## § Proof of projective duality

In projective geometry, we can interchange any statement with "points" and
"lines" and continue to get a true statement. For example, we have the
dual statements:
- Two non-equal lines intersect in a unique point (including a point at infinty for a parallel line).
- Two non-equal points define a unique line.

The proof of the duality principle is simple. Recall that any point in
projective geometry is of the from $[a : b : c] \simeq (b/a, c/a)$. A projective
equation is of the form $px + qy + rz = 0$ for coefficients $p, q, r \in \mathbb C$.
- if we have a fixed point $[a : b : c]$, we can trade this to get a line $ax + by + cz = 0$.
- If we have a line $ax + by + cz = 0$, we can trade this to get a point $[a:b:c]$.
- The reason we need projectivity is because this correspondence is only well defined upto scaling: the line $x + 2y + 3$ is the same as the line $2x + 3y + 6$.
- In using our dictionary, we would get $[1:2:3]$, $[2:4:6]$. Luckily for us, projectivity, these two points are the same! $(2/1, 3/1) = (4/2, 6/2)$.
- The "projective" condition allows us to set points and lines on equal footing: lines can be scaled, as can points in this setting.