Probability that two points are on opposite sides of a line

In summary: So from your definition of ##\psi##, it looks like ##\theta## is uniform between ##-\pi## and ##\pi##, not between ##0## and ##2\pi##.Yes, you are correct. I will make the change in the original problem.In summary, we are given two points ($x_1, y_1$) and ($x_2, y_2$) and a line passing through the origin ($o = (0,0)$) with an angle $\psi$ that is uniformly distributed between 0 and $\pi$. We want to find the probability that the two points lie on opposite sides of this line, with the angle between them being $\theta$, which
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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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  • #6
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.
 
  • #7
##\theta## being uniform between ##0## and ##2\pi##, it appear the points will be on opposite sides of the line half the time?
 
  • #8
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 $
 
  • #9
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##?
 

What is the definition of "Probability that two points are on opposite sides of a line"?

The probability that two points are on opposite sides of a line refers to the likelihood that two given points will fall on different sides of a straight line when plotted on a coordinate plane.

How is the probability of two points being on opposite sides of a line calculated?

The probability of two points being on opposite sides of a line is calculated by dividing the number of possible outcomes where the points are on opposite sides by the total number of possible outcomes.

What is the relationship between the position of the line and the probability of two points being on opposite sides?

The position of the line has a direct impact on the probability of two points being on opposite sides. As the line moves closer to the points, the probability decreases, and as the line moves farther away, the probability increases.

Can the probability of two points being on opposite sides of a line be greater than 1?

No, the probability of two points being on opposite sides of a line cannot be greater than 1. This would indicate that there are more possible outcomes where the points are on opposite sides than there are total possible outcomes, which is not possible.

How does the number of points on a line affect the probability of two points being on opposite sides?

The number of points on a line does not affect the probability of two points being on opposite sides. The probability is solely determined by the position of the line in relation to the two given points.

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