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Co-ordinate geometry

  1. Sep 17, 2007 #1
    im stuck on an extension question to my hm/wk, hope you can help

    1. The problem statement, all variables and given/known data

    The points A, B and C have co-ordinate (-3, 2) (-1, -2), and (0, k) respectively, where k is constant.
    Given that AC = 5BC, find the possible values of k.

    2. Relevant equations



    3. The attempt at a solution

    I used pythagorus to get an equation for AC in terms of k, then the same for BC - then combined the 2 equations, simplified and ended up with 11 = 13k + 3k^2 then completed the square ending up with k = +-(squareroot)(301/36) - (13/6)

    some how I think I've gone wrong or got the wrong method.

    can someone please guide me :D thnx
     
  2. jcsd
  3. Sep 17, 2007 #2

    Dick

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    The method seems to be right. The answer is wrong. You'll need to show us the details of how you got the quadratic to solve.
     
  4. Sep 17, 2007 #3

    symbolipoint

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    My equation becomes [tex] \[
    0 = 76 + 68k + 15k^2
    \]
    [/tex]
     
  5. Sep 17, 2007 #4

    symbolipoint

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    ... That suggests k = 2 or k= 2.53333333...repeating (two and eight-fifteenths)

    ..seems not to make sense.
     
  6. Sep 17, 2007 #5

    HallsofIvy

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    The distance from A to C, squared, is (-3)2+ (2-k)2= 9+ 4- 4k+ k2. The distance from B to C, squared is (-1)2+ (-2-k)2= 1+ 4+ 4k+ k2. Saying that AC= 5BC is the same as AC2= 25BC2 or 13- 4k+ k2= 25(5+ 4k+ k2)= 125+ 100k+ 25k2. That gives 24k2+ 104k+ 112= 0 Dividing through by 8, that is 3k2+ 13k+ 14= 0. That doesn't look like what you got! And it factors rather easily! That's always a good sign.
     
  7. Sep 19, 2007 #6
    thnx, i made a careless error, lol
     
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