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Math Help - Simple problem X(

  1. #1
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    Lightbulb Simple problem X(

    Good Morning!!!!

    my prof solved this but i didn't get it idk T,T
    i must use division of a line segment.


    "find the endpoint of the section joining A(2,3) & M(6,1) if it is extended to a distance 2/3 fro its own length."




    got midterms coming up , this is the only thing that i didn't understand .please help ^^
    thanks in advance!!!!
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  2. #2
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    Re: Simple problem X(

    Extended in which direction? Does 2/3 mean that it is being extended by 2/3 units? Or is it being stretched by an additional 2/3 of its current length (making it 5/3 its current length)?
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  3. #3
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    Re: Simple problem X(

    Quote Originally Posted by SlipEternal View Post
    Extended in which direction? Does 2/3 mean that it is being extended by 2/3 units? Or is it being stretched by an additional 2/3 of its current length (making it 5/3 its current length)?

    yes I saw 5/3 on the solution but i dont know how did that happen
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  4. #4
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    Re: Simple problem X(

    correction A is (-2,-3)
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  5. #5
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    Re: Simple problem X(

    Start with its length. That is one times its length. Add 2/3 of its length. That is 1+2/3 of its length. 1+2/3 = 5/3.
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  6. #6
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    Re: Simple problem X(

    Quote Originally Posted by SlipEternal View Post
    Start with its length. That is one times its length. Add 2/3 of its length. That is 1+2/3 of its length. 1+2/3 = 5/3.
    so line A-M is 1 i think im getting it now

    so i must do
    for x
    a-m/a-p=5/3=-2-6/-2-x

    for y
    5/3=-3-1/-3-y


    is that it??
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  7. #7
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    Re: Simple problem X(

    If $|AM|:|AP| = 3:5$, then you need to flip the fraction. $\dfrac{-2-x}{-2-6} = \dfrac{5}{3}$ and $\dfrac{-3-y}{-3-1} = \dfrac{5}{3}$.

    Then $|AM| = \sqrt{(6-(-2))^2+(1-(-3))^2} = \sqrt{80}$ and $|AP| = \sqrt{\left(\dfrac{34}{3}-(-2)\right)^2+\left(\dfrac{11}{3}-(-3)\right)^2} = \sqrt{\dfrac{2000}{9}} = \sqrt{\dfrac{25}{9}\cdot 80} = \dfrac{5}{3}\sqrt{80}$ as desired.
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  8. #8
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    Re: Simple problem X(

    oh! thats IT!! thanks sir!!
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  9. #9
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    Re: Simple problem X(

    Hello, mventurina1!

    Find the endpoint of the section joining A(2,3) & M(6,1)
    if it is extended to a distance 2/3 of its own length.

    Code:
          |   A
          |   o(2,3)
          |   :   *
          |   :       *
          | -2:           *
          |   :               *
          |   :       +4          *   M
          |   + . . . . . . . . . . . o(6,1)
          |                           :   *
      ----+---------------------------:-------*--------
          |                           :           *   B
          |                           + . . . . . . . o
          |
    Going from A to M, we move:
    . . 4 units right,
    . . 2 units down.

    Going from M to B, we move:
    . . \tfrac{2}{3}(4) = \tfrac{8}{3} units right,
    . . \tfrac{2}{3}(2) = \tfrac{4}{3} units down.

    Hence: . \begin{Bmatrix} x &=& 6 + \frac{8}{3} &=& \frac{26}{3} \\ y &=& 1 - \frac{4}{3} &=& \text{-}\frac{1}{3} \end{Bmatrix}

    Therefore: . B\left(\frac{26}{3},\;\text{-}\frac{1}{3}\right)

    Thanks from johng and mventurina1
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  10. #10
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    Re: Simple problem X(

    Hi,
    Soroban's excellent solution works because of similar triangles:

    Thanks from mventurina1
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