Re: Finding a point on Line Perpendicular to another Line


"ebresie" <ebresie@xxxxxxxxx> wrote in message
news:1177506403.476421.30810@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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

If you are not comfortable with vectors try this.
Let theta be the angle line 1 makes with x axis.
theta = atan( (y1-y2)/(x1-x2))
Let (mx,my) be the point on line 1 you want perpendicular
to spring from and d be the known distance along line 2
delta_y = d*sin(theta)
delta_x = d*cos(theta)
Your desired point on line 2 is
(mx + delta_x , my + delta_y) or (mx - delta_x , my - delta_y)
hth Mick.


.

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