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Math Help - Coordinate geometry

  1. #1
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    Coordinate geometry



    The problem lies in showing k=3
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  2. #2
    Senior Member nikhil's Avatar
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    Exclamation

    y=2x-12 or y-2x+12=0
    point of intersection with the x axis is (6,0)
    now distance between point (0,k) and line is given by
    [k-2(0)+12]/(5)^(1/2).................1
    but using distance formula we get RO=[36+k^2]^(1/2)................2
    squaring and equating equations 1 and 2 we get
    k^2+144+24k=180+5k^2 or
    4k^2-24k+36=0 or
    k^2-6k+9=0 or
    (k-3)^2=0
    k=3
    second method
    since RQ and QP are perpendicular therefor product of their slopes will be -1
    point Q=(6,0)
    slope of RQ=-k/6
    slope of QP=2
    therfor
    (-k/6)x2=-1 or
    k=3
    FOR PERPENDICULAR BISECTOR slope=-(1/2) and one point is the mid point of QP
    now use y=mx+c (slope(m) already given, just put the coordinates of midpoint to get the value of c)
    Last edited by nikhil; January 15th 2010 at 10:20 PM.
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  3. #3
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    so just use gradient=y1-y2/x1-x2
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