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Analytic geometry proof with triangle.

ghostfirefox

New member
Nov 13, 2019
14
Point D divides side AC, of triangle ABC, so that |AD|: |DC| = 1:2. Prove that vectors [tex]\vec{BD}[/tex] = 2/3 [tex]\vec{BA}[/tex] + 1/3 [tex]\vec{BC}[/tex].
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
Point D divides side AC, of triangle ABC, so that |AD|: |DC| = 1:2. Prove that vectors [tex]\vec{BD}[/tex] = 2/3 [tex]\vec{BA}[/tex] + 1/3 [tex]\vec{BC}[/tex].
Have you made a sketch?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 [tex]\vec{BA}[/tex], [tex]\vec{BC}[/tex]. and [tex]\vec{BD}[/tex]?
 

ghostfirefox

New member
Nov 13, 2019
14
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 [tex]\vec{BA}[/tex], [tex]\vec{BC}[/tex]. and [tex]\vec{BD}[/tex]?
I have a question how you calculated the x coordinate of point D?
 
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