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.
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!
Most books discuss projective geometry as an extension of the familiar euclidean geometry. So readers try to understand the concepts of projective geometry in the framework of euclidean geometry. As euclidean geometry is in reality a subset of projective geometry, this causes a lot of confusion.