Study Guide – Locus of Points



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) 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).

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

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

Example 2

Describe verbally the locus of points x = 5.

Case I - On the number line.

Solution is the point A(5).

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).

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.

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

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 [pic].

b) If A(2) and B(10), then M, the midpoint of [pic], is M(6).

Case II – on the plane

a) The perpendicular bisector of [pic].

b) If A(2, -1) and B(10, 5), then the midpoint is (6, 2) and the slope of [pic] is 6/8 or ¾. Slope of [pic] bisector is -4/3.

Solution is [pic]

Case III – in space

a) Solution is the plane perpendicular to [pic] at its midpoint

b) If A(2, -1, 0) and B(10, 5, 0), then the solution is the plane with equation [pic].

Example 6

Draw and describe verbally the locus of points that are 5 units from a given [pic].

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

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 |In a Plane |In Space |

|The locus of points that are 4 units from the set of|Two circles with center (0,0) radii 1 and |Two baseless cylinders with axis being the z-axis and |

|points described by x2 + y2 = 25 |9 and described by |radii of 1 and 9 and described by |

| |x2 + y2 = 1 and x2 + y2 = 81 |x2 + y2 = 1 and x2 + y2 = 81. |

|The locus of points that are 6 units from the set of|One circle described by |One baseless cylinder with axis the z-axis and radius 11 |

|points described by x2 + y2 = 25 |x2 + y2 = 121 |and described by x2 + y2 = 121 |

|The locus of points that are 5 units from the set of|A circle described by |The z-axis and a baseless cylinder with z-axis as its axis|

|points described by x2 + y2 = 25 |x2 + y2 = 100 and the point (0,0) |and radius 10 and described by x2 + y2 = 100 and the point|

| | |(0,0) |

|The locus of points that are described by x2 + y2 |The circle described by |The baseless cylinder with axis the z-axis and radius 5 |

|[pic] 25 |x2 + y2 = 25 and its interior |and its interior |

|The locus of points that are 5 units from the set of|No meaning |A sphere described by described by |

|points described by x2 + y2 + z2 = 25 | |x2 + y2 + z2 = 100 and the point (0,0,0) |

|The locus of points that are 6 units from the set of|No meaning |The sphere described by |

|points | |x2 + y2 + z2 = 25 and its interior. |

|described by x2 + y2 + z2 [pic] 25 | | |

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.

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Dog leash (13 feet)

Height of Clothesline

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Distance on Ground

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dog

20 ft.

15 ft.

15 ft

30 ft.

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