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    The question in full is:

    15) A line is drawn through the point A(1,2) to cut the line 2y=3x-5 in P and the line x+y=12 in Q.
    If AQ= 2AP, find the coordinates of P and Q.

    -----

    I have done a sketch of the lines 2y=3x-5 and x+y=12 and plotting A can see there are two solutions. One instance A is in the centre of P and Q and in the other it is not (i.e. PAQ for one solution and APQ for the other).

    I let P(a,b) and Q(c,d) and therefore P(a,\frac{3a-5}{2}) and Q(c,12-c). Then doing distance between two points equations produced:

    AP^2=\frac{13a^2}{4}-\frac{31a}{2}+\frac{85}{4}\

AQ^2=2c^2-22c+101
    and as AQ^2=4AP^2 then 2c^2-22c+101=13a^2-62a+85.

    Using fact that gradients must be the same can produce another equation in a and c, \frac{10-c}{c-1} = \frac{9-3a}{2-2a}. The problem I have is that if I then substitute these together I end up with a quartic in c which doesn't seem like the correct path to continue on. I feel I'm likely missing something fundamental. Appreciate any pointers on the routes I should be following.
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    (Original post by Franzen)
    I feel I'm likely missing something fundamental. Appreciate any pointers on the routes I should be following.
    I would consider APQ to be the only valid combination.

    As before you have a general point on the first line, P(a,\frac{3a-5}{2})

    NOTE: IF AQ=2AP, then the difference in x-coordinates of A and Q will be twice the difference in x-coordinates of A and P. Similarly the y-coordinates.

    So, you can work out the coordinates of Q in terms of "a".

    This must lie on the 2nd line, so sub into the eqn of 2nd line and solve for a.

    Edit: I get

    Spoiler:
    Show



    a=4

    P(4,\frac{7}{2})


    Q(7,5)


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    (Original post by ghostwalker)
    I would consider APQ to be the only valid combination.

    As before you have a general point on the first line, P(a,\frac{3a-5}{2})

    NOTE: IF AQ=2AP, then the difference in x-coordinates of A and Q will be twice the difference in x-coordinates of A and P. Similarly the y-coordinates.

    So, you can work out the coordinates of Q in terms of "a".

    This must lie on the 2nd line, so sub into the eqn of 2nd line and solve for a.

    Edit: I get

    Spoiler:
    Show




    a=4

    P(4,\frac{7}{2})


    Q(7,5)



    Tyvm ghostwalker Got through to same solution but back of book gives other case too. Will try to apply similar rationale for those pair of P and Q.
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    (Original post by Franzen)
    Tyvm ghostwalker Got through to same solution but back of book gives other case too. Will try to apply similar rationale for those pair of P and Q.
    Fair enough. I wouldn't have consider the other case valid, unless they said the length of AP, or used |AP|. But same methodology will work.
 
 
 
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