Possible projective duality between two determinantal formulas for triangle area The shoelace formula for the area of a polygon in terms of consecutive vertices is well-known. In the particular case of a triangle, this may be written using a 3-by-3 determinant as
$$\text{area of triangle}=\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}$$
where $(x_k,y_k)_{k=1,2,3}$ are the three vertices. Several proofs have appeared on this site already.
What may be surprising is that there's another determinantal formula for the area in terms of the three lines. Specifically, a triangle with lines $a_k x+b_k y+c_k=0$ for $k=1,2,3$ satisfies
$$\text{area of triangle}=\frac{1}{2C_1C_2C_3}\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}^2$$ where $C_1,C_2,C_3$ are the cofactors of the third column. (An older question has multiple proofs.)
The parallels between these formulas intrigue me: Both express the area of a triangle in terms of a determinant, but one in terms of points and the other in terms of lines. This is reminiscent of the duality of lines and points in projective geometry. Hence my question: Can these the two formulas indeed be understood through projective duality?

I believe the connection between these two formulas is just a consequence of Cramer's rule. 
Let $A=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{pmatrix}$ and $B=\begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1\end{pmatrix}$.
Let us assume that $a_jx_i+b_jy_i+c_j=0$, whenever $i\neq j$, and $d_i=a_ix_i+b_iy_i+c_i$. Since this is a triangle then $d_i\neq 0$, for every $i$.
Hence, $AB=\begin{pmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3\end{pmatrix}$ and  $\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{pmatrix}\begin{pmatrix} \frac{x_1}{d_1} & \frac{x_2}{d_2} & \frac{x_3}{d_3} \\ \frac{y_1}{d_1} & \frac{y_2}{d_2} & \frac{y_3}{d_3} \\ \frac{1}{d_1} & \frac{1}{d_2} & \frac{1}{d_3}\end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.
By Cramer's rule, $\frac{1}{\det(A)}\begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0\end{vmatrix}=\frac{1}{d_1}$, $\frac{1}{\det(A)}\begin{vmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 1 \\ a_3 & b_3 & 0\end{vmatrix}=\frac{1}{d_2}$, $\frac{1}{\det(A)}\begin{vmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 1\end{vmatrix}=\frac{1}{d_3}$.  
So $\det(AB)=d_1d_2d_3=\dfrac{\det(A)^3}{C_1C_2C_3}$. Thus, $\det(B)=\dfrac{\det(A)^2}{C_1C_2C_3}$.
