Finding a point on Line Perpendicular to another Line

Discussion in 'General Math' started by ebresie, Apr 25, 2007.

  1. ebresie

    ebresie Guest

    Okay...I'm sure to kick myself once I find the answer, but I have
    already been beating my head on my desk on this one.

    I have two lines (line1 and line2). One is perpendicular the other
    line, ending and bisecting the other line (basically a T). I know the
    3 out of 4 end points on the two lines, with the missing point being
    the bottom of the T (the point not at the intersection - this is
    line2). So I have line1 with (a,b), (c,d) and (e,f),and line2 with
    (c,d) and an unknown (x,y).

    I know the slope (slope1) of the head of the T which is:

    slope1 = (b - f) / (a - e)

    And the slope of the perpendicular line (slope 2) based on known and
    unknow point which is:

    slope2 = (y - d) / (x - c)

    Alternatively, since line2 is perpendicular to line1, based on slope 1
    I can get (using slope1 = -1 / slope2):

    slope2 = -(a - e) / (b - f)

    I can use these two versions and substitute to get:

    (y - d) / (x - c) = - (a - e ) / (b - f)

    I know the length of line2 given length l, from which I can have:

    l ^ 2 = (x - c) ^ 2 + (y-d) ^ 2

    I think my problem may just amount to being confused as to how the
    best way to solve these two simultaneous equations, but I may have
    made some poor assumptions along the way.

    Can someone suggest how best to approach this (am I going about this
    the wrong way?) or provide some additional suggestion as to what to do
    next?

    Many thanks ahead of time.

    Eric
     
    ebresie, Apr 25, 2007
    #1
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  2. <snip>

    Instead of posting again, why don't you read the answers already
    supplied to the same question you posted in sci.math yesterday?

    --Lynn
     
    [Mr.] Lynn Kurtz, Apr 25, 2007
    #2
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  3. ebresie

    ebresie Guest

    Sorry about that.

    I never saw my original post show up. I thought perhaps it had not
    posted so decided to send it again to a broader audiance.

    Thanks for your help.

    Eric
     
    ebresie, Apr 27, 2007
    #3
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