While calculating the area of a triangle using a determinant, we apply modulus ( since area of a triangle has to be positive), but my textbook says that
"Let |∆| be the area of the triangle, Then if vertices of the triangle are taken in anticlockwise order on the coordinate plane, then ∆>0 And if the vertices are taken in clockwise order on the coordinate plane then ∆<0 ".
No proof has been given for it.
In the question below, let's assume the third vertex as (a.0), then by taking the anticlockwise order of the vertices we get only one value of 'a' but if we solve this by the normal way we get two values of 'a'. The latter approach gives me the right answer.
I am unable to understasnd why is that. We can predict the anticlockwise order of the vertices just by knowing that 'a' lies on the x axis and accirding to what my textbook says we must get only one value of 'a'. Where am I wrong?
I don't see anything wrong. The signed area of the triangle formed by points $A=(0,1)$, $B=(2,2)$ and $C=(x,0)$ is given by $$ \Delta={1\over2}\left| \matrix{0\quad 1\quad 1\\2\quad 2\quad 1\\ x\quad 0\quad 1} \right|=-{1\over2}x-1. $$ If the area of the triangle is $2$, then $\Delta$ can be $+2$ or $-2$. In the first case we get $x=-6$ and vertices $ABC$ are taken counterclockwise. In the second case we get $x=2$ and vertices $ABC$ are taken clockwise.