Analytic geometry proof with triangle.

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Discussion Overview

The discussion revolves around proving a vector relationship in triangle ABC, specifically focusing on point D which divides side AC in a 1:2 ratio. The context includes analytic geometry and vector representation within a coordinate system.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the problem of proving that vectors \(\vec{BD} = \frac{2}{3} \vec{BA} + \frac{1}{3} \vec{BC}\) given the ratio of segments |AD|:|DC| = 1:2.
  • Post 2 prompts for a sketch to aid in understanding the geometric configuration.
  • Post 3 suggests setting up a coordinate system with B at the origin and provides coordinates for points A and C, as well as a calculation for point D based on the given ratio.
  • Post 4 reiterates the coordinate setup and calculations for point D, while also questioning the method used to determine the x-coordinate of point D.

Areas of Agreement / Disagreement

Participants are engaged in a technical exploration of the problem, with some agreement on the coordinate system setup, but there is a question regarding the calculation of point D's coordinates, indicating potential disagreement or uncertainty in that aspect.

Contextual Notes

The discussion includes assumptions about the coordinate system and the specific calculations for point D, which may depend on the chosen coordinates for points A and C. The method of deriving the coordinates for D has not been universally accepted, as indicated by the question raised in Post 4.

ghostfirefox
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Point D divides side AC, of triangle ABC, so that |AD|: |DC| = 1:2. Prove that vectors \vec{BD} = 2/3 \vec{BA} + 1/3 \vec{BC}.
 
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ghostfirefox said:
Point D divides side AC, of triangle ABC, so that |AD|: |DC| = 1:2. Prove that vectors \vec{BD} = 2/3 \vec{BA} + 1/3 \vec{BC}.

Have you made a sketch?
 
You titled this "Analytic geometry proof ..." so I would set up a coordinate system so that the origin is at the vertex "B" of the triangle and the x-axis lies along on side BC. Then B= (0, 0), A= (x, y), and C= (z, 0) for some numbers x, y, and z. Since D lies on AC such that "|AD|:|DC|= 1:2", D= ((2x+ z)/3, 2y/3). Now, what are the vectors \vec{BA}, \vec{BC}. and \vec{BD}?
 
HallsofIvy said:
You titled this "Analytic geometry proof ..." so I would set up a coordinate system so that the origin is at the vertex "B" of the triangle and the x-axis lies along on side BC. Then B= (0, 0), A= (x, y), and C= (z, 0) for some numbers x, y, and z. Since D lies on AC such that "|AD|:|DC|= 1:2", D= ((2x+ z)/3, 2y/3). Now, what are the vectors \vec{BA}, \vec{BC}. and \vec{BD}?

I have a question how you calculated the x coordinate of point D?
 
Last edited:

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