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FP1, random question, something with ratios I think? watch

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    http://www.suffolkmaths.co.uk/pages/...02009%20QP.PDF

    Question 7, a)

    The way I'm looking at it is with ratios. 'r' lies a certain distance from a, and b.

    Distance 'ab' = b - a.

    Ratio of r from a : ratio of r from b

    = r - a : b - r

    So, if you multiply the distance 'ab' by the ratio of r from a, and add a, you should find r?

    r = a + (b - a) \frac{(r - a)}{(r - a) + (b - r)}

    Then I'd consider the ratio in terms of c and d and get some other stuff.

    I dunno if I'm going along the right lines here or if I'm completely wrong.
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    (Original post by ebmaj7)
    [
    r = a + (b - a) \frac{(r - a)}{(r - a) + (b - r)}
    I'm not sure what you're doing with "ratio of r from a", but am somewhat wary, since a ratio is a relationship between two things, and the ratio 1:2 is the same as 2:4, so to use just one part of it by itself doesn't seem right.

    Additionally your equation reduces to r=r, which doesn't get you anywhere.

    What I'd do is:

    Work out the gradient of the straight line in terms of a,b,c,d.

    Then express the y co-ordinate or r (which is 0) in terms of c + the distance between a and r times the gradient; be careful with signs.
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    (Original post by ghostwalker)
    I'm not sure what you're doing with "ratio of r from a", but am somewhat wary, since a ratio is a relationship between two things, and the ratio 1:2 is the same as 2:4, so to use just one part of it by itself doesn't seem right.

    Additionally your equation reduces to r=r, which doesn't get you anywhere.

    What I'd do is:

    Work out the gradient of the straight line in terms of a,b,c,d.

    Then express the y co-ordinate or r (which is 0) in terms of c + the distance between a and r times the gradient; be careful with signs.
    Okay.

    Gradient = (c - d) / (b - a)

    y - c = \frac{(c - d)}{(b - a)} (x - a)

    y = 0, x = r

    \frac{-c}{\frac{(c - d)}{(b - a)}}  = r - a

     r = a - c\frac{(b - a)}{(c - d)}

    Hmm, should be + c...

    That's annoying.
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    (Original post by ebmaj7)
    Gradient = (c - d) / (b - a)

    Hmm, should be + c...

    That's annoying.
    That's because you got the gradient wrong.

    Gradient = (d - c) / (b - a)

    The first terms (d,b) top and bottom, refer to Q, and the others refer to P.
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    (Original post by ghostwalker)
    That's because you got the gradient wrong.

    Gradient = (d - c) / (b - a)

    The first terms (d,b) top and bottom, refer to Q, and the others refer to P.
    Ah right, yes.

    Okay, thankyou muchly.
 
 
 
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