Associated Topics || Dr. Math Home || Search Dr. Math

### Locus and Equations of Lines

```
Date: 01/10/99 at 18:35:56
From: Marguerite
Subject: Locus and equations of lines

I'm currently trying to help my niece who is in 10th grade with locus
and line equations. The concept isn't coming back to me. She has the
following homework:

Describe the locus of points that are 3 units from the line x = -1 and
give the equation of the locus.

Describe the locus of points that are 6 units from the point (3,-1)
and give the equation of the locus.

Describe the locus of points equidistant from the x and y axis and
give the equation.

Describe the locus of points equidistant from the points F (-2,1) and
G (6,3) and give the equation.

Describe the locus of points equidistant from lines y = 4 and y = -2
and give the equation of the line.

We can graph the points on the x and y axis but we don't understand
how to find the locus of points. We know the formulas for slope, the
line equation, and the distance formula, but when do you use each?
What is the process for these problems so I can explain them to her?

Marguerite
```

```
Date: 01/11/99 at 11:59:02
From: Doctor Peterson
Subject: Re: Locus and equations of lines

Hi, Marguerite. The basic concept of a locus in geometry is very
closely related to that of an equation in algebra. We want to describe
all points that satisfy some condition, by choosing an unknown point
(x, y) and seeing what it means, algebraically, to satisfy the
condition. Let's look at your problems one at a time.

1. Describe the locus of points that are 3 units from the line x = -1
and give the equation of the locus.

Pick any point (x, y). What does it mean to be 3 units from the line
x = -1? Since this is a vertical line, the distance is just the
horizontal distance from x to -1:

^   ^
|   |
|  y+   *(x,y)
|   |
|   |
|   |
----------+---+---+--
-1|  0|   x
|   |
|   |
|<----->| d=|x-(-1)|

(We use the absolute value because x could be on either side of -1.)

So the equation of the locus is:

|x - (-1)| = 3

Every point that is 3 units from the line satisfies this equation, and
every point that satisfies this equation is 3 units from the line. So
this is the locus we are looking for. (Incidentally, don't worry that
y isn't in this equation - that's how vertical lines work.)

Can you see what this locus is geometrically? Think in terms of
parallel lines, and remember what I said about being on either side!
The equation can be rewritten as two equations, in the form "x = this
OR x = that."

2. Describe the locus of points that are 6 units from the point (3,-1)
and give the equation of the locus.

This is similar, except here the distance formula you want will be the
Pythagorean theorem. Write an expression for the distance of point
(x, y) from (3, -1), and set it equal to 6 to make the equation of the
locus.

To describe the locus, think about how you would draw it, by making a
string of length 6 and tacking one end of it at (3, -1). Where can the
other end be? Could you also draw this with a compass?

3. Describe the locus of points equidistant from the x and y axis and
give the equation.

This requires two expressions: one for the distance of point (x, y)
from the x axis, and one for the distance from the y axis. Those are
pretty simple, but remember those absolute values! Now set them equal,
and you have an equation describing all points for which the two
distances are equal, which is your locus.

To describe this, either look at the equation in terms of quadrants,
or think geometrically about angle bisectors.

4. Describe the locus of points equidistant from the points F (-2,1)
and G (6,3) and give the equation.

Again, write an expression for the distance of (x, y) from (-2, 1),
and one for the distance of (x, y) from (6, 3), and set them equal.
You can simplify this equation quite a lot; squaring both sides will
be the first step.

You may prefer to do the hard work in geometry rather than algebra.
There is a theorem that tells you that the locus of points equidistant
from two points is the perpendicular bisector of the segment they
define. You will have to find the coordinates of the midpoint of this
segment (think of averaging their coordinates), and write the equation
of the line through this point perpendicular to the segment (the slope
of this line will be the negative reciprocal of the slope between the
two points).

It's fun to do this both ways and see that geometry and algebra agree!
But it does take some work either way.

5. Describe the locus of points equidistant from lines y = 4 and y = -2
and give the equation of the line.

Now we're back to something like the first problem. Write expressions
for the distance of (x, y) from each line, and set them equal. The
absolute values get in the way, but if you think carefully you can
simplify this into the equation for a single line. You may be able to
avoid the absolute values entirely if you think about what is happening
geometrically first. Where can a point be if it is the same distance
from two parallel lines?

I hope these examples demonstrate the meaning of locus and the methods
that can be used to find the equation. If you need more, try searching
for the word "locus" in our archives. If some of my specific steps
didn't make sense, or I went too fast, write back and I'll try again.

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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search