calculating the variance of an orthogonal line

Question

I have three points $A(x,y) , B(x,y)$ and $C(x,y)$ the $x$ and $y$ coordinates follow a normal distribution with a known mean and variance. I don,t know if $A$ and $B$ are correlated with each other. I know $C$ dont have an correlation with $A$ and $B$.

There is a line made bij $AB$ and the point $C$. i want to calculate the distance from $C$ to the line $AB$. that means there is an orthogonal line from $C$ on $AB$. the length of that line is calculed with the next formule : $det(A)/|a-b|$ $A$ is an matrix made bij $(a-c,a-b)$

Atm i am using a monte carlo simulation to find the variance and mean. with the simulation, i found that the distance follow a normal distribution. Anyone can help me out? Is it possible to calculate the variance of the length of the orthogonal line without the use of a simulation?

I am not allowed to upload the plot of the points. the dataframe of the points excist of 4096 points located on the map. i have to find for each point a line with the smallest distence. the points are from an other map. the company want to combine those maps to create an better map

Answer 1

The distributions can be wildly varying depending on the distributions of A, B and C.

So I think that it will be difficult to give a simple solution if there is not a clear specification of the distribution, based on which simplifications can be made.

Example for a funny distribution

Let A and B be two distributions centered around $(0,0)$ with A very small variance and B a very large variance.

This makes that the line between A and B is more or less the line through $(0,0)$ and B, or a line through $(0,0)$ with an angle that is homogeneously distributed.

Let C be a distribution far away from $(0,0)$ with small variance. In the computed example below, we used C centered around $(10,0)$

Then the distance of C from the line is more or less equal to

$$d \approx 10 \sin(\theta)$$

Where $\theta$ is the angle between the line through point B and the x-axis.

If we consider the angle is homogenously distributed from $0$ to $\pi$ then you can compute the distribution of $d$ by transformation:

$$\theta = \sin^{-1}(d/10)$$

and the derivative can be used to compute the distribution of the transformed variable

$$f(d) = \frac{2}{\pi \sqrt{100-d^2}}$$

Computation example:

set.seed(1)

### https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Line_defined_by_two_points
distance <- function(x0,y0,x1,y1,x2,y2) {
  abs((y2-y1)*x0-(x2-x1)*y0+x2*y1-y2*x1)/sqrt((x2-x1)^2+(y2-y1)^2)
}

n <- 10^4
x0 <- rnorm(n,mean = 10,sd = 0.1)
y0 <- rnorm(n,mean = 0,sd = 0.1)
x1 <- rnorm(n,mean = 0,sd = 2)
y1 <- rnorm(n,mean = 0,sd = 2)
x2 <- rnorm(n,mean = 0,sd = 0.1)
y2 <- rnorm(n,mean = 0,sd = 0.1)

d <- distance(x0,y0,x1,y1,x2,y2)

hist(d, breaks = seq(0,11,0.2), freq = 0, main = "histogram of d \n with theoretic density curve")
ds <- seq(0,10,0.1)
lines(ds , (2/pi)/sqrt(100-ds^2))