In not shown, the length of radius Provide the missing statements and reasons in the

At times, we need to describe a set of points that satisfy a given condition or set of con- ditions. The term used to describe the resulting geometric figure is locus pronounced l¯o-k˘us, the plural of which is loci pronounced l¯o-s¯i. The English word location is de- rived from the Latin word locus. Locus of Points in a Plane Locus of Points in Space Locus of Points 7.1 KEY CONCEPTS P r Figure 7.1 In this definition, the phrase “all points and only those points” has a dual meaning: 1. All points of the locus satisfy the given condition. 2. All points satisfying the given condition are included in the locus. The set of points satisfying a given locus can be a well-known geometric figure such as a line or a circle. In Examples 1, 2, and 3, several points are located in a plane and then connected in order to form the locus. EXAMPLE 1 Describe the locus of points in a plane that are at a fixed distance r from a given point P. Solution The locus is the circle with center P and radius r. See Figure 7.1. EXAMPLE 2 Describe the locus of points in a plane that are equidistant from two fixed points P and Q. Solution The locus is the line that is the perpendicular bisector of . In Figure 7.2, for any point X on line t. PX = QX PQ A locus is the set of all points and only those points that satisfy a given condition or set of conditions. DEFINITION 쮿 P X Q t Figure 7.2 쮿 P Figure 7.4 P Q Figure 7.5 Some definitions are given in a locus format; for example, the following is an alternative definition of the term circle. EXAMPLE 3 Describe the locus of points in a plane that are equidistant from the sides of an angle in that plane. Solution The locus is the ray that bisects . See Figure 7.3. ∠ABC BD ∠ABC Each of the preceding examples includes the phrase “in a plane.” If that phrase is omit- ted, the locus is found “in space.” For instance, the locus of points that are at a fixed distance from a given point is actually a sphere the three-dimensional object in Figure 7.4; the sphere has the fixed point as center, and the fixed distance determines the length of the radius. Unless otherwise stated, we will consider the locus to be restricted to a plane. B C A D Figure 7.3 A circle is the locus of points in a plane that are at a fixed distance from a given point. DEFINITION Exs. 1–4 EXAMPLE 4 Describe the locus of points in space that are equidistant from two parallel planes P and Q. Solution The locus is the plane parallel to each of the given planes and midway between them. See Figure 7.5. 쮿 There are two very important theorems involving the locus concept. The results of these two theorems will be used in Section 7.2. When we verify the locus theorems, we must establish two results: 1. If a point is in the locus, then it satisfies the condition. 2. If a point satisfies the condition, then it is a point of the locus. The locus of points in a plane and equidistant from the sides of an angle is the angle bisector. THEOREM 7.1.1 PROOF Note that both parts i and ii are necessary. 쮿 B C E A D F a B C E A D F b i If a point is on the angle bisector, then it is equidistant from the sides of the angle. GIVEN: bisects and PROVE: PROOF: In Figure 7.6a, bisects ; thus, . and , so and are right . By Identity, . By AAS, . Then by CPCTC. ii If a point is equidistant from the sides of an angle, then it is on the angle bisector. GIVEN: such that and PROVE: bisects ; that is, D is on the bisector of PROOF: In Figure 7.6b, and , so and are right angles. by hypothesis. Also, by Identity. Then by HL. Then by CPCTC, so bisects by definition. In locus problems, we must remember to demonstrate two relationships in order to validate results. A second important theorem regarding a locus of points follows. ∠ABC BD ∠ABD ⬵ ∠CBD 䉭DEB ⬵ 䉭DFB BD ⬵ BD DE ⬵ DF ∠DFB ∠DEB DF ⬜ BC DE ⬜ BA ∠ABC ∠ABC BD DE ⬵ DF DF ⬜ BC DE ⬜ BA ∠ABC DE ⬵ DF 䉭DEB ⬵ 䉭DFB BD ⬵ BD ∠s ⬵ ∠DFB ∠DEB DF ⬜ BC DE ⬜ BA ∠ABD ⬵ ∠CBD ∠ABC BD DE ⬵ DF DF ⬜ BC DE ⬜ BA ∠ABC BD PROOF i If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the line segment. GIVEN: and point X not on , so that [See Figure 7.7a.] PROVE: X lies on the perpendicular bisector of AB AX = BX AB AB The locus of points in a plane that are equidistant from the endpoints of a line segment is the perpendicular bisector of that line segment. THEOREM 7.1.2 Exs. 5, 6 a A B X Figure 7.7 b A B M X 1 2 PROOF: Let M represent the midpoint of . Draw [See Figure 7.7b.] Then . Because , we know that . By Identity, ; thus, by SSS. By CPCTC, 1 and 2 are congruent and .. By definition, is the perpendicular bisector of , so X lies on the perpendicular bisector of . AB AB Í MX Í MX ⬜ AB ∠s 䉭AMX ⬵ 䉭BMX XM ⬵ XM AX ⬵ BX AX = BX AM ⬵ MB Í MX AB Figure 7.6 쮿 ii If a point is on the perpendicular bisector of a line segment, then the point is equidistant from the endpoints of the line segment. GIVEN: Point X lies on , the perpendicular bisector of [See Figure 7.8a.] PROVE: X is equidistant from A and B [See Figure 7.8b.] PROOF: X is on the perpendicular bisector of , so 1 and 2 are congruent right angles and . With , AMX and BMX are congruent by SAS; in turn, by CPCTC. Then and X is equidistant from A and B. We now return to further considerations of a locus in a plane. Suppose that a given line segment is to be used as the hypotenuse of a right triangle. How might you locate possible positions for the vertex of the right angle? One method might be to draw 30° and 60° angles at the endpoints so that the remaining angle formed must measure 90° [see Figure 7.9a]. This is only one possibility, but because of sym- metry, it actually provides four permissible points, which are indicated in Figure 7.9b. This problem is completed in Example 5. XA = XB XA ⬵ XB 䉭s XM ⬵ XM AM ⬵ MB ∠s AB AX = XB AB Í MX A B M X 1 2 b A B M X a Figure 7.8 60° 30° a Figure 7.9 b Reminder An angle inscribed in a semicircle is a right angle. Exs. 7, 8 EXAMPLE 5 Find the locus of the vertex of the right angle of a right triangle if the hypotenuse is in Figure 7.10a. Solution Rather than using a “hit or miss” approach for locating the possible vertices as suggested in the paragraph preceding this example, recall that an angle inscribed in a semicircle is a right angle. Thus, we construct the circle whose center is the midpoint M of the hypotenuse and whose radius equals one-half the length of the hypotenuse. Figure 7.10b: First, the midpoint M of the hypotenuse is located. AB AB a A B Figure 7.10 A B b M A B c M 쮿 In Example 5, the construction involves locating the midpoint M of , and this is found by the construction of the perpendicular bisector. The compass is then opened to a radius whose length is MA or MB, and the circle is drawn. When a construction is performed, it falls into one of two categories: 1. A basic construction method 2. A compound construction problem that may require several steps and may involve several basic construction methods as in Example 5 The next example also falls into category 2. Recall that the diagonals of a rhombus are perpendicular and also bisect each other. With this information, we can locate the vertices of the rhombus whose diagonals lengths are known. AB Exs. 9, 10 EXAMPLE 6 Construct rhombus ABCD given its diagonals and . See Figure 7.11. Solution Figure 7.11a: To begin, we construct the perpendicular bisector of ; we know that the remaining vertices B and D must lie on this line. As shown, M is the midpoint of . Figure 7.11b: To locate the midpoint of , we construct its perpendicular bisector as well. The midpoint of is also called M. Figure 7.11c: Using an arc length equal to one-half the length of such as MB , we mark off this distance both above and below on the perpendicular bisector determined in Figure 7.11a. AC BD BD BD AC AC BD AC a A C A C M B D Figure 7.11 b B D M c A C D M B d A C D M B Figure 7.10c: With the length of the radius of the circle equal to one-half the length of the hypotenuse such as MB, the circle with center M is drawn. The locus of the vertex of the right angle of a right triangle whose hypotenuse is given is the circle whose center is at the midpoint of the given segment and whose radius is equal in length to half the length of the given segment. Every point except A and B on is the vertex of a right triangle with hypotenuse ; see Theorem 6.1.9. AB } M 쮿 쮿 Figure 7.11d: Using the marked arcs to locate determine points B and D, we join A to B, B to C, C to D, and D to A. The completed rhombus is ABCD as shown. Exercises 7.1

1. In the figure, which of the points A, B,

C, D , and E belong to “the locus of points in the plane that are at distance r from point P”? In Exercises 11 to 22, sketch and describe each locus in the plane.

11. Find the locus of points that are at a given distance from

a fixed line.

12. Find the locus of points that are equidistant from two

given parallel lines.

13. Find the locus of points that are at a distance of 3 in.

from a fixed point O.

14. Find the locus of points that are equidistant from two

fixed points A and B.

15. Find the locus of points that are equidistant from three

noncollinear points D, E, and F.

16. Find the locus of the midpoints of the radii of a circle

O that has a radius of length 8 cm.

17. Find the locus of the midpoints of all chords of circle

Q that are parallel to diameter .

18. Find the locus of points in the interior of a right triangle

with sides of 6 in., 8 in., and 10 in. and at a distance of 1 in. from the triangle.

19. Find the locus of points that are equidistant from two

given intersecting lines.

20. Find the locus of points that are equidistant from a fixed

line and a point not on that line. NOTE: This figure is known as a parabola.

21. Given that lines p and q intersect, find the locus of points

that are at a distance of 1 cm from line p and also at a distance of 2 cm from line q.

22. Given that congruent circles O and P have radii of length

4 in. and that the line of centers has length 6 in., find the locus of points that are 1 in. from each circle. In Exercises 23 to 30, sketch and describe the locus of points in space.

23. Find the locus of points that are at a given distance from

a fixed line.

24. Find the locus of points that are equidistant from two

fixed points.

25. Find the locus of points that are at a distance of 2 cm

from a sphere whose radius is 5 cm.

26. Find the locus of points that are at a given distance from

a given plane.

27. Find the locus of points that are the midpoints of the

radii of a sphere whose center is point O and whose radius has a length of 5 m.

28. Find the locus of points that are equidistant from three

noncollinear points D, E, and F.

29. In a room, find the locus of points that are equidistant

from the parallel ceiling and floor, which are 8 ft apart. PR P E B C D A r B C A

2. In the figure, which of the

points F, G, H, J, and K belong to “the locus of points in the plane that are at distance r from line ”? r r J F G H K In Exercises 3 to 8, use the drawing provided.

3. Given:

Obtuse Construct: The bisector of ∠ABC 䉭ABC Exercises 3–8

4. Given:

Obtuse Construct: The bisector of

5. Given:

Obtuse Construct: The perpendicular bisector of

6. Given:

Obtuse Construct: The perpendicular bisector of

7. Given:

Obtuse Construct: The altitude from A to HINT: Extend .

8. Given:

Obtuse Construct: The altitude from B to

9. Given:

Right Construct: The median from S to

10. Given: Right

Construct: The median from R to ST 䉭RST RT 䉭RST AC 䉭ABC BC BC 䉭ABC AC 䉭ABC AB 䉭ABC ∠BAC 䉭ABC S T R Exercises 9–10

30. Find the locus of points that are equidistant from all

points on the surface of a sphere with center point Q. In Exercises 31 and 32, use the method of proof of Theorem 7.1.1 to justify each construction method. 31. The perpendicular bisector method. 32. The construction of a perpendicular to a line from a point outside the line. In Exercises 33 to 36, refer to the line segments shown.

33. Construct an isosceles right triangle that has hypotenuse

. AB

39. Use the following theorem to locate the

center of the circle of which is a part. Theorem: The perpendicular bisector of a chord passes through the center of a circle. RT ¬

34. Construct a rhombus whose sides are equal in length

to AB, and so that one diagonal of the rhombus has length CD.

35. Construct an isosceles triangle in which each leg has

length CD and the altitude to the base has length AB.

36. Construct an equilateral triangle in which the altitude to

any side has length AB.

37. Construct the three angle bisectors and then the inscribed

circle for obtuse . 䉭RST A B C D Exercises 33–36 S T R Exercises 37, 38 S T R W X Y Z B C A

40. Use the following theorem to construct the geometric

mean of the numerical lengths of the segments and . Theorem: The length of the altitude to the hypotenuse of a right triangle is the geometric mean between the lengths of the segments of the hypotenuse. YZ WX

41. Use the following theorem to construct a triangle similar

to the given triangle but with sides that are twice the length of those of the given triangle. Theorem: If the three pairs of sides for two triangles are in proportion, then those triangles are similar SSS . Concurrent Lines Incenter Incircle Circumcenter Circumcircle Orthocenter Centroid Concurrence of Lines 7.2 KEY CONCEPTS In this section, we consider lines that share a common point. A number of lines are concurrent if they have exactly one point in common. DEFINITION

38. Construct the three perpendicular bisectors of sides and

then the circumscribed circle for obtuse . 䉭RST

42. Verify this locus theorem:

The locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining those points. The three lines in Figure 7.12 are concurrent at point A. The three lines in Figure 7.13 are not concurrent even though any pair of lines such as r and s do intersect. Parts of lines rays or segments are concurrent if they are parts of concurrent lines and the parts share a common point. Discover A computer software program can be useful in demonstrating the concurrence of the lines described in each theorem in this section. n m A p m, n, and p are concurrent Figure 7.12 s r t r, s, and t are not concurrent Figure 7.13 The three angle bisectors of the angles of a triangle are concurrent. THEOREM 7.2.1 For the informal proofs of this section, no Given or Prove is stated. In more advanced courses, these parts of the proof are understood. Exs. 1, 2 Reminder A point on the bisector of an angle is equidistant from the sides of the angle. EXAMPLE 1 Give an informal proof of Theorem 7.2.1. Proof In Figure 7.14a, the bisectors of and intersect at point E. Because the bisector of is the locus of points equidistant from the sides of , we know that in Figure 7.14b. Similarly, because E is on the bisector of . ∠ABC EM ⬵ EP EM ⬵ EN ∠BAC ∠BAC ∠ABC ∠BAC a B A C E Figure 7.14 b B A C E M P N By the Transitive Property of Congruence, it follows that . Because the bisector of an angle is the locus of points equidistant from the sides of the angle, E is also on the bisector of the third angle, . Thus, the angle bisectors are concurrent at point E. ∠ACB EP ⬵ EN 쮿 The point E at which the angle bisectors meet in Example 1 is the incenter of the triangle. As the following example shows, the term incenter is well deserved because this point is the center of the inscribed circle of the triangle.