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Thread: 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 $\displaystyle 2y = 3x - 5 $ and $\displaystyle x+y=12$. A line passing through a pt A $\displaystyle (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:

    $\displaystyle
    AQ = 2 \cdot AP $.

    So,

    $\displaystyle AQ^2 = 4AP^2 $.

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

    $\displaystyle AQ^2 = (c - 1)^2 + (d - 2)^2 $

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

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

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

    and $\displaystyle d $ as

    $\displaystyle d = 12 - c $

    which we can then use to derive an equation in terms of $\displaystyle a $ and $\displaystyle 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 $\displaystyle 2b=3a-5~\&~c+d=12$.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Well you also know that $\displaystyle 2b=3a-5~\&~c+d=12$.
    I thought I'd already mentioned using these results...?
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