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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

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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