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A Question related to Co-ordinate Geometry

  1. Sep 28, 2009 #1
    Hello. I have this question related to Co-ordinate geometry that I have trouble solving.

    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 co-ordinates of P and Q.

    2. Relevant equations

    3. The attempt at a solution

    It apparently has 2 solutions, according to this website:


    I'm supposed to find out the exact co-ordinates without having to do any kind of graphical work. I couldn't make any reasonable attempt.
  2. jcsd
  3. Sep 28, 2009 #2
    give equations for all the lines that go through (1,2). These are all the lines of the form y = ax + b where b = ......... ? There's also a vertical line.

    a line y = ax + b goes through a point (p,q) if x = p and y= ax+b = q

    find where each of that family of lines intersects the other 2 lines. you know how to find the intersection of 2 lines? Also don't forget the vertical line.

    now compute the distances from these intersection points to A. set one of them equal to twice the other and solve the equation you get
  4. Sep 30, 2009 #3
    Wait, what about 'a', that's also unknown isn't it?

    Okay, understood.

    I lost you here. What family are you talking about? I know how to find the intersection of 2 lines (by simultaneously solving them), but how am I supposed to yield a value when I don't know what y=ax+b is? If you could please provide me with a solution I'd be really grateful.
  5. Oct 1, 2009 #4
    you'll get a line that depends on a. (you can find b if a is given because the line
    must go through (1,2))

    it really isn't a problem if you don't know a. The point of intersection will
    of course also depend on a. an example:

    if you want to find the intersection of the lines y = ax - 2a and x+y = a - 8
    solfve one of the equations for x or y and substitute in the other:

    substitute x = a - 8 - y in y = ax -2a to get y = a(a - 8 - y) -2a

    y + ya = a^2 - 10 a

    (1+a)y = a^2 - 10 a

    y = (a^2 - 10 a) / (1+a)

    substitute the answer for y in x = a - 8 - y to get x

    x = a - 8 -y = (a^2 + a - 8a - 8 - a^2 - 10a)/(a+1) = - (17a + 8)/(a+1)

    so the lines y = ax - 2a and x+y = a - 8 intersect in the point

    (- (17a + 8)/(a+1), (a^2 - 10 a) / (1+a))

    if (a = -1) the lines don't intersect.

    you'll get an intersection point of the mystery line with 2y=3x-5 and one with the line x+y=12
    the postition of both points will also depend on a, and so AQ and AP will also depend on a.

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