The Line Passing Through Two Distinct Pts. in the Real Proj. Plane

The Equation of the Line Passing Through Two Distinct Points in the Real Projective Plane

Recall from the The Existence of a Unique Line Through Two Distinct Points In The Real Projective Plane page that if $\mathbf{p} = [p_1, p_2, p_3], \mathbf{q} = [q_1, q_2, q_3] \in \mathbb{P}^2 (\mathbb{R})$ then there exists a unique line $<a_1, a_2, a_3>$, $a_1, a_2, a_3$ not all zero, that passes through the points $\mathbf{p}$ and $\mathbf{q}$.

We will now see how to determine the equation of these lines. Suppose that $<a_1, a_2, a_3>$ is the unique line that passes through the distinct points $\mathbf{p}, \mathbf{q} \in \mathbb{P}^2(\mathbb{R})$. Then the line $<a_1, a_2, a_3>$ is the set of points $\mathbf{x} = [x_1, x_2, x_3] \in \mathbb{P}^2 (\mathbb{R})$ which satisfy:

(1)
\begin{align} \quad a_1x_1 + a_2x_2 + a_3x_3 = 0 \end{align}

In particular, the points $\mathbf{p}, \mathbf{q}$ satisfy the equation above. Now consider the determinant equation:

(2)
\begin{align} \quad \begin{vmatrix} x_1 & x_2 & x_3\\ p_1 & p_2 & p_3\\ q_1 & q_2 & q_3 \end{vmatrix} = 0 \\ \begin{vmatrix} p_2 & p_3\\ q_2 & q_3 \end{vmatrix} x_1 + \begin{vmatrix} p_3 & p_1\\ q_3 & q_1 \end{vmatrix} x_2 + \begin{vmatrix} p_1 & p_2\\ q_1 & q_2 \end{vmatrix} x_3 = 0 \end{align}

The equation above is the equation of a line. Notice that if we plug in the point $\mathbf{p}$ we get:

(3)
\begin{align} \quad (p_2q_3 - p_3q_2)p_1 + (p_3q_1 - p_1q_3)p_2 + (p_1q_2 - p_2q_1)p_3 \\ \quad = p_1p_2q_3 - p_1p_3q_2 + p_2p_3q_1 - p_1p_2q_3 + p_1p_3q_2 - p_2p_3q_1 \\ \quad (p_1p_2q_3 - p_1p_2q_3) + (p_1p_3q_2 - p_1p_3q_3) + (p_2p_3q_1 - p_2p_3q_1) \\ \quad = 0 \end{align}

Also, if we plug in the point $\mathbf{q}$ we get:

(4)
\begin{align} \quad (p_2q_3 - p_3q_2)q_1 + (p_3q_1 - p_1q_3)q_2 + (p_1q_2 - p_2q_1)q_3 \\ \quad p_2q_1q_3 - p_3q_1q_2 + p_3q_1q_2 - p_1q_2q_3 + p_1q_2q_3 - p_2q_1q_3 \\ \quad (p_2q_1q_3 - p_2q_1q_3) + (p_3q_1q_2 - p_3q_1q_2) + (p_1q_2q_3 - p_1q_2q_3) \\ \quad = 0 \end{align}

Therefore the points $\mathbf{p}, \mathbf{q}$ lie on the line given by the determinant equation above. Since the line that passes through the points $\mathbf{p}$ and $\mathbf{q}$ is unique, we have that:

(5)
\begin{align} \quad <a_1, a_2, a_3> = \left < \begin{vmatrix} p_2 & p_3\\ q_2 & q_3 \end{vmatrix}, \begin{vmatrix} p_3 & p_1\\ q_3 & q_1 \end{vmatrix}, \begin{vmatrix} p_1 & p_2\\ q_1 & q_2 \end{vmatrix} \right > \end{align}

Example 1

Find the equation of the line that passes through the points $[2, 1, 1], [4, 3, 1] \in \mathbb{P}^2(\mathbb{R})$.

Using the formula from above, we have that the line that passes through these points is:

(6)
\begin{align} \quad \left < \begin{vmatrix} 1 & 1\\ 3 & 1 \end{vmatrix}, \begin{vmatrix} 1 & 2\\ 1 & 4 \end{vmatrix}, \begin{vmatrix} 2 & 1\\ 4 & 3 \end{vmatrix} \right > = <-2, 2, 2> = <-1, 1, 1> \end{align}

Therefore the equation of the line that passes through $[2, 1, 1]$ and $[4, 3, 1]$ is:

(7)
\begin{align} \quad -x_1 + x_2 + x_3 = 0 \end{align}

We can verify that $[2, 1, 1]$ and $[4, 3, 1]$ are on the line $<-1, 1, 1>$ by verifying that $[2, 1, 1] \cdot <-1, 1, 1> = 0$ and $[4, 3, 1] \cdot <-1, 1, 1> = 0$ (where $\cdot$ represents the dot product). We have that:

(8)
\begin{align} \quad [2, 1, 1] \cdot <-1, 1, 1> = (2)(-1) + (1)(1) + (1)(1) = 0 \end{align}

And also:

(9)
\begin{align} \quad [4, 3, 1] \cdot <-1, 1, 1> = (4)(-1) + (3)(1) + (1)(1) = 0 \end{align}