Point-Line Duality

In this post, we will develop the theory of homogeneous coordinates. We will use capital letters to denote lines and lowercase letters to denote points. When using homogeneous coordinates in two dimensions, we create a bijection between the set of points of a projective plane and the coordinates $(x,y,z)$ where $x + y + z = 1$ and a bijection between the set of lines of the projective plane and the coordinates $[x,y,z]$ where $x + y + z = 1$, such that a point $p = (x,y,z)$ and a line $L = [X,Y,Z]$ are incident if and only if their dot product $p \cdot L = xX + yY+ zZ$ is equal to 0. Any multiples of these coordinates are equivalent to each other. Also, we are allowed to have the coordinates sum to $0$ (since we can "multiply by infinity") as long as it's not $(0,0,0)$ or $[0,0,0]$. Barycentric coordinates is a homogeneous coordinate system on the real projective plane. The coordinates which sum to 0 would form the line at infinity when using barycentric coordinates. We see an immediate corrollary from the definition.

Corollary 1. The equation of a line $X = [A,B,C]$ is $$Ax + By + Cz = 0$$ where $(x,y,z)$ is a point on the line.

This should be familiar if you have studied barycentric coordinates. Also, the bracket notation for a line isn't particularly common, so I would suggest against using it on an olympiad, but I will use it in this post to show the point line duality of projective geometry.

Now, let us find the coordinates for the line through two points. Let $x = (x_1, x_2, x_3)$ and $y = (y_1, y_2, y_3)$. We wish to find the line $Z = [Z_1, Z_2, Z_3]$ such that $x \cdot Z = y \cdot Z = 0$. Then we have a system of equation $$x_1Z_1 + x_2Z_2 + x_3Z_3 = 0$$$$y_1Z_1 + x_2Z_2 + x_3Z_3 = 0$$ We can represent the two equation as an equality between matrices and factor to get $$\begin{bmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{bmatrix} \begin{bmatrix} Z_1 \\ Z_2 \\ Z_3 \end{bmatrix} = 0$$ The solution to this equation is $$Z_1 = k\cdot \det\begin{bmatrix} x_2 & x_3 \\ y_2 & y_3\end{bmatrix}$$$$Z_2 = k\cdot \det\begin{bmatrix} x_3 & x_1 \\ y_3 & y_1\end{bmatrix}$$$$Z_3 = k\cdot \det\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2\end{bmatrix}$$ This gives us a somewhat nice result.

Theorem 1. The line through $x = (x_1, x_2, x_3)$ and $y = (y_1,y_2,y_3)$ is $$\left[\begin{vmatrix} x_2 & x_3 \\ y_2 & y_3\end{vmatrix},\begin{vmatrix} x_3 & x_1 \\ y_3 & y_1\end{vmatrix},\begin{vmatrix} x_1 & x_2 \\ y_1 & y_2\end{vmatrix}\right]$$

We notice that we could solve the intersection of two lines the same way we did just right now.

Corollary 2. The intersection of $X =[X_1, X_2, X_3]$ and $Y = [Y_1,Y_2,Y_3]$ is $$\left(\begin{vmatrix} X_2 & X_3 \\ Y_2 & Y_3\end{vmatrix},\begin{vmatrix} X_3 & X_1 \\ Y_3 &Y_1\end{vmatrix},\begin{vmatrix} X_1 & X_2 \\ Y_1 &Y_2\end{vmatrix}\right)$$

We can use similar techniques to easily find formulas for collinearity and concurrency.

Theorem 2. Three points $x = (x_1,x_2,x_3)$, $y = (y_1,y_2,y_3)$, $z = (z_1,z_2,z_3)$ are collinear if and only if$$\begin{vmatrix}x_1 & x_2 & x_3 \\y_1 & y_2 & y_3 \\z_1 & z_2 & z_3 \end{vmatrix} = 0$$

Proof. Suppose the three points were collinear. Let $A = [A_1, A_2, A_3]$ be the line containing the three. Then we have that $$x_1A_1 + x_2A_2 + x_3A_3 = 0$$$$y_1A_1 + y_2A_2 + y_3A_3 = 0$$$$z_1A_1 + z_2A_2 + z_3A_3 = 0$$ We express both sides as matrices. $$\begin{bmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{bmatrix} \begin{bmatrix} A_1 \\ A_2 \\ A_3 \end{bmatrix} = 0$$ We easily see that it is necessary that $$\begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{vmatrix} = 0.$$ If not, then the matrix on the left side is invertible and our line is $[0,0,0]$, which implies that there exists no such line. $\Box$

Exercise. Finish the proof by showing that the converse is true.

The proof for concurrency is literally the same, so we will omit it.

Corollary 3. Three lines $X = [X_1,X_2,X_3]$, $Y = [Y_1,Y_2,Y_3]$, $Z = [Z_1,Z_2,Z_3]$ are concurrent if and only if $$\begin{vmatrix} X_1 & X_2 & X_3 \\ Y_1 & Y_2 & Y_3 \\ Z_1 & Z_2 & Z_3 \end{vmatrix} = 0$$

Hopefully, you now understand the point-line duality of projective geometry, which, ultimately, can be used with barycentric coordinates.