Existence: A Unique Line Through Two Distinct Pts in the R. Proj. Plane
The Existence of a Unique Line Through Two Distinct Points In The Real Projective Plane
Recall from the Lines in the Real Projective Plane page that a line $<a_1, a_2, a_3>$ in the real projective plane is a set of points $\mathbf{x} = [x_1, x_2, x_3] \in \mathbb{P}^2 (\mathbb{R})$ which satisfy the equation $\mathbf{a}^T \mathbf{x} = 0$ or equivalently:
(1)\begin{align} \quad a_1x_1 + a_2x_2 + a_3x_3 = 0 \end{align}
We will now show that if $\mathbf{p}$ and $\mathbf{q}$ are two points in the real projective plane then there exists a unique line $<a_1, a_2, a_3>$ that contains both of these points.
| Proposition 1: Let $\mathbf{p} = [p_1, p_2, p_3], \mathbf{y} = [q_1, q_2, q_3] \in \mathbf{P}^2 (\mathbb{R})$. Then there exists a unique line $<a_1, a_2, a_3>$, $a_1, a_2, a_3$ not all $0$, through the points $\mathbf{p}$ and $\mathbf{q}$. |
- Proof: Let $\mathbf{p} = [p_1, p_2, p_3]$ and $\mathbf{q} = [q_1, q_2, q_3]$. For $a_1, a_2, a_3 \in \mathbb{R}$, not all zero, consider the system of equations:
\begin{align} \quad a_1p_1 + a_2p_2 + a_3p_3 = 0 \\ \quad a_1q_1 + a_2q_2 + a_3q_3 = 0 \end{align}
- We want to show that there exists unique real numbers $a_1, a_2, a_3$ which satisfy the equation above which will ensure that the line $<a_1, a_2, a_3>$ contains $\mathbf{p}$ and $\mathbf{q}$ and is unique. Consider the system above written in terms of matrices:
\begin{align} \quad \begin{bmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3 \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} \end{align}
- Since $\mathbf{p}$ and $\mathbf{q}$ are distinct, we have that the rows of the matrix $\begin{bmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3 \end{bmatrix}$ are linearly independent, there must exist a unique nontrivial solution $\mathbf{a}$ to this system, i.e., there exists a unique line $<a_1, a_2, a_3>$ where $a_1, a_2, a_3$ are not all zero that passes through the points $\mathbf{p}$ and $\mathbf{q}$. $\blacksquare$