Results 1 to 5 of 5

Math Help - vecror problem for the best only

  1. #1
    Newbie
    Joined
    Jul 2013
    From
    algeria
    Posts
    3

    vecror problem for the best only

    vecror problem for the best only-captures.png
    ps : sorry for my english ; i don't speak english very well it obvious .so sorry
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1

    Re: vecror problem for the best only

    Quote Originally Posted by hacen View Post
    Click image for larger version. 

Name:	Captures.PNG 
Views:	19 
Size:	44.0 KB 
ID:	28743
    As written (or labelled) I don't think it is true.

    Consider that \overleftrightarrow {DE} \cap \overleftrightarrow {CE} = \left\{ E \right\}, but if G \in \overleftrightarrow {DE} \wedge G \in \overleftrightarrow {CE} then that means
    G=E but G is the mid-point of \overline{CE}.

    Have I miss-read the problem?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2013
    From
    algeria
    Posts
    3

    Re: vecror problem for the best only

    sorry it my fault ;it is not line (DE) but is (DB)
    so : Prove that lines (DB) and (AF) met in G.
    vecror problem for the best only-captures.png
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    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.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum