Finding if a point belongs to a line in space.

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Discussion Overview

The discussion revolves around determining whether a given point belongs to a line defined in a two-dimensional space using parametric equations. Participants explore the method of checking point inclusion based on the equations derived from a point and a direction vector.

Discussion Character

  • Exploratory, Mathematical reasoning

Main Points Raised

  • One participant presents a method for determining if a point lies on a line by solving the parametric equations for specific values of t.
  • Another participant provides an example with specific coordinates to illustrate the method, demonstrating how to solve for t and check for consistency.
  • Responses include expressions of appreciation for the clarity of the explanation and personal comments unrelated to the mathematical content.

Areas of Agreement / Disagreement

Participants generally agree on the method for checking if a point lies on a line, as demonstrated through examples. However, there is no explicit discussion of alternative methods or disagreement on the approach.

Contextual Notes

No limitations or unresolved mathematical steps are noted in the discussion.

Who May Find This Useful

This discussion may be useful for individuals interested in understanding parametric equations of lines and methods for verifying point inclusion in a geometric context.

smithnya
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Hello everyone,

I have a simple question, but I am unsure. I know from a point p0 = (x1, y1) and a vector v = <a, b>, I can obtain a parametrized set of equations for a line in space such that x = x1 + at and y = y1 + bt. How can I check that any other point, not p0, is/isn't in that line?
 
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For example, let's say you have x=2+3t, y=3+4t

Then a number (x,y) lies on the lies if and only if there exist a t such that x=2+3t AND y=3+4t

Let's see if (0,0) is on the line. If (0,0) were on the line, then there would exist a t such that 0=2+3t AND 0=3+4t. Solving the equations gets us that t=-2/3 AND t=-3/4. This is clearly false (t can not be two values at once). Thus (0,0) is not on the line.

Take (8,11). If this were on the line, then there would exist a t such that 8=2+3t AND 11=3+4t. Solving the equations gets us t=2 AND t=2. So such a t exists (and equal 2). Thus (8,11) is on the line.
 
Thank you so much. That was very simple and it explained what I needed to know. Nice avatar by the way. I am listening to "The Great Gig in the Sky" as I type.
 
smithnya said:
Thank you so much. That was very simple and it explained what I needed to know. Nice avatar by the way. I am listening to "The Great Gig in the Sky" as I type.

You have a great taste in music! :approve:
 

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