areatriangles
According to the Shoelace formula :
If the points are labeled sequentially in the counterclockwise direction, then the area is positive, if they are labeled in the clockwise direction, the area will be negative.
Can anyone please explain why the area is negative, with the change in the order of the points? How does this work? I want to know the logic behind this.
I can answer how your question about how this is proved. It assumes the counterclockwise direction and uses vector calculus.
What the proof means by positive angles $is$ that it is counterclockwise direction. In this diagram:
$\theta$ is positive only if you are measuring from $p_1$ to $p_2$, the counterclockwise direction.
As for the proof: there may be other ways, but the way I know uses Green's Theorem:
$$\iint_D \frac{\,dQ}{\,dx}-\frac{\,dP}{\,dy}\,dA=\int_C F\cdot\,dC$$
where $C$ is a parametrized curve in 2 dimensions and $F$ is a vector field that can be represented by $F=(P, Q)$. If $\frac{\,dQ}{\,dx}-\frac{\,dP}{\,dy}=1$, the left hand side of this equation will give you the area. However, as it is impossible to find bounds for this double integral (because there are up to $\infty$ points in the polygon), we use Green's Theorem.
We will use the vector field $(-y, x)$; you will see why in a moment. In this case, $\frac{\,dQ}{\,dx}-\frac{\,dP}{\,dy}=1+1=2$. This gives the equation:
$$\iint_D 2\,dA=\int_C F\cdot\,dC$$
Pulling the 2 out of the double integral and dividing gives $$\iint_D 1 \,dA=A=\frac{1}{2}\int_C F\cdot \,dC$$
Hence the $\frac{1}{2}$ in the Shoelace Formula.
Now the goal is to parametrize the polygon. The line integral previously stated can be broken up into $n$ line integrals along each of the lines connecting two points. In other words:
$$\frac{1}{2}\int_C F\cdot \,dC=\frac{1}{2}\bigg[\sum_{i=1}^{n} \int_{C_i} F\cdot\,dC\bigg]$$
If we can show that one of these line integrals gives ${b_{2}a_1-b_{1}a_2}$, we will have proved the shoelace formula: the $\frac{1}{2}$ distributes out over the sum of the $n$ line integrals.
We will say that $p_1$ is $(a_1, b_1)$ and $p_2$ is $(a_2, b_2)$. A parametrization of this line can be $(a_1, b_1)+t(a_2-a_1, b_2-b_1)$, or
$$x=ta_2-ta_1+a_1\\y=tb_2-tb_1+b_1$$ For $t\in[{0, 1}]$. Good. Now we have our parametrization. If we plug into $F=(-y, x)$, we get $F(t)=(tb_1-tb_2-b_1, ta_2+a_1-ta_1)$. We also need $\,dC$ in terms of $\,dt$ so we can integrate in respect to $t$. If we take the derivatives of our parametrization in respect to t we get: $$\frac{\,dx}{\,dt}=\frac{\,d}{\,dt}\big(ta_2-ta_1+a_1\big)=(a_2-a_1)\,dt\\ \frac{\,dy}{\,dt}=\frac{\,d}{\,dt}\big(tb_2-tb_1+b_1\big)=(b_2-b_1)\,dt$$ The dot product is a little tedious, but you get the following: $$F\cdot\,dC_1=(tb_1-tb_2-b_1, ta_2+a_1-ta_1)\cdot(a_2-a_1,b_2-b_1)\,dt\\=(ta_2b_1-ta_1b_1-ta_2b_2+ta_1b_2-a_2b_1+a_1b_1\\+ta_2b_2-ta_2b_1+a_1b_2-a_1b_1-ta_1b_2+ta_1b_1)\,dt\\ =(a_1b_2-a_2b_1)\,dt$$ At this point you might be able to see where the formula comes from. Taking the line integral: $$\int_0^1 (a_1b_2-a_2b_1)\,dt=a_1b_2-a_2b_1.$$ Finally, plugging back into the previous equation gives the shoelace formula: $$\frac{1}{2}\bigg[\sum_{i=1}^{n}a_ib_{i+1}-a_{i+1}b_i\bigg].\tag*{$\blacksquare$}$$
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Best Answer
The shoelace formula is a sum of signed triangle areas.
The area of a triangle is (half) a $3 \times 3$ determinant.
Exchanging two rows or columns of a matrix changes the sign of its determinant.