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Math Help - Co ordinate Geometry Help urgently needed - exam in two days!!

  1. #1
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    Exclamation Co ordinate Geometry Help urgently needed - exam in two days!!

    ok so i was working through a past paper earlier today and i came across this question:

    An orienteering course has been marked by four points on a graph O(0,0), A(-200,500), B(300,600) and C(800,0)

    a) a drink station is to be positioned at the intersection on the diagonals of the quadrilateral OABC
    Find the coordinates of the drink station

    ok so i looked through my maths resourse book and worked out that the diagonal lines are perpendicular to each other. i know this because i worked out the gradients of each diagonal and the equation, they are:

    OB: (0,0) and (300,600)

    600 - 0
    300 - 0 = m = 2

    midpoint is (150,300) and m= 2

    y - 300 = 2 (x - 150)
    = 2x - 300
    y = 2x

    AC: (-200,500) and (800,0)

    0 - 500
    800 - -200 = m = -0.5

    midpoint is (300,250) and m= -0.5

    y - 250 = -0.5 (x - 300)
    = -0.5x + 150
    y = -0.5x + 400

    if we times the gradients we get -1 so the two diagonal lines are perdendicular. This is where i am stick: i dont know how to work out where perpendicular lines meet because they dont meet at their midpoints.
    can someone help me please??? its really urgent as my maths exam is this thursday and i really need help!!!

    please help me
    ToriBori
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  2. #2
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    Lexington, MA (USA)
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    Hello, ToriBori!

    You're correct: the diagonals are perpendicular.
    But that's not relevant to the problem . . .


    An orienteering course has been marked by four points on a graph:
    . . O(0,0),\;A(-200,500),\;B(300,600),\; C(800,0)

    a) A drink station is to be positioned at the intersection
    of the diagonals of the quadrilateral OABC.
    Find the coordinates of the drink station.
    A quick sketch reveals that AC and OB are the diagonals.


    The slope of AC is: . m_{AC} \:=\:\frac{0-500}{800-(-200)} \:=\:-\frac{1}{2}

    The line through C(800,0) with slope -\frac{1}{2}
    . . has the equation: . y - 0 \:=\:-\frac{1}{2}(x-800)\quad\Rightarrow\quad y \:=\:-\frac{1}{2}x + 400 .[1]


    The slope of OB is: . m_{OB} \:=\:\frac{600-0}{300-0} \:=\:2

    The line through O(0,\,0) with slope 2

    . . has the equation: . y -0 \:=\:2(x -0)\quad\Rightarrow\quad y \:=\:2x .[2]


    To find their intersection, equate [1] and [2]:
    . . 2x \:=\:-\frac{1}{2}x + 400\quad\Rightarrow\quad \frac{5}{2}x \:=\:400\quad\Rightarrow\quad x \:=\:160

    Substitute into [2]: . y \:=\:2x\:=\:320


    Therefore, the drink station is at: \bf{(160,\,320)}

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