Calculate third point from 2 points and an angle

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Discussion Overview

The discussion revolves around calculating a third point based on two given points and an angle. Participants explore methods for determining this third point through geometric transformations, specifically rotation.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in formulating a method to find a third point given two points and an angle.
  • Another participant suggests that the third point can be found by rotating the second point around the first point by the specified angle.
  • A method for calculating the coordinates of the third point is proposed, involving trigonometric functions and the coordinates of the first two points.
  • A later reply confirms that the proposed method correctly describes the rotation of the second point about the first point through the specified angle.

Areas of Agreement / Disagreement

Participants appear to agree on the method of rotation to find the third point, but there is no explicit consensus on the correctness of the proposed formula.

Contextual Notes

The discussion does not clarify any assumptions regarding the angle's measurement (degrees or radians) or the specific context in which the points are defined.

GamerVSL
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I'm having a hard time putting together a formula. I have 2 points (x0, y0) and (x1, y1) and an angle (k).
Using this information I need to calculate a third point that is k degrees from the previous 2 points.
View attachment 7880
Is it possible to do that? Thank you for your attention.
 

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From your diagram, it looks as though the point C is found by rotating (x1,y1) about (x0,y0) by an angle of x degrees. Is this what you want? If so, there are standard ways of doing this problem. If you need further help on this, post again.
 
I found a method:

Cx=cos(θ)⋅(X1−X0)−sin(θ)⋅(Y1−Y0)+X0
Cy=sin(θ)⋅(X1−X0)+cos(θ)⋅(Y1−Y0)+Y1

Is it right?
 
Yes, this rotates (x1,y1) about (x0,y0) through an angle of $\theta$.
 

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