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    Hey can anyone help me answer this question, 5 marks
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    Infact don't worry I know how to solve it.

    The vector AR can be found in two different ways in which they are both equal in each other, therefore I can re-arrange for r

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    Correct me if I'm wrong tho
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    Since R divides AB into the ration m:n, \frac{AR}{RB} = \frac{m}{n}. Then rearrange to find r
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    (Original post by CraigBackner)
    Infact don't worry I know how to solve it.

    The vector AR can be found in two different ways in which they are both equal in each other, therefore I can re-arrange for r

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    Correct me if I'm wrong tho

    I don't think that will work. since the vectors when you find AR in the other way will cancel to leave your original AR
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    (Original post by h3rmit)
    Since R divides AB into the ration m:n, \frac{AB}{RB} = \frac{m}{n}. Then rearrange to find r
    should't the equation be \frac{AR}{RB} = \frac{m}{n}?
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    (Original post by CraigBackner)
    should't the equation be \frac{AR}{RB} = \frac{m}{n}?
    Yes, and that's what I typed first time correctly with 100% definitely no mistakes.
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    h3rmit is correct although  \displaystyle \frac{AB}{RB} = \frac{m+n}{n}

    edit: im so confused with edits
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    (Original post by h3rmit)
    I don't think that will work. since the vectors when you find AR in the other way will cancel to leave your original AR
    The vector AR is r-a and the second vector is  \frac{m}{m+n}\left(AB\right) which equals \frac{mb-ma}{m+n}


    when these two are equal and you rearange for r you get \frac{mb+na}{m+n}

    which is the given expression
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    (Original post by CraigBackner)
    The vector AR is r-a and the second vector is (m)/(m+n)(AB) which equals (mb-ma)/(m+n)

    when these two are equal and you rearange for r you get (bm+an)/(m+n)

    which is the given expression
    Ah, I thought you meant using the triangle and a, r, and b only
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    It's not even a proof question but a show me
 
 
 
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