Equivalent parametrizations of the Radon transform 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?
 A: 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 $t$.
