# Proof: The point $\mathbf{x}$ lies on the line $\mathbf{l}$ if and only if $\mathbf{x}^T \mathbf{l} = 0$.

My textbooks provides the following result:

The point $$\mathbf{x}$$ lies on the line $$\mathbf{l}$$ if and only if $$\mathbf{x}^T \mathbf{l} = 0$$.

Which was immediately preceded by the following information:

Homogeneous representation of points. A point $$\mathbf{x} = (x, y)^T$$ lies on the line $$\mathbf{l} = (a, b, c)^T$$ if and only if $$ax + by + c = 0$$. This may be written in terms of an inner product of vectors representing the point as $$(x, y, 1)(a, b, c)^T = (x, y, 1)\mathbf{l} = 0$$; that is the point $$(x, y)^T$$ in $$\mathbb{R}^2$$ is represented as a 3-vector by adding a final coordinate of 1. Note that for any non-zero constant $$k$$ and line $$\mathbf{l}$$ the equation $$(kx, ky, k)\mathbf{l} = 0$$ if and only if $$(x, y, 1)\mathbf{l} = 0$$. It is natural, therefore, to consider the set of vectors $$(kx, ky, k)^T$$ for varying values of $$k$$ to be a representation of the point $$(x, y)^T$$ in $$\mathbb{R}^2$$. Thus, just as with lines, points are represented by homogeneous vectors. An arbitrary homogeneous vector representative of a point is of the form $$\mathbf{x} = (x_1, x_2, x_3)^T$$, representing the point $$(x_1/x_3, x_2/x_3)^T$$ in $$\mathbb{R}^2$$. Points, then, as homogeneous vectors are also elements of $$\mathbb{P}^2$$.

I want to prove the aforementioned result.

Since we need to prove logical equivalence, we will need two proofs (one for either direction).

Proof 1:

I begin by assuming that the point $$\mathbf{x}$$ lies on the line $$\mathbf{l}$$.

The point $$\mathbf{x} = (x_1 / x_3, x_2 / x_3)^T$$ lies on the line $$\mathbf{l} = (a, b, c)^T$$. $$\mathbf{x} = (x_1/x_3, x_2/x_3)^T \in \mathbb{R}^2$$ represents an arbitrary homogeneous vector representative of a point $$\mathbf{x} = (x_1, x_2, x_3)^T \in \mathbb{R}^3$$.

Therefore, the equation of the point $$\mathbf{x}$$ on the $$\mathbf{l}$$ is

\begin{align} & a \dfrac{x_1}{x_3} + b \dfrac{x_2}{x_3} + c = 0 \\ &\Rightarrow a \dfrac{x_1}{x_3} + b \dfrac{x_2}{x_3} + \left( \dfrac{x_3}{x_3} \right)c = 0 \\ &\Rightarrow a \dfrac{x_1}{x_3} + b \dfrac{x_2}{x_3} = - \left( \dfrac{x_3}{x_3} \right)c \\ &\Rightarrow ax_1 + bx_2 = (-x_3)c \\ &\Rightarrow a x_1 + b x_2 + c x_3 = 0 \\ &\Rightarrow \mathbf{x}^T \mathbf{l} = 0 \end{align}

Proof 2:

I now begin by assuming that $$\mathbf{x}^T \mathbf{l} = 0$$.

Let $$\mathbf{x} = (x_1, x_2, x_3) \in \mathbb{R}^3$$ be a point and $$\mathbf{l} = (a, b, c)^T$$ be a line.

\begin{align} & \mathbf{x}^T \mathbf{l} = 0 \\ &\Rightarrow (x_1, x_2, x_3) \cdot (a, b, c) = 0 \\ &\Rightarrow ax_1 + bx_2 + cx_3 = 0 \\ &\Rightarrow ax_1 + bx_2 = -cx_3 \\ &\Rightarrow a \left( \dfrac{x_1}{x_3} \right) + b \left( \dfrac{x_2}{x_3} \right) + c = 0 \end{align}

$$a \left( \dfrac{x_1}{x_3} \right) + b \left( \dfrac{x_2}{x_3} \right) + c = 0$$ is the equation of a line, where the point $$\mathbf{x} = (x_1/x_3, x_2/x_3)^T \in \mathbb{R}^2$$ lies on the line $$\mathbf{l} = (a, b, c)$$. Note that $$\mathbf{x} = (x_1, x_2, x_3)^T$$ is an arbitrary homogeneous vector representation of a point in $$\mathbb{R}^3$$, and this also represents the point $$(x_1 / x_3, x_2/ x_3)^T$$ in $$\mathbb{R}^2$$.

And so the proof is done.

I would greatly appreciate it if people could please take the time to review my proof for correctness.

• Your proof is fine. Since there is an implicit assumption $x_3 \neq 0$, you can multiply or divide any equation by $x_3$. – Maxim Jun 10 '19 at 14:07
• @Maxim Yes, I was counting on that. Thanks for the review! – The Pointer Jun 10 '19 at 14:15
• Out of curiosity, what book is this from? – YiFan Jun 14 '19 at 6:41
• @YiFan Multiple View Geometry in Computer Vision by Hartley and Zisserman. – The Pointer Jun 14 '19 at 6:44

While your proof is correct for the Euclidean plane $$\mathbb{R^2}$$ in which $$x_3$$ is not permitted to be 0, it is not valid in the projective plane $$\mathbb{P^2}$$ where $$x_3$$ can be 0 (making the point an ideal point). In fact in $$\mathbb{P^2}$$ there is no proof of this statement (also called the incidence relation) because this equation is an algebraic model of a subset of the axioms of projective geometry which by definition can not be proved!