1. Properties or Parallel Lines
GEOMETRY LESSON 3-1
(For help, go to page 24 or Skills Handbook, page 720.)
Solve each equation.
1. x + 2x + 3x = (w + 23) + (4w + 7) = 180
3. 90 = 2y – – 5y = 135
Write an equation and solve the problem.
5. The sum of m 1 and twice its complement is 146. Find m 1.
6. The measures of two supplementary angles are in the ratio 2 : 3. Find their measures.
3-1

2. Properties or Parallel Lines
GEOMETRY LESSON 3-1
1. Combine like terms: 6x = 180; divide both sides by 6: x = 30
2. Combine like terms: 5w + 30 = 180; Subtract 30 from both sides: 5w = 150; divide both sides by 5: w = 30
3. Add 30 to both sides: 120 – 2y; divide both sides by 2: 60 = y
4. Subtract 180 from both sides: – 5y = –45; divide both sides by –5: y = 9
5. The complement of is 90 – m 1; m (90 – m 1) = 146; distribute:
m – 2m = 146; combine like terms: 180 – m 1 = 146; subtract
180 from both sides: –m 1 = –34; multiply both sides by –1: m 1 = 34
6. Let 2x represent the measure of one of the angles and 3x represent the measure
of the other angle. Then 2x + 3x = 180; combine like terms: 5x = 180; divide
both sides by 5: x = 36. Then 2x = 72 and 3x =
Solutions
3-1

3. Properties or Parallel Lines
GEOMETRY LESSON 3-1
Use the diagram above. Identify which angle forms a pair of same-side
interior angles with Identify which angle forms a pair of corresponding
angles with 1.
Same-side interior angles are on the same side of transversal t
between lines p and q.
4, 8, and 5 are on the same side of the transversal as 1,
but only 1 and 8 are interior.
So 1 and 8 are same-side interior angles.
3-1

4. Properties or Parallel Lines
GEOMETRY LESSON 3-1
(continued)
Corresponding angles also lie on the same side of the transversal.
The angle corresponding to 1 must lie in the same position relative to
line q as 1 lies relative to line p. Because 1 is an interior angle, 1 and 5
are corresponding angles.
One angle must be an interior angle, and the other must be an exterior angle.
3-1

5. Properties or Parallel Lines
GEOMETRY LESSON 3-1
Compare 2 and the vertical angle of Classify the angles
as alternate interior angles, same–side interior angles, or
corresponding angles.
The vertical angle of 1 is between the parallel runway segments.
2 is between the runway segments and on the opposite side of
the transversal runway.
Because alternate interior angles are not adjacent and lie between
the lines on opposite sides of the transversal, 2 and the vertical angle
of 1 are alternate interior angles.
3-1

6. Properties or Parallel Lines
GEOMETRY LESSON 3-1
Which theorem or postulate gives the reason that m m 2 = 180?
3 and 2 are adjacent angles that form a straight angle.
m m 2 = 180 because of the Angle Addition Postulate.
3-1

7. Properties or Parallel Lines
GEOMETRY LESSON 3-1
In the diagram above, || m. Find m 1 and then m 2.
If you substitute 42 for m 1, the equation becomes 42 + m 2 = 180.
Subtract 42 from each side to find m 2 = 138.
Because 1 and 2 are adjacent angles that form a straight angle,
m m 2 = 180 by the Angle Addition Postulate.
1 and the 42° angle are corresponding angles. Because || m,
m 1 = 42 by the Corresponding Angles Postulate.
3-1

8. Properties or Parallel Lines
GEOMETRY LESSON 3-1
In the diagram above, || m. Find the values of a, b, and c.
a = Alternate Interior Angles Theorem
c = Alternate Interior Angles Theorem
a + b + c = Angle Addition Postulate
65 + b + 40 = Substitution Property of Equality
b = Subtraction Property of Equality
3-1

9. Properties or Parallel Lines
GEOMETRY LESSON 3-1
Pages Exercises
1. PQ and SR with transversal SQ; alt. int.
2. PS and QR with transversal SQ; alt. int.
3. PS and QR with transversal PQ; same-side int.
4. PS and QR with transversal SR; corr. s
and 2: corr.
3 and 4: alt. int.
5 and 6: corr.
and 2: same-side int.
3 and 4: corr.
and 2: corr.
3 and 4: same-side int.
5 and 6: alt. int.
8. alt. int.
9. a. 2
b. 1
c. corr.
10. a. Def. of
b. Def. of right
c. Corr. of i lines are .
d. Subst.
e. Def. of right
f. Def. of
3-1

10. Properties or Parallel Lines
GEOMETRY LESSON 3-1
11. m 1 = 75 because corr. of || lines are ; m 2 = 105 because same-side int. of || lines are suppl.
12. m 1 = 120 because corr. of || lines are ; m 2 = 60 because same-side int. of || lines are suppl.
13. m 1 = 100 because same-side int. of || lines are suppl.; m 2 = 70 because alt. int. of || lines have = measure. s
14. 70; 70, 110
15. 25; 65
16. 20; 100, 80
17. m 1 = m 3 = m 6 = m 8 = m 9 = m = m 13 = m 15 = 52; m 2 = m 4 = m 5 = m 7 = m 10 = m 12 = m 14 = 128
18. You must find the measure of one . All that are vert., corr., or alt. int. to that will have that measure. All other will be the suppl. of that measure.
19. two
20. four
21. two
22. four
23. 32 s
3-1

11. Properties or Parallel Lines
GEOMETRY LESSON 3-1
28. Answers may vary.
Sample: E illustrates corr. ( 1 and 3, and 4) and same-side int. ( and 2, 3 and 4); I illustrates alt. int. ( 1 and 4, and 3) and same-side int ( 1 and 3, 2 and 4).
29. a. alt. int.
b. He knew that alt. int. of || lines are .
30. a. 57
b. same-side int.
24. x = 76, y = 37, v = 42, w = 25
25. x = 135, y = 45
26. The labeled are corr. and should be If you solve 2x – 60 = 60 – 2x, you get x = 30. This would be impossible since 2x – 60 and 60 – 2x would equal 0.
27. Trans means across or over. A transversal cuts across other lines. s s s s s s s s s
3-1

12. Properties or Parallel Lines
GEOMETRY LESSON 3-1
31. a. If two lines are || and cut by a transversal, then same-side ext. are suppl.
b. Given: a || b
Prove: 4 and 5 are suppl.
1. a || b (Given) 2. m m 6 = 180 ( Add. Post.)
(Corr. are ) 4. m m 4 = 180 (Subst.) and 5 are suppl. (Def. of suppl.)
a || b (Given) (Vert. are .)
(Corr. are .) (Trans. Prop.)
33. Never; the two planes do not intersect.
34. Sometimes; if they are ||.
35. Sometimes; they may be skew.
36. Sometimes; they may be ||.
37. D
38. G
39. D
40. I s s s s
3-1

13. Properties or Parallel Lines
GEOMETRY LESSON 3-1
41. [2] a.  First show that Then show that Finally, show that (OR other valid solution plan).
b.  because vert. are because corr. of || lines are Finally, by the Trans. Prop. of ,
[1] incorrect sequence of steps OR incorrect logical argument s
43. 59
45. (0.5, 7)
46. (–0.5, 3.5)
47. (3, 3)
48. add 4; 20, 24
49. multiply by –2; 16, –32
50. subtract 7; –5, –12
3-1

14. Properties or Parallel Lines
GEOMETRY LESSON 3-1
In the diagram below, m || n. Use the diagram for Exercises 1–5.
1. Complete: and are alternate interior angles.
2. Complete: and are corresponding angles.
3. Suppose that m 3 = 37. Find m 6.
4. Suppose that m 1 = x + 12 and m 5 = 3x – 36. Find x.
5. If a transversal intersects two parallel lines, then same-side
exterior angles are supplementary. Write a Plan for Proof.
Given: m || n Prove: and are supplementary. 6 4
143 24
Show that m 2 = m Then show that m m 7 = 180,
and substitute m 2 for m 6.
3-1

15. Proving Lines Parallel
GEOMETRY LESSON 3-2
(For help, go to page 24 and Lesson 2-1.)
Solve each equation.
1. 2x + 5 = a – 12 = 20
3. x – x + 80 = x – 7 = 3x + 29
Write the converse of each conditional statement.
Determine the truth value of the converse.
5. If a triangle is a right triangle, then it has a 90° angle.
6. If two angles are vertical angles, then they are congruent.
7. If two angles are same-side interior angles, then they are supplementary.
3-2

16. Proving Lines Parallel
GEOMETRY LESSON 3-2
Solutions
1. Subtract 5 from both sides: 2x = 22; divide both sides by 2: x = 11
2. Add 12 to both sides: 8a = 32; divide both sides by 8: a = 4
3. Combine like terms: 5x + 50 = 180; subtract 50 from both sides: 5x = 130; divide both sides by 5: x = 26
4. Add –3x + 7 to both sides: 6x = 36; divide both sides by 6: x = 6
5. Reverse the hypothesis and conclusion: If a triangle has a 90° angle,
then it is a right triangle. By definition of a right triangle, it is true.
6. Reverse the hypothesis and conclusion: If two angles are congruent, they are
vertical angles. A counterexample is the congruent base angles of an isosceles
triangle, which are not vertical angles. The converse is false.
7. Reverse the hypothesis and conclusion: If two angles are supplementary, then
they are the same-side interior angles of parallel lines. A counterexample is
two angles that form a straight angle. The converse is false.
3-2

17. Proving Lines Parallel
GEOMETRY LESSON 3-2
If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel.
Prove:  || m
Given: 
Write the flow proof below of the Alternate Interior Angles
Theorem as a paragraph proof.
By the Vertical Angles Theorem, , so by the
Transitive Property of Congruence. Because 3 and 2 are corresponding
angles, || m by the Converse of the Corresponding Angles Postulate.
3-2

18. Proving Lines Parallel
GEOMETRY LESSON 3-2
Use the diagram above. Which lines, if any, must be parallel if 3 and 2
are supplementary?
It is given that and are supplementary.
The diagram shows that and are supplementary.
Because supplements of the same angle are congruent
(Congruent Supplements Theorem),
Because and are congruent corresponding angles, EC || DK
by the Converse of the Corresponding Angles Postulate.
3-2

19. Proving Lines Parallel
GEOMETRY LESSON 3-2
Use the diagram above. Which angle would you use with 1 to prove
the theorem In a plane, if two lines are perpendicular to the same line,
then they are parallel to each other (Theorem 3-6) using the Converse
of the Alternate Interior Angles Theorem instead of the Converse of the
Corresponding Angles Postulate?
By the Vertical Angles Theorem, 2 is congruent to its vertical angle.
Because , 1 is congruent to the vertical angle of 2 by
the Transitive Property of Congruence.
Because alternate interior angles are congruent, you can use the vertical
angle of and the Converse of the Alternate Interior Angles Theorem to
prove that the lines are parallel.
3-2

20. Proving Lines Parallel
GEOMETRY LESSON 3-2
Find the value of x for which || m.
The labeled angles are alternate interior angles.
If || m, the alternate interior angles are congruent,
and their measures are equal.
Write and solve the equation 5x – 66 = x.
5x – 66 = x
5x = x Add 66 to each side.
2x = Subtract 3x from each side.
x = Divide each side by 2.
3-2

21. Proving Lines Parallel
GEOMETRY LESSON 3-2
Suppose that the top and bottom pieces of a picture frame are cut to make 60° angles with the exterior sides of the frame. At what angle should the two sides be cut to ensure that opposite sides of the frame will be parallel?
In order for the opposite sides of the frame to be parallel,
same-side interior angles must be supplementary.
Two 90° angles are supplementary, so find an adjacent angle that,
together with 60°, will form a 90° angle: 90° – 60° = 30°.
3-2

22. Proving Lines Parallel
GEOMETRY LESSON 3-2
Pages Exercises
1. BE || CG; Conv. of Corr. Post.
2. CA || HR; Conv. of Corr. Post.
3. JO || LM; if two lines and a transversal form same-side int. that are suppl., then the lines are ||. s
4. a || b; if two lines and a transversal form same-side int. that are suppl., then the lines are ||.
5. a || b; if two lines and a transversal form same-side int. that are suppl., then the lines are ||.
6. none
7. none
8. a || b; Conv. of Corr. Post.
9. none
10. a || b; Conv. of Alt. Int.
Thm.
|| m; Conv. of Corr. Post.
12. none
13. a || b; Conv. of Corr. Post.
14. none
3-2

23. Proving Lines Parallel
GEOMETRY LESSON 3-2
24. When the frame is put together, each of the frame is a right . Two right are suppl. By the Conv. of the Same-Side Int. Thm., opp. sides of the frame are ||.
25. The corr. are , so the lines are || by the Conv. of Corr. Post.
26. a. Corr.
b–c. 1, 3 (any order)
d. Conv. of Corr. s
|| m; Conv. of Alt. Int. Thm.
16. a. Def. of
b. Given
c. All right are .
d. Conv. of Corr. Post.
17. a. 1
b. 1
c. 2
d. 3
e. Conv. of Corr. s
18. 30
19. 50
20. 59
21. 31
22. 5
23. 20
3-2

24. Proving Lines Parallel
GEOMETRY LESSON 3-2
27. 10; m 1 = m 2 = 70
28. 5; m 1 = m 2 = 50
; m 1 = m 2 = 30
; m 1 = m 2 = 10
31. The corr. he draws are .
32. PL || NA and PN || LA by Conv. of Same-Side Int. Thm. s
33. PL || NA by Conv. of Same-Side Int. Thm.
34. none
35. PN || LA by Conv. of Same-Side Int. Thm.
36. Answers may vary. Sample: In the diagram, AB BH and AB BD, but BH || BD. They intersect.
37. Reflexive: a || a; false; any line intersects itself.
Symmetric: If a || b, then b || a; true; b and a are coplanar and do not intersect.
Transitive: In general, if a || b, and b || c, then a || c; true; however, when a || b, and b || a, it does not follow that a || a. s
3-2

25. Proving Lines Parallel
GEOMETRY LESSON 3-2
38. Reflexive: a a; false; lines are two lines that intersect to form right .
Symmetric: If a b, then b a; true; b and a intersect to form right .
Transitive: If a b, and b c, then a c; false; in a plane, two lines to the same line are ||.
39. The corr. are , and the oars are || by the Conv. of Corr. Post.
40. Answers may vary. Sample: ; j || k by Conv. of the Alt. Int. Thm.
41. Answers may vary. Sample: ; j || k by Conv. of the Alt. Int. Thm. and || m by Conv. of Same-Side Int. Thm. s
42. Answers may vary. Sample: ; || m by Conv. of the Alt. Int. Thm. and j || k by Conv. of Corr.
Post.
43. Answers may vary. Sample: 3 and 12 are suppl.; j || k by the Conv. of Corr. Post.
44. Vert. Thm. and Conv. of Corr. Post.
3-2

26. Proving Lines Parallel
GEOMETRY LESSON 3-2
45. It is given that || m, so by Corr.
Post. It is also given that , so by Trans. Prop. of .
So, j || k by the Conv. of Corr. Post. s
46.
47.
3-2

27. Proving Lines Parallel
GEOMETRY LESSON 3-2
48.
49.
50. a. Answers may vary. Sample:
b. Given: a || b with transversal e, c bisects AOB, d bisects AXZ.
c. Prove: c || d
3-2

28. Proving Lines Parallel
GEOMETRY LESSON 3-2
50. (continued)
d. To prove that c || d, show that
if AOB
OXZ. AOB
OXZ by the Corr. Post. s
e. 1. a || b (Given)
2. AOB AXZ (Corr. Post.)
3. m AOB = m AXZ (Def. of )
4. m AOB = m m 2; m AXZ = m m 4 ( Add. Post.)
5. c bisects AOB; d bisects AXZ. (Given)
6. m 1 = m 2; m 3 = m 4 (Def. of bisector)
7. m m 2 = m m 4 (Trans. Prop. of )
8. m m 1 = m m 3 (Subst.)
9. 2m 1 = 2m 3 (Add. Prop.)
10. m 1 = m 3 (Div. Prop.)
11. c || d (Conv. of Corr. Post.) s
3-2

29. Proving Lines Parallel
GEOMETRY LESSON 3-2
54. (continued)
b. x + 21 = 2x so x = Lines c and d are not || because x cannot = both 21 and 23 (OR equivalent explanation).
[1] incorrect equations OR incorrect solutions
55. [4] a.
51. C
52. F
53. B
54.  [2] a. 136 + (x + 21) = 180 so x = 23 (OR equivalent equation resulting in x = 23).
   
55. (continued)
b. m m 3 = If 2x – x = 180, then x = 25. The measures are 2x – 38 = 12 and 25, but So a can’t be || to b.
[3]  appropriate methods, but with one computational error
[2]  incorrect diagram solved correctly OR correct diagram solved incorrectly = /
3-2

30. Proving Lines Parallel
GEOMETRY LESSON 3-2
55. (continued) 
[1] correct answer (lines a and b are not ||), without work shown
56. m 1 = 70 since it is a suppl. of the 110° . m 2 = 110 since same-side int. are suppl.
57. m 1 = 66 because alt. int. are . m 2 = 180 – 94 = 86 because same-side int. are suppl.
58. If you are west of the Mississippi River, then you are in Nebraska. Original statement is true, converse is false.
59. If a circle has a radius of 4 cm, then it has a diameter of 8 cm. Both are true.
60. If same-side int. are suppl., then a line intersects a pair of ||
lines. Both are true.
61. If you form the past tense of a verb, then you add ed to the verb. Original statement is false, converse is false.
62. If there are clouds in the sky, then it is raining. Original statement is true, converse is false.
in.2
cm2 s s s s
3-2

31. Proving Lines Parallel
GEOMETRY LESSON 3-2
ft2
in.2 m2 m2
ft2 m2
3-2

32. Proving Lines Parallel
GEOMETRY LESSON 3-2
Use the diagram and the given information to
determine which lines, if any, are parallel. Justify your
answer with a theorem or postulate.
and are supplementary.
4. Find the value of x for which a || b.
5. Find the value of x for which m || n.
m || n by Converse of Corresponding Angles Post.
a || b by Converse of Same-Side Interior Angles Theorem.
No lines must be parallel.
Suppose that m 1 = 3x + 10, m 2 = 3x + 14,
and m 6 = x + 58 in the diagram above. 24 26
3-2

33. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
(For help, go to Lesson 1-4.)
Classify each angle as acute, right, or obtuse.
Solve each equation.
x = x = 180
6. x + 58 = x = 90
3-3

34. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Solutions
1. The measure of the angle is 90°, so it is a right angle.
2. The measure of the angle is between 0° and 90°, so it is an acute angle.
3. The measure of the angle is between 0° and 90°, so it is an acute angle.
4. Combine like terms: x = 180; subtract 120 from both sides: x = 60
5. Combine like terms: x = 180; subtract 160 from both sides: x = 20
6. Subtract 58 from both sides: x = 32
7. Subtract 32 from both sides: x = 58
3-3

35. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Find m Z.
m Z = 180 Triangle Angle-Sum Theorem
115 + m Z = 180 Simplify.
m Z = 65 Subtract 115 from each side.
3-3

36. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
In triangle ABC, ACB is a right angle, and CD AB.
Find the values of a, b, and c.
Find c first, using the fact that ACB is a right angle.
m ACB = 90 Definition of right angle
c + 70 = 90 Angle Addition Postulate
c = 20 Subtract 70 from each side.
3-3

37. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
To find a, use ADC.
(continued)
a + m ADC + c = 180 Triangle Angle-Sum Theorem
m ADC = 90 Definition of perpendicular lines
a = 180 Substitute 90 for m ADC and 20 for c.
a = 180 Simplify.
a = 70 Subtract 110 from each side.
To find b, use CDB.
70 + m CDB + b = Triangle Angle-Sum Theorem
m CDB = Definition of perpendicular lines
b = Substitute 90 for m CDB.
160 + b = Simplify.
b = Subtract 160 from each side.
3-3

38. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Classify the triangle by its sides and its angles.
The three sides of the triangle have three different lengths,
so the triangle is scalene.
One angle has a measure greater than 90, so the triangle is obtuse.
The triangle is an obtuse scalene triangle.
3-3

39. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Find m 1.
m = 125 Exterior Angle Theorem
m 1 = 35 Subtract 90 from each side.
3-3

40. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Explain what happens to the angle formed by the back of the chair and the armrest as you lower the back of the lounge chair.
The exterior angle and the angle formed by the back of the chair and
the armrest are adjacent angles, which together form a straight angle.
As one measure increases, the other measure decreases.
The angle formed by the back of the chair and the armrest increases as
you lower the back of the lounge chair.
3-3

41. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Pages Exercises
1. 30
3. 90
4. 71
5. 90
6. x = 70; y = 110; z = 30
7. t = 60; w = 60
8. x = 80; y = 80
9. 70
10. 30
11. 60
12. acute, isosceles
13. acute, equiangular, equilateral
14. right, scalene
15. obtuse, isosceles
16.
17. Not possible; a right will always have one longest side opp. the right .
18.
3-3

42. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
19.
20.
21.
22.
23.
24. a. 5, 6, 8
b and 3 for 5
1 and 2 for 6
1 and 2 for 8
c. They are vert. .
25. a. 2
b. 6 s
28. m 3 = 92; m 4 = 88
29. x = 147, y = 33
30. a = 162, b = 18
31. x = 52.5; 52.5, 52.5, 75; acute
32. x = 7; 55, 35, 90; right
33. x = 37; 37, 65, 78; acute
3-3

43. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
34. x = 38, y = 36, z = 90;
ABD: 36, 90, 54; right; BCD: 90, 52, 38; right; ABC: 74, 52, 54; acute
35. a = 67, b = 58, c = 125, d = 23, e = 90; FGH: 58, 67, 55; acute;
FEH: 125, 32, 23; obtuse; EFG: 67, 23, 90; right
36. x = 32, y = 62, z = 32, w = 118; ILK: 118, 32, 30; obtuse
37. 60; 180  3 = 60
38. Yes, an equilateral is isosc. Because if three sides of a are , then at least two sides are . No, the third side of an isosc. does not need to be to the other two.
39. eight
3-3

44. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
41. Check students’ work. Answers may vary. Sample: The two ext.
Formed at vertex A are vert. and thus have the same measure.
and 60
43. a. 40, 60, 80
b. acute s
47. 32
48. a. 90
b. 180
c. 90
d. compl.
e. compl.
49. a. Add.
b Sum
c. Trans.
d. Subtr. s
; since the missing
is 68, the largest ext. Is 180 – 48 = 132.
51. Check students’ work.
52. a. 81
b. 45, 63, 72
c. acute
or 60
or 45
55. 90
3-3

45. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
56. Greater than, because there are two with measure 90 where the meridians the equator.
57.
58.
59. 1
60.
61. 0 s 1 3 7 19
63. Answers may vary. Sample: The measure of the ext. is = to the sum of the measures of the two remote int Since these are , the formed by the bisector of the ext. are to each of them. Therefore, the bisector is || to the included side of the remote int. by the Conv. of the Alt Int. Thm.
64. B
65. G
66. B
67. H s
3-3

46. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
68. [2] a.
b. The correct equation is 2x = (2x – 40) + (x – 15) with the sol. x = 55. The three int measure 70, 40, and 70.
[1] incorrect sketch, equation, OR solution
69. [2] a. 159; the sum of the three of the is 180, so m Y + m M + m F = 180. Since m F = 21, m Y + m M + 21 = 180. Subtr. 21 from both sides results in m Y + m M = 159.
69. (continued)
b. 1 to 68 ; since Y is obtuse, its whole number range is from 91 to 158, allowing the measure of 1 for m M when m Y = When m Y = 91, then m M = 68.
[1] incorrect answer to part (a) or (b) OR incorrect computation in either part s s
3-3

47. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
70. 53
71. 46
72. 7
73.
74.
3-3

48. Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
1. A triangle with a 90° angle has sides that are 3 cm, 4 cm,
and 5 cm long. Classify the triangle by its sides and angles.
Use the diagram for Exercises 2–6.
2. Find m 3 if m 2 = 70 and m 4 = 42.
3. Find m 5 if m 2 = 76 and m 3 = 90.
4. Find x if m 1 = 4x, m 3 = 2x + 28, and m 4 = 32.
5. Find x if m 2 = 10x, m 3 = 5x + 40, and m 4 = 3x – 4.
6. Find m 3 if m 1 = 125 and m 5 = 160.
scalene right triangle 68
166 30 8
105
3-3

49. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
(For help, go to Lesson 1-4 and 3-3.)
Find the measure of each angle of quadrilateral ABCD. 3.
3-4

50. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Solutions
1. m DAB = = 77; m B = 65; m BCD = = 131; m D = 87
2. m DAC = m ACD = m D and m CAB = m B = m BCA; by the Triangle
Angle-Sum Theorem, the sum of the measures of the angles is 180,
so each angle measures , or 60. So, m DAB = = 120,
m B = 60, m BCD = = 120, and m D = 60.
3. By the Triangle Angle-Sum Theorem m A = 180, so m A = 70.
m ABC = = 85; by the Triangle Angle-Sum Theorem,
m C = 180, so m C = 125; m ADC = = 80
180 3
3-4

51. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Name the polygon. Then identify its vertices, sides, and angles.
Its angles are named by the vertices, A (or EAB or BAE),
B (or ABC or CBA), C (or BCD or DCB),
D (or CDE or EDC), and E (or DEA or AED).
Its sides are AB or BA, BC or CB, CD or DC, DE or ED, and EA or AE.
Its vertices are A, B, C, D, and E.
The polygon can be named clockwise or counterclockwise,
starting at any vertex.
Possible names are ABCDE and EDCBA.
3-4

52. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Starting with any side, count the number of sides clockwise
around the figure. Because the polygon has 12 sides,
it is a dodecagon.
Classify the polygon below by its sides. Identify it as convex or concave.
Think of the polygon as a star. If you draw a diagonal
connecting two points of the star that are next to each other,
that diagonal lies outside the polygon, so the dodecagon
is concave.
3-4

53. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Find the sum of the measures of the angles of a decagon.
A decagon has 10 sides, so n = 10.
Sum = (n – 2)(180) Polygon Angle-Sum Theorem
= (10 – 2)(180) Substitute 10 for n.
= 8 • Simplify.
= 1440
3-4

54. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Find m X in quadrilateral XYZW.
The figure has 4 sides, so n = 4.
m X + m Y + m Z + m W = (4 – 2)(180) Polygon Angle-Sum Theorem
m X + m Y = Substitute.
m X + m Y = Simplify.
m X + m Y = Subtract 190 from each side.
m X + m X = Substitute m X for m Y.
2m X = Simplify.
m X = Divide each side by 2.
3-4

55. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
A regular hexagon is inscribed in a rectangle. Explain how you know that all the angles labeled have equal measures.
Sample: The hexagon is regular, so all its angles are congruent.
An exterior angle is the supplement of a polygon’s angle because
they are adjacent angles that form a straight angle.
Because supplements of congruent angles are congruent, all the angles
marked have equal measures.
3-4

56. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
,000
19. 37
Pages Exercises
1. yes
2. No; it has no sides.
3. No; it is not a plane figure.
4. No; two sides intersect between endpoints.
5. MWBFX; sides: MW, WB, BF, FX, XM ;
: M, W, B, F, X s
6. KCLP; sides: KC, CL, LP, PK; : K, C,
L, P
7. HEPTAGN; sides: HE, EP, PT, TA, AG, GN, NH; : H, E, P,
T, A, G, N
8. pentagon; convex
9. decagon; concave
10. pentagon; concave
3-4

57. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
20. 60, 60, 120, 120
, 119
; 72
; 30
; 20
; 3.6
26. 45, 45, 90
27.
28.
29.
30.
31.
32. 3
33. 8
34. 13
35. 18
36. a. 3; 6
b. 4; 8
c. 5; 10
3-4

58. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
37. octagon; m 1 = 135; m 2 = 45
38. If you solve
= 130,
you get n = 7.2. This number is not an integer. ;
; 5
; 10
; 20
(n – 2) 180 n
; 12
– x;
45.
46. a. n • 180
b. (n – 2)180
c. 180n – 180(n – 2) = 360
d. Polygon Ext Sum Thm.
47. y = 103; z = 70; quad.
360 x 4 5
48. w = 72, x = 59, y = 49, z = 121;
49. x = 36, 2x = 72, 3x = 108, 4x = 144; quad.
50–53. Answers may vary. Samples are given
50.
51.
3-4

59. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
52.
53.
54. Yes; the sum of the measures of at the int. point is 360. The sum of the measures of all the is 180n. 180n – 360 = (n – 2)180
55. Answers may vary. Sample: The figure is a convex equilateral quad. The sum of its is 2 • 180 or 360.
56. octagon
57. a. (20, 162), (40, 171), (60, 174), (80,175.5), (100, 176.4), (120, 177), (140, 177.4), (160, ), (180, 178), (200, 178.2) b.
c. It is very close to 180
d. No, two sides cannot be collinear. s s s
3-4

60. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
180n – 360 n
360 n
58. a. [180(n – 2)] ÷ n = = 180 –
b. As n gets larger, the size of the angles get closer to 180. The more sides it has, the closer the polygon is to a circle.
59. 36
60–63. Answers may vary. Samples are given.
60.
61. Not possible; opp. sides would overlap.
62.
63. Not possible; opp. and adj. sides would overlap.
65. 20
3-4

61. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
69. 27
70. 10
, 25
72. 50, 40
, 76, 35, 69
74. Distr. Prop.
75. Subst. Prop.
76. Reflexive Prop. of
77. Symm. Prop. of
78. Div. Prop.
79. Trans. Prop.
80. RT, RK
BRT, BRK
82. Answers may vary. Sample: BR and TK.
83. Answers may vary. Sample: BRM
TRM
TRK
86. R
3-4

62. The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
For Exercises 1 and 2, if the figure is a polygon, name it by its
vertices and identify its sides. If the figure is not a polygon,
explain why not.
3. Find the sum of the measures of the angles in an octagon.
4. A pentagon has two right angles, a 100° angle and a
120° angle. What is the measure of its fifth angle?
5. Find m ABC.
6. XBC is an exterior angle at vertex B. Find m XBC.
not a polygon because
two sides intersect at
a point other than
endpoints
quadrilateral ABCD;
AB, BC, CD, DA
1080
140
ABCDEFGHIJ is a regular decagon.
144 36
3-4

63. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
(For help, go to Page 151.)
Find the slope of the line that contains each pair of points.
1. A(–2, 2), B(4, –2) 2. P(3, 0), X(0, –5)
3. R(–3, –4), S(5, –4) 4. K(–3, 3), T(–3, 1)
5. C(0, 1), D(3, 3) 6. E(–1, 4), F(3, –2)
7. G(–8, –9), H(–3, –5) 8. L(7, –10), M(1, –4)
3-5

64. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Solutions
y2 – y1
x2 – x1
–2 – 2
4 – (–2) –4 6 2 3
1. m = = = = –
2. m = = = =
3. m = = = = 0
4. m = = = , which is undefined; There is no slope.
5. m = = =
6. m = = = = –
7. m = = =
8. m = = = = – 1
y2 – y1
x2 – x1
–5 – 0
0 – 3 –5 –3 5 3
y2 – y1
x2 – x1
–4 – (–4)
5 – (–3) 8
y2 – y1
x2 – x1
1 – 3
– 3 – (–3) –2
y2 – y1
x2 – x1
3 – 1
3 – 0 2 3
y2 – y1
x2 – x1
–2 – 4
3 – (–1) –6 4 3 2
y2 – y1
x2 – x1
–5 – (–9)
– 3 – (–8) 4 5
y2 – y1
x2 – x1
–4 – (–10)
1 – 7 6 –6
3-5

65. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Use the slope and y-intercept to graph the line y = –2x + 9.
When an equation is written in the form y = mx + b, m is the slope
and b is the y-intercept.
In the equation y = –2x + 9, the slope is –2 and the y-intercept is 9.
3-5

66. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Use the x-intercept and y-intercept to graph 5x – 6y = 30.
To find the x-intercept, substitute
0 for y and solve for x.
To find the y-intercept, substitute
0 for x and solve for y.
5x – 6y = 30
5x – 6(0) = 30
5x – 0 = 30
5x = 30
x = 6
5x – 6y = 30
5(0) – 6y = 30
0 – 6y = 30
–6y = 30
y = –5
The x-intercept is 6.
A point on the line is (6, 0).
The y-intercept is –5.
A point on the line is (0, –5).
3-5

67. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Plot (6, 0) and (0, –5). Draw the line containing the two points.
(continued)
3-5

68. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Transform the equation –6x + 3y = 12 to slope-intercept form, then graph the resulting equation.
Step 1: Transform the equation
to slope-intercept form.
Step 2: Use the y-intercept and
the slope to plot two points
and draw the line containing them.
–6x + 3y = 12
3y = 6x + 12 Add 6x to each side.
= Divide each side by 3.
y = 2x + 4 3y 3 6x 12
The y-intercept is 4 and the slope is 2.
3-5

69. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Write an equation in point-slope form of the line with slope –8 that contains P(3, –6).
y – y1 = m(x – x1) Use point-slope form.
y – (–6) = –8(x – 3) Substitute –8 for m and (3, –6) for (x1, y1).
y + 6 = –8(x – 3) Simplify.
3-5

70. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Write an equation in point-slope form of the line that contains the points G(4, –9) and H(–1, 1).
Step 1: Find the slope.
Use the formula for slope.
Substitute (4, –9) for (x1, y1) and (–1, 1) for (x2, y2).
Simplify.
m = –2
m =
y2 – y1
x2 – x1
1 – (–9)
–1 – 4 10 –5
3-5

71. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
(continued)
Step 2: Select one of the points.
Write the equation in point-slope form.
y – y1 = m(x – x1) Point-slope form
y – (–9) = –2(x – 4) Substitute –2 for m and (4, –9) for (x1, y1).
y + 9 = –2(x – 4) Simplify.
3-5

72. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Write equations for the horizontal line and the vertical line that contain A(–7, –5).
Every point on the horizontal line through A(–7, –5)
has the same y-coordinate, –5, as point A.
The equation of the line is y = –5.
It crosses the y-axis at (0, –5).
Every point on the vertical line through A(–7, –5)
has the same x-coordinate, –7, as point A.
The equation of the line is x = –7.
It crosses the x-axis at (–7, 0).
3-5

73. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5 1. 2. 3. 4. 5. 6. 7. 8.
3-5

74. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5 9.
10.
11. y = 2x + 1
12. y = x + 1
13. y = –2x + 4
14. y = –2x + 4
3-5

75. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
15. y = – x + 1
16.
17. y – 3 = 2(x – 2)
18. y + 1 = 3(x – 4) 1 3
19. y – 5 = –1(x + 3)
20. y + 6 = –4(x + 2)
21. y – 1 = (x – 6)
22. y – 4 = 1(x – 0) or
y – 4 = x
23–28. Equations may vary from the pt. chosen. Samples are given.
23. y – 5 = (x – 0)
24. y – 2 = – (x – 6) 2 5
25. y – 6 = 1(x – 2)
26. y – 4 = 1(x + 4)
27. y – 0 = (x + 1)
28. y – 10 = (x – 8)
29. a. y = 7
b. x = 4
30. a. y = –2
b. x = 3
31. a. y = –1
b. x = 0
3-5

76. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
32. a. y = 4
b. x = 6
33.
34.
35.
36.
37.
38. a. 0.05
b. the cost per min
c. 4.95
d. the initial charge for a call
39. No; a line with no slope is a vertical line. 0 slope is a horizontal line.
40. a. m = 0; it is a horizontal line.
b. y = 0
41. a. Undefined; it is a vertical line.
b. x = 0
3-5

77. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
42–44. Answers may vary. Samples are given.
42. The eq. is in standard form; change to slope-intercept form, because it is easy to graph the eq. from that form.
43. The eq. is in slope-int. form; use slope-int. form, because the eq. is already in that form.
44. The eq. is in point-slope form; use point-slope form, because the eq. is already in that form.
45.
The slopes are the same and the y-intercepts are different.
46.
The slopes are all different and the y-intercepts are the same.
47. Check students’ work.
48.
3-5

78. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
49.
50.
51.
= 0.3, = 0.083; > ; it is possible only if the ramp zigzags.
53. The y-intercepts are the same, and the lines have the same steepness. One line rises from left to right while the other falls from left to right.
54. Answers may vary. Sample: x = 5, y – 6 = 2(x – 5), y = x + 1
55. (2, 0), (0, 4); m = = = –2
y – 0 = –2(x – 2), 2x + y = 4 or y = –2x + 4
56. a. y – 0 = (x – 0) or y = x
b. y – 5 = – (x – 2) or y = – x + 10 3 10 1 12 3 10 1 12
0 – 4
2 – 0 –4 2 5 2 5 2 5 2 5 2
3-5

79. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
67. C
68. C
56. (continued)
c. The abs. value of the slopes is the same, but one slope is pos. and the other is neg. One y-int. is at (0, 0) and the other is at (0, 10).
57. Yes; the slope of AB = the slope of BC.
58. No; the slope of DE the slope of EF.
59. Yes; the slope of
GH = the slope of HI.
60. Yes; the slope of JK = the slope of KL.
61. y – 2 = 3(x + 2); 3x – y = –8
62. y – 5 = (x – 5); x – 2y = –5
63. y – 6 = (x – 2); 2x – 3y = –14
64. D
65. G
66. B 1 2 3 = /
3-5

80. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
74. yes
75. No; || lines never intersect, but they are not skew.
76. No; all obtuse have two acute .
77. a = 5; m MPR = 30
78. a = 4; m MPR = 21
69. [2] a.  Line a: y = x + 12 OR equivalent equation;
Line b: 3y + 4x = 19 OR equivalent equation
b. 
point of intersection: (22, 9)
[1] at least one correct eq. or graph 3 2 s
3-5

81. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
79. a = 2; m QPR = 8
80. a = 6; m MPQ = 21
3-5

82. Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
1. Find the x-intercept and the y-intercept of the line 5x + 4y = –80.
2. Write an equation in point-slope form of the line with slope –1
that contains point C.
3. Write an equation in point-slope form of the line that contains
points A and B.
4. Write an equation of the line that contains B and C.
5. Graph and label the equations of the lines in
Exercises 2–4 above.
Three points are on a coordinate plane: A(1, 5), B(–2, –4), and C(6, –4).
x-intercept: –16, y-intercept: –20
y + 4 = –1(x – 6)
y – 5 = 3(x – 1) or y + 4 = 3(x + 2)
y = –4
3-5

83. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
(For help, go to page 151 and Lesson 3-5.)
Find the slope of the line through each pair of points.
1. F(2, 5), B(–2, 3) H(0, –5), D(2, 0)
3. E(1, 1), F(2, –4)
4. y = 2x – x + y = 20
6. 2x – 3y = x = y
8. y = y = x + 7 2 3
Find the slope of each line.
3-6

84. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Solutions
y2 – y1
x2 – x1
3 – 5
– 2 – 2
– 2
– 4 1 2
1. m = = = =
y2 – y1
x2 – x1
0 – (–5)
2 – 0 5 2
2. m = = =
y2 – y1
x2 – x1
– 4 – 1
2 –1
– 5 1
3. m = = = = –5
4. The equation is in slope-intercept form and the slope is the coefficient
of x, which is 2.
5. Rewrite the equation in slope-intercept form: y = – x + 20; the slope
is the coefficient of x, which is – 1.
3-6

85. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Solutions (continued) 2 3
6. Rewrite the equation in slope-intercept form: y = x – 2; the slope
is the coefficient of x, which is .
7. The equation in slope-intercept form is y = x + 0; the slope is the
coefficient of x, which is 1.
8. The equation in slope-intercept form is y = 0x + 7; the slope is the
coefficient of x, which is 0.
9. The equation is in slope-intercept form and the slope is the coefficient
of x, which is . 2 3 2 3
3-6

86. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Line 1 contains P(0, 3) and Q(–2, 5). Line 2 contains R(0, –7)
and S(3, –10). Are lines and 2 parallel? Explain.
Find and compare the slopes of the lines.
Slope of line = = = = –1
y2 – y1
x2 – x1
5 – 3
–2 – 0 2 –2
Slope of line = = = = –1
y2 – y1
x2 – x1
–10 –(–7)
3 – 0 –3 3
Each line has slope –1.
The y-intercepts are 3 and –7.
The lines have the same slope and different y-intercepts, so they are parallel.
3-6

87. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Are the lines y = –5x + 4 and x = –5y + 4 parallel? Explain.
The equation y = –5x + 4 is in slope-intercept form.
Write the equation x = –5y + 4 in slope-intercept form.
x = –5y + 4
x – 4 = –5y Subtract 4 from each side.
– x = y Divide each side by –5.
y = – x + 1 5 4
The line x = –5y + 4 has slope – . 1 5
The line y = –5x + 4 has slope –5.
The lines are not parallel because their slopes are not equal.
3-6

88. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Write an equation in point-slope form for the line
parallel to 6x – 3y = 9 that contains (–5, –8).
Step 1: To find the slope of the line, rewrite the equation
in slope-intercept form.
6x – 3y = 9
–3y = –6x Subtract 6x from each side.
y = 2x – Divide each side by –3.
The line 6x – 3y = 9 has slope 2.
Step 2: Use point-slope form to write an equation for the new line.
y – y1 = m(x – x1)
y – (–8) = 2(x – (–5)) Substitute 2 for m and (–5, –8) for (x1, y1).
y + 8 = 2(x + 5) Simplify.
3-6

89. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Line 1 contains M(0, 8) and N(4, –6). Line 2 contains P(–2, 9) and Q(5, 7). Are lines 1 and 2 perpendicular? Explain.
Step 1: Find the slope of each line.
m1 = slope of line = = = = –
y2 – y1
x2 – x1
–6 – 8
4 – 0
–14 4 7 2
m2 = slope of line = = = = –
y2 – y1
x2 – x1
7 – 9
5 – (–2) –2 7 2
Step 2: Find the product of the slopes.
m1 • m2 = – • – = 1 2 7
Lines 1 and 2 are not perpendicular because the product
of their slopes is not –1.
3-6

90. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Write an equation for a line perpendicular to
5x + 2y = 1 that contains (10, 0).
Step 1: To find the slope of the given line, rewrite the equation
in slope-intercept form. 5x + 2y = 1
2y = –5x Subtract 5x from each side.
y = – x Divide each side by 2. 5 2 1
The line 5x + 2y = 1 has slope – . 5 2
Step 2: Find the slope of a line perpendicular to 5x + 2y = 1.
Let m be the slope of the perpendicular line.
– m = – The product of the slopes of perpendicular lines is –1.
m = –1 • ( – ) Multiply each side by – .
m = Simplify. 5 2
3-6

91. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Step 3: Use point-slope form, y – y1 = m(x – x1), to write an equation
for the new line.
(continued)
y – 0 = (x – 10) Substitute for m and (10, 0) for (x1, y1).
y = (x – 10) Simplify. 2 5
3-6

92. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
The equation for a line containing a lead strip is y = x – 9.
Write an equation for a line perpendicular to it that contains (1, 7). 1 2
Step 1: Identify the slope of the given line.
slope
y = x – 9 1 2
Step 2: Find the slope of the line perpendicular to the given line.
Let m be the slope of the perpendicular line.
m = –1 The product of the slopes of perpendicular lines is –1.
m = –2 Multiply each side by 2. 1 2
3-6

93. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
(continued)
Step 3: Use point-slope form to write an equation for the new line.
y – y1 = m(x – x1)
y – 7 = –2(x – 1) Substitute –2 for m and (1, 7) for (x1, y1).
Step 4: Write the equation in slope-intercept form.
y – 7 = –2(x – 1)
y – 7 = –2x + 2 Use the Distributive Property.
y = –2x + 9 Add 7 to each side.
3-6

94. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Pages Exercises
1. Yes; both slopes = – .
2. No; the slope of 1 = , and the slope of 2 = .
3. No; the slope of 1 = , and the slope of 2 = 2.
4. Yes; both slopes = 4.
5. Yes; both slopes = 0. 1 2 3 3 4
6. Yes; the lines both have a slope of 2 but different y-intercepts.
7. Yes; the lines both have a slope of but different y-intercepts.
8. Yes; the lines both have a slope of –1 but different y-intercepts.
9. No; one slope = 7 and the other slope = –7.
10. No; one slope = – and the other slope = –3.
11. Yes; the lines both have a slope of – but different y-intercepts.
12. y – 3 = –2(x – 0) or y – 3 = –2x
13. y – 0 = (x – 6) or y = (x – 6) 2 5 3 4 1 3 1 3
3-6

95. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
25. No; • 2 –1.
26. Yes; 1 • (–1) = –1.
27. Yes; one is vertical and the other is horizontal.
28. No; – • (–3) –1.
29. Yes; – • = –1.
30. No; • – –1.
14. y – 4 = (x + 2)
15. y + 2 = – (x – 6)
16. Yes; the slope of 1 = – , and the slope of
2 = 2; – • 2 = –1.
17. Yes; the slope of 1 = – , and the slope of
2 = ; – • = –1.
18. No; the slope of 1 = –1, and the slope of = ; –1 • –1. 1 2 3 4 5
19. Yes; the slope of 1 = –1, and the slope of 2 = 1; –1 • 1 = –1.
20–23. Answers may vary. Samples are given.
20. y – 6 = – (x – 6)
21. y = –2(x – 4)
22. y – 4 = (x – 4)
23. y = x
24. y = – x 7 = /
3-6

96. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
31. slope of AB = slope of CD = ; AB || CD
slope of BC = slope of AD = –3; BC || AD
32. slope of AB = slope of CD = – ; AB || CD
slope of BC = slope of AD = 1; BC || AD
33. slope of AB = ; slope of CD = ; AB || CD
slope of BC = –1; slope of AD 5 – ; BC || AD 2 3 4 1
34. slope of AB = slope of CD = 0; AB || CD
slope of BC = 3 and slope of AD = ; BC || AD
35. Answers may vary. Sample: y = x + 5, y = – x + 5
36. No; two II lines with the same y-intercept are the same line. 3 2 4 5
37. RS and VU are horizontal with slope = 0; RS II VU;
slope of RW = slope of UT = 1; RW || UT;
slope of WV = slope of ST = –1; WV || ST
38. No; because no pairs of slopes have a product of –1.
39. The lines will have the same slope.
3-6

97. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
44. neither
45.
46.
47. AC: d = (7 – 9)2 + (11 – 1)2 =
BD: d = (13 – 3)2 + (7 – 5)2 =
AC BD
48. slope of AC = –5; slope of BD = ; since –5 • = –1,
AC BD; midpoint AC = (8, 6); midpoint BD = (8, 6); since the midpoints are the same, the diagonals bisect each other.
40. When lines are , the product of their slopes is –1. So, two lines to the same line must have the same slope.
41. a. y + 20 = (x – 35)
b. because you are given a point and can quickly find the slope
42. ||
43. 3 4 1 5
3-6

98. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
49. a–b. Answers may vary.
Sample:
c. The other possible locations for S are (–2, 3) and (8, 7).
50. y – 5 = (x – 4)
51. B
52. I
53. C
54. [2] a. slope of line c:
b. 0
[1] at least one correct slope
55. y – 3 = – (x – 0) or y – 3 = – x
56. y – 2 = (x + 4) 1 3
1 – (–2)
–4 – 2 –6 = 2
= –
; slope of line to c: 2 5
3-6

99. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6 3 4
57. y + 2 = (x – 3)
58. Refl. Prop. of
59. Mult. Prop. of =
60. Dist. Prop.
61. Symm. Prop. Of
62. If you are in geometry class, then you are at school.
63. If you travel to Switzerland, then you have a passport.
3-6

100. Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
1. Are lines 1 and 2 parallel? Explain.
2. Are the lines x + 4y = 8 and
2x + 6y = 16 parallel? Explain.
3. Write an equation in point-slope form for the line parallel
to –18x + 2y = 7 that contains (3, 1).
4. Are the lines y = x + 5 and 3x + 2y = 10 perpendicular? Explain.
5. Write an equation in point-slope form for the line
perpendicular to y = – x – 2 that contains (–5, –8). 1 6
Yes; the lines have the same slope
and different y-intercepts.
No; their slopes are not equal.
y – 1 = 9(x – 3) 2 3
Yes; the product of their slopes is –1.
y + 8 = 6(x + 5)
3-6

101. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
(For help, go to Lesson 1-5.)
Use a straightedge to draw each figure. Then use a straightedge
and compass to construct a figure congruent to it.
1. a segment 2. an obtuse angle 3. an acute angle
4. a segment 5. an acute angle 6. an obtuse angle
Use a straightedge to draw each figure. Then use a straightedge
and compass to bisect it.
3-7

102. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Solutions
3-7

103. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Examine the diagram below. Explain how to construct congruent to H.
Use the method learned for constructing congruent angles.
Step 1: With the compass point
on point H, draw an arc that
intersects the sides of H.
Step 2: With the same compass
setting, put the compass point
on point N. Draw an arc.
Step 3: Put the compass point below point N where the arc intersects HN.
Open the compass to the length where the arc intersects line .
Keeping the same compass setting, put the compass point above point N
where the arc intersects HN. Draw an arc to locate a point.
Step 4: Use a straightedge to draw line m through the point you located and point N.
3-7

104. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Construct a quadrilateral with both pairs of sides parallel.
Step 1: Draw point A and two rays with
endpoints at A. Label point B on
one ray and point C on the other ray.
Step 2: Construct a ray parallel to AC
through point B.
3-7

105. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
(continued)
Step 3: Construct a ray parallel to AC
through point C.
Step 4: Label point D where the ray parallel
to AC intersects the ray parallel to AB.
Quadrilateral ABDC has both pairs of
opposite sides parallel.
3-7

106. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
In constructing a perpendicular to line at point P, why must you open the compass wider to make the second arc?
With the compass tip on A and B, the same compass setting would
make arcs that intersect at point P on line . Without another point,
you could not draw a unique line.
With the compass tip on A and B, a smaller compass setting would make arcs
that do not intersect at all. Once again, without another point, you could not
draw a unique line.
3-7

107. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Examine the construction. At what special point does RG meet line ?
Point R is the same distance from point E as it is from point F
because the arc was made with one compass opening.
Point G is the same distance from point E as it is from point F
because both arcs were made with the same compass opening.
This means that RG intersects line at the midpoint of EF, and RG
is the perpendicular bisector of EF.
3-7

108. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
5–7. Constructions may vary. Samples using the following segments are shown. 5. 6.
Pages Exercises 1.
|| AB 2. 3. 4.
3-7

109. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
11. RS
12. 7. 8. AB 9.
10.
3-7

110. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
17.
18.
13. RS
14.
15. Construct a alt. int. ; then draw the || line.
16. a–b
17–25. Constructions may vary. Samples are given.
3-7

111. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
23.
24. a.
19.
20.
21. a.
b. The sides are || and .
c. Check students’ work.
22.
3-7

112. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
24. (continued)
b. Answers may vary. Sample:
c. One; if the lengths for the 3 sides are given, only one is possible; many different quad. are possible because the formed by the sides can vary.
25. a–c.
d. The sides of the smaller are half the length of the sides of the larger that they are || to.
e. Check students’ work.
26. a–b. Check students’ work.
c. p || m; in a plane, two lines to a third line are ||.
27–28. Answers may vary. Samples are given.
27.
The quad. is a rectangle. s
3-7

113. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
32.
33.
28.
The quad. is a square.
29.
30.
31.
3-7

114. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
34.
35. Not possible; if a = 2b = 2c, then 2a = 2b + 2c or a = b + c. The smaller sides would meet at the midpoint of the longer side, forming a segment.
36. Not possible; the smaller sides would meet at the midpoint of the longer side, forming a segment.
37. A
38. I
39. [2] a.
39. (continued)
b.  All points on the arc with center G are the same dist. from G, so GR = GT. The is isosc. because an isosc. must have at least two sides of the same length.
[1]  incorrect sketch OR incorrect classification
3-7

115. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
40. [2] a. I, IV, III, I
b.  (III): location of compass at points C and G; (I): same
as c and the intersection points of with arcs drawn in (III)
[1] incorrect sequence OR incorrect location of compass point
41. No; the slopes are different.
42. No; the slopes are different.
43. Yes; the slopes are both – .
44. 10
46. EB
47. DF 1 3
3-7

116. Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Draw a figure similar to the one given. Then complete the construction.
1. Construct a line through D that is parallel to XY.
2. Construct a quadrilateral with one pair of parallel sides of lengths p and q.
3. Construct the line perpendicular Construct the perpendicular
to line m at point Z to line n through point O.
Answers may vary. Sample given:
Answers may vary. Sample given:
3-7

117. Parallel and Perpendicular Lines
GEOMETRY CHAPTER 3
Page 176
1. acute isosceles
2. obtuse scalene
3. m 1 = 65 because corr. are ; m 2 = 65 because vert. are .
4. m 1 = 85 because alt. int. are ; m 2 = 110 because same-side int. are suppl. s
5. m 1 = 85 because corr. are ; m 2 = 95 because same-side int. are suppl.
6. m 1 = 70 because corr. are ; m 2 = 110 because same-side int. are suppl.
7. yes
8. yes
9. no
10. no s
11. 5
12. 25
13. 6
14. 75
15.
3-A

118. Parallel and Perpendicular Lines
GEOMETRY CHAPTER 3
20. y + 1 = –5(x – 3) or y = –5x + 14
21. Draw segments connecting nonconsecutive vertices. If no points of the segments are outside the polygon, then it is convex. Otherwise, the polygon is concave.
23. x = 85; y = 100; z = 100
24.
25. ||
26. neither
27.
28. 30
16.
17. Answers may vary. Sample: Z
18. Check students’ work.
19. a. Given
b. Corr. are .
c. Given
d. Trans. Prop.
e. If corr. are , then the lines are ||. s
3-A