Name _____________________
Worksheet 13.1
Locus Definition
A locus of points can be thought of as a subset of the Cartesian plane. In other words, if we consider every possible pair of numbers ( , ) , only some pairs will satisfy a given condition.
Often the collection of all such pairs will form an interesting shape that merits further study.
As an example, consider the line y x 3 . This equation can be viewed as a function (for each input x, there exists an output y), though it can also be viewed as a way to define a locus of points. This can be written more formally as a set of points.
{( , ) | y
 
3}
This is read as “the set of all ordered pairs ( , ) such that y is three greater than x.” Thus (4, 7) is in this set, but (6, 5) is not. In fact, another way to write the above set would be to simply list every possible pair in the set.
Unfortunately, we do not have enough space to write down every pair in the set, so we draw a picture instead. Points on the line satisfy the equation y x 3 , points not on the line do not.
The restriction does not need to be an equation, but can also be some sort of verbal description.
Furthermore, the locus does not need to be a line or even a function. All that really matters is that, given the condition, there is no ambiguity as to whether a particular point is in the locus.
{ P

( , ) | P is in the third quadrant }
This defines a subset of the Cartesian plan which is represented by the following diagram.

Equidistant From Two Points
Consider to fixed points in space A and B. Consider the following subset.
{ P

( , ) | P is equidistant from A and B}
The locus of points that is equidistant from two fixed points is a _________________
Example
Consider two points A

(1, 4) and B

(5, - 2) .
(a) Use the distance formula to give the equation of the locus of points P such that P is equidistant from A and B. In more formal terms, find ( ) such that the following two subsets are equivalent.
{ | ( , )

( , )}

{( , ) | y

( )}
(b) Sketch your solution.

Equidistant from One Point
The locus of points that is equidistant from a single fixed point is a ____________________
Examples
(a) Consider the point A

(2, 1) . Sketch the locus of points P such that ( , ) 3
(b) Find the equation of this locus. You may find the distance formula useful.
(c) A circle is given by the equation x
2 
10 x
 y
2 
6 y

30

0 . Rearrange this equation such that you can easily state the center and radius of the circle. (Ask if you do not remember how to
“complete the square”).
(d) How many points of intersection are there between the y-axis and the circle defined by x y x

4)
2 
( y

2)
2 
9}

Not Quite Equidistant
Let A

(0, 2) and B

(0,8) . Consider the locus of points such that the distance to B is twice the distance to A.
{ P

( , ) | ( , )

2 ( , )}
(a) Sketch the locus of P on the bullet paper below. (The Cartesian coordinates will need to be drawn on).
(b) Work out the equation of P (simplify to find the center and radius of this circle)

Another Circle
A ( 2, 0) , B

(4, 0) , and k = 2. Consider the locus of points P such that ( , )

( , )
(a) There are two distinct points C and D on the locus that also fall on the x-axis. Find the points
C and D (use your intuition, not any equation).
(b) Assume you are confident the locus defines above is a circle. Use the location of C and D
(and the fact that they are part of the locus) to find the equation of the circle.

Interesting Problems
(a) Sketch the locus of points which are all one unit from a square (just four segments – not the interior) with side length of 4 units.
(b) Consider a robotic arm that consists of two line segments AB and BC. A = (0,0) is a fixed point. AB has a length of 4 meters, while BC
(clearly B is some sort of robot joint) has a length of 1 meter. Sketch all the places in the
Cartesian plane that point C can travel. This space is called the workspace of the robot.
(c) Now assume that both joints only have 180 degrees of freedom. Consider all possible places C can move to given that B must remain above the x-axis, and that the angle between
AB and BC is between -90° and 90°. Clearly sketch and label the workspace with relevant dimensions, and trust that this problem is trickier than it may first appear.