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Math Help - Intersecting lines; distance ratios

  1. #1
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    Intersecting lines; distance ratios

    I'm really stumped on this question, but I'm sure the answer is really simple!

    "There exist two lines 2y = 3x - 5 and x+y=12. A line passing through a pt A  (1, 2) intersects these lines at the pts P and Q, respectively. The distance AQ is twice that of AP. What are the co-ordinates of P and Q?"

    So far, I have said:

    <br />
AQ = 2 \cdot AP  .

    So,

     AQ^2 = 4AP^2 .

    If we denote the coordinates of P as  (a, b) and those of Q,  (c, d) , then by Pythagoras' Theorem,

     AQ^2 = (c - 1)^2 + (d - 2)^2

    and  AP^2 = (a - 1)^2 + (b - 1)^2 .

    We can therefore substitute these two equations into  AQ^2 = 4AP^2 to give an equation in a, b, c and  d . Then, using the equations of the lines we can see that we can rewrite  b as

     b = \frac{3a - 5}{2}

    and  d as

     d = 12 - c

    which we can then use to derive an equation in terms of  a and  c . But how to eliminate one of these variables eludes me! Your help is much appreciated.
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  2. #2
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    Well you also know that 2b=3a-5~\&~c+d=12.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Well you also know that 2b=3a-5~\&~c+d=12.
    I thought I'd already mentioned using these results...?
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