# Finding a point on Line Perpendicular to another Line

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

1. ### ebresieGuest

Okay...I'm sure to kick myself once I find the answer, but I have

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.

the wrong way?) or provide some additional suggestion as to what to do
next?

Eric

ebresie, Apr 25, 2007

2. ### [Mr.] Lynn KurtzGuest

<snip>

supplied to the same question you posted in sci.math yesterday?

--Lynn

[Mr.] Lynn Kurtz, Apr 25, 2007

3. ### ebresieGuest

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.