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Coordinate Geometry


Date: 09/23/98 at 16:32:56
From: Peter
Subject: Coordinate geometry

Given two known coordinate points in cartesian plane pt1, given by 
(x1, y1), and pt2, given by (x2, y2), how can I locate a third point 
pt3, given by (x3, y3), which is perpendicular to the line joining pt1 
and pt2 and is of certain distance d from either pt1 or pt2?

I've established two equations with two unknowns as following:

   m = (y2-y1)/(x2-x1), which implies (y3-y1)/(x3-x1) = -1/m  (eqn. 1)

   (y3-y1)^2 + (x3-x1)^2 = d^2  (eqn. 2)

Since x1, y1, x2, y2, and d are known, I can solve for x3 and y3 
variables using either the substitution or the elimination method. 
There will also be two values for both x3 and y3, since the equation 
becomes a quadratic equation.

To simplify the whole thing, is there a standard equation I can use to 
solve x3 and y3?  

Your response will be greatly appreciated.


Date: 09/24/98 at 12:04:16
From: Doctor Peterson
Subject: Re: Coordinate geometry

Hi, Peter. Your first statement of the problem sounded as if you wanted 
a point on the perpendicular bisector of P1P2 at distance d from each 
of P1 and P2. But your equations tell me you want it a distance d from 
P1 along the perpendicular at P1, which is much easier. You get an easy 
quadratic with no linear term. It can also be solved geometrically:

      A
    +----+P2
    |   /
   B|  /
    | /L
    |/      C
    +---------------+
    P1  \           |
            \       |D
           d    \   |
                    +P3

The two triangles are similar, and:

    A = x2-x1
    B = y2-y1
    L = sqrt(A^2+B^2)
    C = x3-x1
    D = y3-y1

so we get:

     C     B     D     A
    --- = ---   --- = ---
     d     L     d     L

from which we get:

                                             d(y2-y1)
    x3 = x1 + C = x1 + Bd/L = x1 +/- ---------------------------
                                     sqrt((x2-x1)^2 + (y2-y1)^2)

                                             d(x2-x1)
    y3 = y1 + D = y1 + Ad/L = y1 +/- ---------------------------
                                     sqrt((x2-x1)^2 + (y2-y1)^2)

(The +or- allows the perpendicular to extend in either direction. Both 
are + for one point, - for the other.)

If what you really wanted was a point on the perpendicular bisector 
(say, the center of a circle of radius d containing P1 and P2), you 
can apply this formula to the midpoint P4 of P1P2 and the distance 
sqrt(d^2 - (s/2)^2), where s is the distance sqrt((x2-x1)^2+(y2-y1)^2) 
between P1 and P2:

    +-----+P2
    |    /  \
    | s /     \
    |  +        \d
    | /P4  \      \
    |/         \    \
    +-----________ \  \
    P1       d    -----+P3

This gives you:

     x1+x2    sqrt(d^2 - (((x2-x1)^2+(y2-y1)^2)/4) * (y2 - (y1+y2)/2)
x3 = ----- +/- -------------------------------------------------------
       2           sqrt((x2 - (x1+x2)/2)^2 + (y2 - (y1+y2)/2)^2)

which I will let you simplify.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry

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