Computed tomography (CT) requires integration of attenuation values along x-ray linear paths within the body at multiple different rotation angles of the x-ray tube, i.e. Radon transform. In this video, the linear paths, $L(\theta, t),$ are simplified as being parallel to each other and parametrized by a point $t$ along a perpendicular line to the path of the x-ray beam, as well as the angle of rotation $\theta:$
The question is the definition of $L(\theta,t)$ given as:
$$L(\theta, t) = \{(x,y) \in \mathbb R \times \mathbb R: x\cos\theta + y \sin\theta=t\}$$
I don't get the geometry behind this.
On the other hand, the parametrization given in this post makes total sense:
$$L(\theta,z)=(x(z),y(z))=\{(x,y) \in \mathbb R \times \mathbb R: x= r\cos\theta - z \sin \theta \; ; \; y = r \sin \theta + z \cos \theta\}$$
as a change of coordinate basis from $(x,y)$ to $(r,z):$
$$\begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin \theta & \cos\theta \end{bmatrix} \begin{bmatrix}r\\z \end{bmatrix}$$
How to reconcile both expressions?
UPDATE:
The initial source describes a parameterization of the receptor line in Hesse normal form, and assuming the x-ray beams are parallel.
Considering the modified diagram from the OP:
$$\begin{align} t & \overset{?}= x\, \cos\theta + y\, \sin \theta \\[2ex] &=x\,\cos\left (\phi +\frac\pi 2\right) + y\,\sin\left (\phi +\frac\pi 2\right)\\[2ex] &= -x \,\sin \phi + y\, \cos \phi\\[2ex] & = -x \frac y t + y \frac x t\\[2ex] &=0 \end{align}$$
On the other hand, parametrizing with $ \varphi $ could work out:
$$\begin{align} t &= x \cos(\varphi) + y \sin(\varphi)\\[2ex] &=x \frac x t + y \frac y t\\[2ex] &=\frac{x^2}{t}+ \frac{y^2}{t}\\[3ex] \implies & t^2 = x^2 + y^2 \end{align}$$
This seems to be reflected here (page 21 of 167):
And now the second parametrization in the OP matches the first parametrization only for one of the multiple parallel lines in the diagram - specifically, the line going through the origin, which would make $r=0$:
$$\begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin \theta & \cos\theta \end{bmatrix} \begin{bmatrix}0\\z \end{bmatrix}=\begin{bmatrix} -z\sin\theta\\ z\cos\theta \end{bmatrix} $$
Now, by PT, and considering the angle from the $x$ axis to the $z$ axis, $\phi = \theta+\frac \pi 2:$
$$\begin{align}z^2 &= x^2 + y^2\\[2ex] &=x\left( -z \sin\theta \right) + y \left(z\cos \theta \right)\\[2ex] &=x \left( -z \sin\left(\phi - \frac \pi 2 \right) \right) + y \left( z \cos\left(\phi - \frac \pi 2 \right) \right)\\[2ex] &=x \left( z \cos \phi\right) + y\left( z \sin \phi \right)\\[2ex] &\implies z = x\cos\phi + y \sin \phi \end{align}$$
And $\phi$ is exactly the equivalent of $\varphi$ in the diagram at the beginning of this answer, and $z$ is the same as $t$.