Distribution of a slice of a line with normally distributed slope Give the random variable $X \sim N(\mu,\sigma)$  consider the process $Y_t = tX$ with $t \in I = [t_1,t_2]$, informally a line with a random slope.
I would like to find the distribution of the points in the process $\{Y_t\}$ that follow the constraint of being on a line, say $l = tS + b$. Informally, the distribution of the intersection of the process $\{Y_t\}$ and the line $l$.
On the picture that would be the points lying on the blue line, where the red line is the mean of the process and the gray lines mark the variance (not the standard deviation).

$\forall t \in I$ I used the fact that $E[Y_t] = tE[X]$ and $Var[Y_t] = t^2Var[X]$.
Running a quick experiment delivers the following histogram of the (x-coordinate of the) intersection points

I would assume it is some sort of skewed normal, as it looks like from the histogram, but I don't know how to derive it analytically.
Is it actually possible to derive this distribution analytically? Or, is this following a known physical process?
 A: The line $y=tX$ intersects the line $y=tS+b$ at the value $t$ satisfying
$$
tX=tS+b.
$$
Solving for $t$ gives
$$
t=\frac b{X-S}.\tag1
$$
If $S$ and $b$ are constants and $X$ has normal$(\mu,\sigma)$ distribution, then the denominator of (1) has normal$(\mu-S,\sigma)$ distribution. So if we denote by $T$ the $x$-coordinate of this point of intersection, the random variable $T:=\frac b{X-S}$  is distributed like ($b$ times) the reciprocal of a normal distribution. In terms of a standard normal $Z$ we can write
$$
T=\frac b{\sigma Z+(\mu-S)}.
$$
If unrestricted, $T$ will take both positive and negative values. In general the mean of $T$ does not exist. You can determine the density of $T$ analytically via its cumulative distribution function. For example in the simple case $\mu=S$ and $\sigma=b=1$ you can compute the cdf of $1/Z$ as follows: If $u>0$ then
$$
\begin{align}
P\left(\frac1Z\le u\right)&=P\left(\frac1Z\le u, Z<0\right) + P\left(\frac1Z\le u, Z>0\right)\\
&=\frac12 +P\left(Z\ge \frac1u\right)\\
&=\frac12 +\int_{1/u}^\infty \phi(z)\,dz,
\end{align}
$$
where $\phi$ denotes the standard normal density. You can differentiate this last expression using the fundamental theorem of calculus and the chain rule to obtain
$$
f_{1/Z}(u) = \frac1{u^2}\phi\left(\frac1u\right).
$$ The case $u<0$ is handled similarly. The density in this simple case is bimodal.
