5

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?

Community
  • 1
Semiclassical
  • 18,592
  • Note: I am not interested in additional proofs of such, since they're already available on this site. I'm also not interested in seeing an algebraic proof that the two are equivalent unless it gives insight into how they formulas are related. – Semiclassical Jan 17 '17 at 18:45
  • In homogeneous coordinates for $\mathbb R\mathbb P^2$, the line through a pair of points can be represented by their cross product, which can be written as a determinant. You might find a connection between the formulas that way. – amd Jan 17 '17 at 20:17

1 Answers1

2

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, $\det(B)=\dfrac{\det(A)^2}{C_1C_2C_3}$.

Daniel
  • 6,057
  • 1
    The (correct) way you use Cramer's formulas can be a little misleading. You should say that you transform your matrix equality $$\begin{pmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3\end{pmatrix}\begin{pmatrix} \frac{x_1}{d_1} & \frac{x_2}{d_2} & \frac{x_3}{d_3} \ \frac{y_1}{d_1} & \frac{y_2}{d_2} & \frac{y_3}{d_3} \ \frac{1}{d_1} & \frac{1}{d_2} & \frac{1}{d_3}\end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}$$ into 3 linear systems (ctd...) – Jean Marie Jan 17 '17 at 20:12
  • 1
    (...ctd) the first one being $$\begin{pmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3\end{pmatrix}\begin{pmatrix} \frac{x_1}{d_1} \ \frac{y_1}{d_1} \ \frac{1}{d_1} \end{pmatrix}=\begin{pmatrix} 1 \ 0\0\end{pmatrix}$$ etc. and that you apply Cramer's rule to these 3 different systems (call them (1) (2) (3)) for getting the first element in (1), the second element in (2), the third element in (3). – Jean Marie Jan 17 '17 at 20:12
  • besides, very interesting presentation. – Jean Marie Jan 18 '17 at 14:25