown, prove that the area of triangle OAB is equal to.
area(AOB) = sin(w)*(x*y'-x'*y)/2.
Here A(x,y), B(x',y') are the coordinates of the points with respect  to the oblique axes shown, w being the angle of the coordinate axes. Deduce that the same area is expressed through.
area(AOB) =  (u'*v-v'*u)/(2*sin(w)).
Here (u,v) are the normal coordinates w.r. to the axes: i.e. signed  distances of the point A from the axes.

In particular, when w = 90 degrees, then sin(w)=1 and the formula gives the area as the determinant of the corresponding column vectors, representing now the cartesian coordinates with respect to these orthogonal axes:

Taking the origin as the third point C of a triangle and denoting the cartesian coordinates with C(c1, c2), B(b1, b2), A(a1, a2) we have:

For a couple of similar formulas, giving the area in barycentrics and trilinears, see the file AreaInBarycentrics.html .

## 2. Triangle Area in various coordinate systems

Some additional formulas for polygons whose vertices are expressed in orthogonal coordinates (Salmon p. 32). If the vertices of the polygon are {(xi,yi)}, then rearanging the terms we obtain.
area(Pn) = (1/2)( (x1y2-x2y1) + ... + (xny1-x1yn) ).
area(Pn) = (1/2)( x1(y2-yn) + x2(y3-y1) + ... + xn(y1-yn-1) ).
area(Pn) = (1/2)( y1(xn-x2) + y2(x1-x3) + ... + yn(xn-1-x1) ).
The area of the triangle ABC bounded by the lines (see ThreeLines.html ).
ax+by+c=0,            a'x+b'y+c'=0,              a''x+b''y+c''=0.