MHB Analytic geometry proof with triangle.

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Point D divides side AC of triangle ABC in a 1:2 ratio, leading to the coordinates D = ((2x + z)/3, 2y/3). To prove the vector equation \vec{BD} = 2/3 \vec{BA} + 1/3 \vec{BC}, a coordinate system is established with B at the origin. The coordinates for points A and C are defined as A = (x, y) and C = (z, 0). The discussion includes inquiries about the calculation of D's x-coordinate, emphasizing the importance of understanding vector representations in analytic geometry. The proof hinges on correctly establishing the coordinates and vector relationships within the triangle.
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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?
 
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