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Math Help - Midpoint formula question

  1. #1
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    Midpoint formula question

    The questions asks to find the point that divides the interval AB ( A(-7, 5), B(3, 1) ) into the ratio 3:1

    I'm looking at the answer but I can't work out why they changed 3:1 to 3:-1 and therefore get a different answer to me
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  2. #2
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    Quote Originally Posted by chancey View Post
    The questions asks to find the point that divides the interval AB ( A(-7, 5), B(3, 1) ) into the ratio 3:1...
    Hi,

    you are looking for a unknown point U so that the distances
    \frac{AU}{UB}=\frac{3}{1}

    AU=3\cdot UB

    UB = AB - AU . So you get:

    AU=3\cdot (AB-AU). Thus:
    AU=3AB-3AU
    AU=\frac{3}{4}\cdot AB

    Because B(x_A+10,y_A+(-4)) your unknown point is now well-known:
    U(x_A+\frac{3}{4}\cdot 10,y_A+\frac{3}{4}\cdot (-4)). That means: U(0.5,2)

    I've attached a diagram to show you the situation. In the textbox you find the ratio.

    EB
    Attached Thumbnails Attached Thumbnails Midpoint formula question-teilstrecke1.gif  
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  3. #3
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    Hello, chancey!

    You could baby-talk your way through this one . . .


    Find the point that divides the line segment AB into the ratio 3:1
    given: A(-7, 5),\;B(3, 1)

    We want a point P which is \frac{3}{4} the distance from A to B.

    . . A . . . . . . . . . . . P . . . B
    . . * - - + - - + - - \bullet - - *

    To go from A(-7,5) to B(3,1), we'd move: \begin{Bmatrix}\text{10 units right} \\ \text{4 units down}\end{Bmatrix}

    To go three-fourths the distance, we'd move: \begin{Bmatrix}\frac{3}{4}\cdot10 = \frac{15}{2}\text{ units right} \\ \frac{3}{4}\cdot4 = \text{3 unit down}\end{Bmatrix}

    Hence, P is: . \begin{Bmatrix}\text{-}7 + \frac{15}{2} = \frac{1}{2} \\ 5 + (\text{-}3) = 2\end{Bmatrix} \quad \Rightarrow\quad P\left(\frac{1}{2},\,2\right)


    Of course, this is basically Earboth's solution.
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