Study Guide – Locus of Points
The word “locus” comes from a Latin word meaning “location” or “place.” In geometry, a locus
is a set of all points that satisfies a given condition or a set of conditions – we could think of a
locus as a set of locations that satisfies a given condition or a set of conditions. We can describe
locations verbally (in words) and algebraically (with coordinates or an equation). In order to
determine the locus of points, the sample space must be known. Answers vary according to the
sample space used. Model the locus first so that you have a firm idea of what you are trying to
describe.
These are the steps to solving a locus problem.
1) Where is it? Is the locus on the number line, on a plane, or in space? Note: The answers
will be different for each environment.
2) Model the given information.
3) Model the locus condition(s).
4) Describe the locus verbally.
5) Describe the locus algebraically if required.
Example 1
a) Describe verbally the locus of points that are 7 cm. from a point P.
b) Describe algebraically the locus of points that are 7 cm. from a point P
Solution:
Case I - on the number line (in cm.)
●
●
A(a-7)
P(a)
a) Solution is two points A and B each seven units from P
b) If P(a), then solution is two points A(a-7) and B(a+7).
●
B(a+7)
Case II - on the plane
a) Solution is a circle with center P and radius 7 cm.
b) If P(a,b), then solution is (x-a)2 + (y-b)2 = 49
●
P(a,b)
Case III - in space
a) Solution is a sphere with center P and radius 7 cm.
b) If P(a,b,c), then solution is (x-a)2 + (y-b)2 +(z-c)2 = 49
●
P
Example 2
Describe verbally the locus of points x = 5.
Case I - On the number line.
Solution is the point A(5).
●
●
P(0)
●
A(5)
y
Case II – On the plane
Solution is a line perpendicular to the x-axis at (5,0) or,
worded another way, solution is a line parallel to the
y-axis that intersects the x-axis at (5,0).
1
Case III – In Space
Solution is a plane perpendicular to the x-axis at (5,0,0)
or, worded another way, solution is a plane parallel to the
y-z plane that intersects the x-axis at (5,0,0).
Note: A 2-dimensional drawing of a 3-dimensional
solution is not always very clear. Because of this,
we will not require drawing the locus of points in
space.
●5
Example 3
a) Describe verbally the locus of points 3 units from x = 5.
b) Describe algebraically the locus of points 3 units from x = 5.
Case I – On the number line
a) Two points 3 units on either side of P(5).
b) Two points A(2) and B(8)
Case II – On the plane
a) Two lines parallel to and 3 units on either side of the line x = 5
b) Two lines with equations x = 2 and x = 8.
Case III – In space
a) Two planes parallel to and 3 units on either side of the plane x = 5.
b) Two planes with equations x = 2 and x = 8
x
Example 4
a) Describe verbally the locus of points 5 units from the y-axis.
b) Describe algebraically the locus of points 5 units from the y-axis.
Case I – on the number line
This problem has no meaning on the number line.
Case II – on the plane
a) Two lines parallel to the y-axis and 5 units on either side of it
b) Two lines x = 5 and x = -5
Case III – in space
a) A baseless cylinder with axis being the y-axis and radius 5.
b) A baseless cylinder x2 + z2 = 25
Example 5
a) Describe verbally the locus of points equidistant from points A and B
b) Describe algebraically the locus of points equidistant from points A and B.
Case I – on the number line
a) Solution is the midpoint of AB .
b) If A(2) and B(10), then M, the midpoint of AB , is M(6).
Case II – on the plane
a) The perpendicular bisector of AB .
b) If A(2, -1) and B(10, 5), then the midpoint is (6, 2) and the slope of AB is 6/8 or ¾.
Slope of  bisector is -4/3.
4
(x  6)
Solution is y  2 
3
Case III – in space
a) Solution is the plane perpendicular to AB at its midpoint
b) If A(2, -1, 0) and B(10, 5, 0), then the solution is the plane with equation
4
y2
(x  6) .
3
Example 6
Draw and describe verbally the locus of points that are 5 units from a given AB .
Case I – on the number line
The problem has no meaning.
Case II – on the plane
Solution is the union of two segments parallel to the original segment and five units away on
either side with two semicircles of radius five and centers at the endpoints of the given segment
Case III – in space
Solution is the union of a baseless cylinder with axis the given segment, radius of 5, and height =
given segment’s length and two hemispheres with radius five and centers at the endpoints of the
given segment.
Example 7
A dog is tied to a clothesline with a 13-foot leash so that the leash can move along the
clothesline, but stops at each end. If the clothesline is 39 feet long and is 5 feet above the dog’s
collar, find the locus of the boundary of the dog’s running area.
Top View
End View
Dog leash (13 feet)
Height of Clothesline
(5 feet)
Distance on Ground
(12 feet)
Solution is the union of a rectangle that is 24 feet by 39 feet and two semicircles that have
centers directly below the endpoints of the clothesline and radii of 12 feet and the interiors of
each.
Example 8
Problem
The locus of points that are 4
units from the set of points
described by x2 + y2 = 25
The locus of points that are 6
units from the set of points
described by x2 + y2 = 25
The locus of points that are 5
units from the set of points
described by x2 + y2 = 25
The locus of points that are
described by x2 + y2  25
The locus of points that are 5
units from the set of points
described by x2 + y2 + z2 = 25
The locus of points that are 6
units from the set of points
described by x2 + y2 + z2  25
In a Plane
Two circles with center
(0,0) radii 1 and 9 and
described by
x2 + y2 = 1 and x2 + y2 = 81
One circle described by
x2 + y2 = 121
A circle described by
x2 + y2 = 100 and the point
(0,0)
The circle described by
x2 + y2 = 25 and its interior
No meaning
No meaning
In Space
Two baseless cylinders with axis
being the z-axis and radii of 1 and 9
and described by
x2 + y2 = 1 and x2 + y2 = 81.
One baseless cylinder with axis the zaxis and radius 11 and described by x2
+ y2 = 121
The z-axis and a baseless cylinder
with z-axis as its axis and radius 10
and described by x2 + y2 = 100 and the
point (0,0)
The baseless cylinder with axis the zaxis and radius 5 and its interior
A sphere described by described by
x2 + y2 + z2 = 100 and the point (0,0,0)
The sphere described by
x2 + y2 + z2 = 25 and its interior.
dog
Homework Problem
A dog is on a 20-ft leash. The leash is attached to a pipe at the
midpoint of the back wall of a 30 ft-by-30 ft. house, as shown in
the diagram. Sketch and use shading to indicate the region in
which the dog can play while attached to the leash. Include
measurements to describe the region.
15 ft.
15 ft
30 ft.
20 ft.