Results 1 to 5 of 5

- July 6th 2013, 03:56 AM #1

- Joined
- Jul 2013
- From
- algeria
- Posts
- 3

- July 6th 2013, 04:28 AM #2

- Joined
- Apr 2005
- Posts
- 15,696
- Thanks
- 1467

## Re: vecror problem for the best only

Tedious, but I think the most straightforward way is to set up a coordinate system so that B is at the origin and C is on the x-axis, at (c, 0). You can take A to be at (a, b) and then find the coordinates of the other points.

- July 6th 2013, 04:51 AM #3

- July 6th 2013, 08:02 AM #4

- Joined
- Jul 2013
- From
- algeria
- Posts
- 3

- August 14th 2013, 05:52 PM #5

- Joined
- Nov 2012
- From
- Southern California
- Posts
- 18
- Thanks
- 1

## Re: vecror problem for the best only

Put A at (0,0), B at (0,3), C at (4,0).

According to the problem statement, D is at (3,0), E is at (0,2) and G (midpoint of CE) is at (2,1).

x - 2y = 0 defines the line AG.

3x + 4y = 12 defines the line BC.

Solve for the intersection and you get F at (12/5,6/5) which is 2/5 of the way from C to B.

So AF and CE intersect at G for the 3-4-5 triangle with A at the origin.

Stretching the axes and varying the angle between them preserves the proportional placement of the points and demonstrates the relation for any triangle.