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}