Probability that two points are on opposite sides of a line

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SUMMARY

The discussion focuses on calculating the probability that two points, represented as ($x_1, y_1$) and ($x_2, y_2$), lie on opposite sides of a line through the origin that makes an angle $\psi$, uniformly distributed in $[0, \pi]$, with the x-axis. The angle between the two points is denoted as $\theta$, which has a probability density function $f_{\theta}(\rho)$ for $\theta \in (0, 2\pi)$. Key insights include that if $\theta$ is uniformly distributed between $0$ and $2\pi$, the points will be on opposite sides of the line half the time, leading to a probability of $\theta/\pi$ for a fixed $\theta$. The discussion emphasizes the need for clarity in problem definition and notation.

PREREQUISITES
  • Understanding of probability density functions, specifically $f_{\theta}(\rho)$.
  • Familiarity with angles and their distributions, particularly uniform distributions.
  • Basic knowledge of geometric concepts related to lines and points in a Cartesian plane.
  • Proficiency in using LaTeX for mathematical notation.
NEXT STEPS
  • Explore the implications of uniform distributions in probability theory.
  • Learn about the properties of angles in geometry, particularly in relation to Cartesian coordinates.
  • Study the application of probability density functions in real-world scenarios.
  • Investigate the use of LaTeX for mathematical expressions and its best practices.
USEFUL FOR

This discussion is beneficial for mathematicians, statisticians, physics students, and anyone interested in probability theory and geometric analysis. It is particularly relevant for those working with angular distributions and probability density functions.

LCDF
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I want to find the probability that the two points ($x_1, y_1$) and ($x_2, y_2$) lie on the opposite sides of a line passing through the origin $o = (0, 0)$ and makes an angle $\psi$ that is uniformly distributed in $ [0, \pi]$ with the $x$ axis when the angle is measured in clockwise direction. The angle between the two vectors corresponding two points $(x_1, y_1)$ and ($x_2, y_2$) is $\theta$ in clockwise direction. The angle $\theta$ has the probability density function $f_{\theta}(\rho)$ for $\theta \in (0, 2\pi)$. I have
geogebra-export.png
 
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With that information alone you can't tell anything. You'll need another angle (at least its distribution) or some other information.

Measuring angles clockwise is against mathematical convention by the way, better to measure in the other direction.
 
mfb said:
With that information alone you can't tell anything. You'll need another angle (at least its distribution) or some other information.

Measuring angles clockwise is against mathematical convention by the way, better to measure in the other direction.
What if we assume the angle $\psi$ is uniformly distributed between $[0, \pi]$? I understand measuring angle clockwise is against the convention, but I need it due to the setup of a bigger problem.
 
For a given choice of ##\theta## between 0 and##\pi## isn't that probability just ##\theta/\pi##?

And then if ##\theta## has some distribution you just have to do some calculus or algebra. The geometry is totally stripped out.
 
First, writing equations in TeX format but without actually putting them in TeX makes it extremely hard to read.

Second, like many problems, this is more about specifying what you want than calculating anything. Office Shredder has a very good answer to a question, but it is so simple, I'm not as sure it is the answer to your question. Especially as there is a rho in the question.
 
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And third, it is good to make the diagram as simple as possible. As far as I can tell from the question, one could define the x-axis to be a line from the origin to Point 1. If so, draw it that way. And draw a mirror image, so that the angle convention is the same as the rest of the universe's.
 
##\theta## being uniform between ##0## and ##2\pi##, it appear the points will be on opposite sides of the line half the time?
 
LCDF said:
($x_1, y_1$)
I think you need to write this as (##x_1, y_1##) and not ($x_1, y_1$) ie use ## instead of $
 
mathman said:
##\theta## being uniform between ##0## and ##2\pi##, it appear the points will be on opposite sides of the line half the time?

I think θ is intended to be fixed and the answer is given by @Office_Shredder . This leaves ρ kind of left hanging, though. This is part of my point that we need a well-defined problem to make progress. The actual calculation is likely to be easy.
 
  • #10
Thanks for suggestions. I am new to Physics Forum and its protocols. I was following the typical Latex notation. Taking angles in clockwise is due to the setup in the bigger problem that I am trying to solve. ##\rho## is just a dummy variable representing the random variable ##\theta##. I think the answer is probably ##\theta/\pi##, but I am confused by the fact that ##\theta \in (0, 2\pi)##. To make things clearer, ##\theta## has a probability density function ##f_{\theta}##, and ##\psi## is uniformly distributed in ##0## and ##\pi##.
 
  • #11
You need to use ## i e Two hashes at each end. The latex guide is available below the frame. $$gives new line. I don't know other TEX . Welcome.
 
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  • #12
LCDF, if you have an angle that is bigger than ##\pi##, do you know how to translate that into an angle that is smaller than ##\pi##?
 

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