Area of a Triangle, If three vertices are given taken in anticlockwise direction. Three vertices are given. We can find the area using the determinant. Can someone explain it to me why the number will be a positive number, if vertices are chosen in anticlockwise direction.
 A: The basis for the idea would be this picture (slightly modified from Wikipedia):

Here we would picture the given triangle as the one formed by $a,b$ and the heavy black line. The area of the parallelogram is twice the area of the triangle, so you would have to halve it at the end.
Suppose $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the three points given counterclockwise from the base of $a$. Then the vector for $a$ is $[x_2-x_1,y_2-y_1]$, the vector for $b$ is $[x_3-x_1,y_3-y_1]$, and so the "determinant" picture of the cross product yields:
$$a\times b=\left|\begin{array}&i&j&k\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0\end{array}\right|=\\  ((x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1))k$$
Thus the area of the triangle would be $\frac12((x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1))$
By keeping the points in counterclockwise order, it's guaranteed that the angle between $a$ and $b$ will be between $0$ and $\pi$, so the coefficient next to $k$ is positive. If the order is reversed, you'd just have to strip the negative sign off of the result.
A: 
$$\begin{align}\Delta&=ar(BFDA)+ar(BFEC)-ar(ACEF)\tag{1}\\\end{align}$$
Does that give you a hint?
$$\Delta=\frac12\left|\begin{array}&x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|$$
Since B is above AC, which is when ABC are clockwise.
When B is below AC, which is when ABC are anticlockwise.
A: Using rschwieb's diagram, change the coordinate system so that $(x_1,y_1)$ is at the origin, and so that $x_2>0$ and $y_2 = 0$. (This obviously doesn't change the area). Then the three points are $(0,0)$, $(x_2,0)$, and $(x_3,y_3)$. Since the angle between $\mathbf{a}$ and $\mathbf{b}$ is strictly between $0$ and $\pi$, we know that $y_3 > 0$. Then
$$\begin{vmatrix} x_2 & 0 \\ x_3 & y_3\end{vmatrix} = x_2y_3 > 0.$$
