Chapter 3 ISM

56 Copyright © 2014 Pearson Education, Inc.

Chapter 3 Section 3.1 Practice

1. a. , , and BC FG EH are parallel to .AD

b. BF and EF are skew to AD and also pass through point F.

c. , , , and AB DC AE DH are perpendicular

to .AD d. Plane EFGH is parallel to plane .ABCD 2. a. The alternate exterior angle pairs are

7 and 6,∠ ∠ and 1 and 4.∠ ∠ b. The same-side interior angle pairs are

8 and 3,∠ ∠ and 2 and 5.∠ ∠ 3. a. 3 and 7∠ ∠ are alternate interior angles. b. 4 and 6∠ ∠ are corresponding angles.

Vocabulary & Readiness Check 3.1 1. Planes that do not intersect are called parallel

planes. 2. Coplanar lines that do not intersect are called

parallel lines. 3. A line that intersects two or more coplanar

lines at different points is called a transversal. 4. Skew lines are not coplanar; they are not

parallel and do not intersect. 5. Angles 3, 4, 5, and 6 are called interior

angles. 6. Angles 1, 2, 7, and 8 are called exterior

angles. 7. Angles 2 and 6 are called corresponding

angles. 8. Angles 4 and 5 are called alternate interior

angles. 9. Angles 1 and 8 are called alternate exterior

angles.

10. Angles 3 and 5 are called same-side interior angles.

Exercise Set 3.1

2. EH and EA are perpendicular.

4. HD and BC are skew.

6. FB and GC are parallel. 8. Plane ABCD and plane EFGH are parallel.

10. Answers may vary. Sample: Plane ELH is parallel to plane .JCD

12. Answers may vary. Sample: GB is parallel

to .DH

14. Answers may vary. Sample: JE is skew to

.BH 16. Blue angles are corresponding angles. Red angles are same-side interior angles. 18. Blue angles are alternate exterior angles. Red angles are same-side interior angles. 20. 2∠ and 3∠ are alternate interior angles. 22. The alternate interior angles are 9 and 7.∠ ∠

24. The alternate exterior angles are

8 and 10.∠ ∠

ISM: Geometry Chapter 3: Parallel and Perpendicular Lines

Copyright © 2014 Pearson Education, Inc. 57

26. The two pairs of corresponding angles are 4 and 14,∠ ∠ and 1 and 3.∠ ∠

28. The same-side interior angles are

1 and 5.∠ ∠

30. 4∠ and 8∠ are alternate interior angles. 32. 3∠ and 7∠ are corresponding angles. 34. 13∠ and 11∠ are corresponding angles. 36. 15∠ and 4∠ are alternate interior angles. 38. 7∠ and 12∠ are alternate exterior angles.

40. False. ED and HG cannot be parallel because they are noncoplanar. These are skew lines.

42. False. Plane ABH and plane CDF are not

parallel. If extended, they eventually intersect at a line, and therefore cannot be parallel.

44. False. AE and BC are not skew lines because, if they are extended, they will eventually intersect at one point. By definition, if they intersect, then they cannot be skew.

46.

There are 4 pairs of corresponding angles.

48.

There are 4 pairs of vertical angles. 50. Never. By definition, skew lines are

noncoplanar. 52. Never. By definition, parallel lines are

coplanar. If two lines are in two separate planes, they can not be coplanar.

54. Never. In order for two objects to be skew,

they must be noncoplanar, not parallel, and must not intersect. Planes are either parallel or they intersect so they can never be skew. Since the line does not intersect with the plane, then it is on a plane that is parallel to the other plane.

56. No; parallel means that they never touch, and

the rug and the wall meet at the base of the wall. This is similar to two parallel planes intersecting at a line.

58. a. The intersection of planes A and C will be

a line and the intersection of planes B and C will be a line. Since A and B are parallel, the lines of intersection will be parallel.

b. Answers may vary. Sample: Two opposite walls in a classroom are parallel where the ceiling intersects both of them.

60.

62. Yes, this is true for any line that is in planes

A or B that is parallel to CD and has a plane going through them. In the image below, plane P intersects planes A and B with lines

EF and GH being parallel to .CD

Chapter 3: Parallel and Perpendicular Lines ISM: Geometry


64. Use the Linear Pair Theorem. 180

180

180 121

59

m YDA m YDF

m YDA m YDF

∠ + ∠ = °∠ = °− ∠

= °− °= °

66. Use the Vertical Angle Theorem.

59m FDI m YDA∠ = ∠

= °

Use definition of an angle bisector.

1

21

(59 ) 29.52

m FDR m FDI∠ = ∠

= ° = °

68. Use the Linear Pair Theorem. 180

180

180 29.5

150.5

m ADR m FDR

m ADR m FDR

∠ + ∠ = °∠ = ° − ∠

= ° − °= °

70. 6 and 3∠ ∠ are alternate interior angles. 72. 6 and 7∠ ∠ are same-side interior angles.

Section 3.2 Practice

1. and m are parallel if 6 2,∠ ≅ ∠ because 6∠ and 2∠ are alternate interior angles

formed by lines and m transversal b. 2.

3∠ and 6∠ are supplementary

Given

5∠ and 6∠ are supplementary Linear pairs are supplementary

3 6 180 ; 5 6 180m m m m∠ + ∠ = ° ∠ + ∠ = °

Definition of supplementary angles ↓

3 6 5 6m m m m∠ + ∠ = ∠ + ∠ Substitution

↓ 3 5m m∠ = ∠

Subtraction Property of Equality ↓

3 5∠ ≅ ∠ Definition of congruent angles

↓ m

Alternate Interior Angles Theorem

3. It is given that 1 2.∠ ≅ ∠ 2 3∠ ≅ ∠ because they are vertical angles. So, 1 3∠ ≅ ∠ by the Transitive Property of Congruence. Therefore, r s by the Corresponding Angles Theorem.

4. c d if ( )3 2 55w− ° = ° by the

Corresponding Angles Theorem.

( )3 2 55

3 57

19

w

w

w

− ° = °==

Vocabulary & Readiness Check 3.2

1. 5∠ and 7∠ are vertical angles.

2. 10∠ and 12∠ are vertical angles.

3. 4∠ and 10∠ are alternate interior angles.

4. 2∠ and 12∠ are alternate exterior angles.

5. 10∠ and 3∠ are same-side interior angles.

6. 1∠ and 9∠ are corresponding angles.

7. 5∠ and 15∠ are alternate exterior angles.

8. 8∠ and 14∠ are alternate interior angles.

9. 14∠ and 6∠ are corresponding angles.

10. 7∠ and 14∠ are same-side interior angles.

Exercise Set 3.2

2. m n by the Corresponding Angles Theorem.

4. The lines are not parallel. The marked

congruent angles are not corresponding angles, alternate interior angles, or alternate exterior angles.

6. m n by the Same-Side Interior Angles

Theorem. 8. The lines are not parallel. The marked

congruent angles are not corresponding angles, alternate interior angles, or alternate exterior angles.



10. AC BD by the Alternate Interior Angles Theorem.

12. PS QT by the Corresponding Angles

Theorem.

14. AQ TR by the Corresponding Angles

Theorem.

16. KR MT by the Corresponding Angles Theorem.

18. Since 85 95 180 ,° + ° = ° p q by the Same-

Side Interior Angles Theorem. 20. m n by the Alternate Interior Angles

Theorem. 22. Since 66 35 101 ,° + ° = ° and

68 33 101 ,° + ° = ° n p by the

Corresponding Angles Theorem. 24. 41 41 84 ,° + ° ≠ ° so the alternate exterior

angles are not congruent.

26. By the Alternate Interior Angles Theorem, if the angles that measure 95° and ( )2 5x − °

are equal in measure, then .l m

( )2 5 95

2 100

50

x

x

x

− ° = °==

28. By the Vertical Angles Theorem, the same-

side interior angle as the ( )3 18x − ° angle

has a measure of 105 .° If the same-side interior angles are supplementary, then .l m

( )3 18 105 180

3 87 180

3 93

31

x

x

x

x

− ° + ° = °+ ° = °

==

30. The lines are not parallel. There is no theorem that states that the lines are parallel if 1 3.∠ ≅ ∠

32. The angles are congruent by the

Corresponding Angles Theorem. 60 2 70 4

60 2 70

2 10

5

x x

x

x

x

− = −+ =

==

Use substitution to find 1m∠ and 2.m∠ 1 60 2 5

60 10

50

m∠ = − ⋅= −=

2 70 4 5

70 20

50

m∠ = − ⋅= −=

34. The angles are congruent by the

Corresponding Angles Theorem. 20 8 30 16

20 8 30

8 10

1.25

x x

x

x

x

− = −+ =

==

Use substitution to find 1m∠ and 2.m∠ 1 20 8 1.25

20 10

10

m∠ = − ⋅= −=

2 30 16 1.25

30 20

10

m∠ = − ⋅= −=

36. If 1 7,∠ ≅ ∠ then n by the Alternate

Exterior Angles Theorem.



38. 8 3 180m m∠ + ∠ = °

Given 8 9 180m m∠ + ∠ = °

Given 8 3 8 9m m m m∠ + ∠ = ∠ + ∠

Substitution ↓

3 9m m∠ = ∠ Subtraction Property of Equality

↓ 3 9∠ ≅ ∠

Definition of congruent angles ↓

j k

Alternate Interior Angles Theorem

8 3 180m m∠ + ∠ = ° Given

↓ 8∠ and 3∠ are supplementary.

Definition of Supplementary Angles ↓

l n Same-Side Interior Angles Theorem

40. a b by the Same-Side Interior Angles

Theorem. 42. No lines can be proven parallel. 44. m by the Corresponding Angles

Theorem. 46. m by the Alternate Interior Angles

Theorem. 48. No lines can be proven parallel. 50. By the Vertical Angles Theorem, the same-

side interior angle as the 2x° angle has a

measure of ( )5 40 .x + ° If the same-side

interior angles are supplementary, then .l m

( )5 40 2 180

7 40 180

7 140

20

x x

x

x

x

+ °+ ° = °+ =

==

52. Answers may vary. Sample answer: Both types of proofs can be used to draw logical conclusions. A flow proof uses arrows to show logical connections between the statements. Two-column proofs use columns for the statements and reasons.

54. Never. Skew lines do not lie in the same

plane, and cannot intersect. 56. Sometimes. Rays can be parallel, but they do

not have to be. For example, rays in an angle are not parallel.


1. a. 1 3 48m m∠ = ∠ = ° by the Vertical Angles Theorem.

b. 2 180 3 180 48 132m m∠ = °− ∠ = °− ° = ° by the Linear Pair Theorem.

c. 8 2 132m m∠ = ∠ = ° by the Alternate Interior Angles Converse.

d. 6 8 132m m∠ = ∠ = ° by the Vertical Angles Theorem.

2. 2 180 1 180 97 83m m∠ = °− ∠ = °− ° = ° by the

Linear Pair Theorem. 3 1 97m m∠ = ∠ = ° by the Vertical Angles

Theorem. 4 2 83m m∠ = ∠ = ° by the Corresponding

Angles Converse. 5 1 97m m∠ = ∠ = ° by the Alternate Exterior

Angles Converse. 6 4 83m m∠ = ∠ = ° by the Vertical Angles

Theorem. 7 5 97m m∠ = ∠ = ° by the Vertical Angles


Theorem. 3. a. 1 4 75m m∠ = ∠ = ° by the Alternate

Interior Angles Converse. b. 2 1 75m m∠ = ∠ = ° by the Vertical Angles

Theorem. c. 5 3 105m m∠ = ∠ = ° by the Vertical

Angles Theorem. d. 6 105m∠ = ° by the Alternate Interior

Angles Converse. e. 7 6 105m m∠ = ∠ = ° by the

Corresponding Angles Converse. f. 8 6 105m m∠ = ∠ = ° by the

Corresponding Angles Converse



4. 5 135m∠ = ° by the Corresponding Angles Converse. 5∠ is supplementary with the unknown angle. 180 135 ( 10)180 145

35

° = ° + + °° = ° + °=

yy

y


1. Given line and point P, how many lines can

be drawn through P parallel to line ? One 2. The postulate that gives us the answer to

Exercise 1 above is called the Parallel Postulate.

Exercise Set 3.3

2. 1 45m∠ = ° by the Corresponding Angles Converse.

2 1 45m m∠ = ∠ = ° by the Vertical Angles Theorem.

4. 1 139m∠ = ° by the Alternate Interior Angles

Converse. 2 180 139 41m∠ = °− ° = ° by the Same-Side

Interior Angles Converse. 6. 1 120m∠ = ° by the Corresponding Angles

Converse. 2 180 120 60m∠ = °− ° = ° by the Same-Side

Interior Angles Converse. 8. 7∠ is congruent to the angle whose measure

is given by the Vertical Angles Theorem. 4∠ is congruent to the angle whose measure

is given by the Alternate Interior Angles Converse.

5∠ is congruent to the angle whose measure is given by the Corresponding Angles Converse.

10. 3∠ is congruent to the angle whose measure

is given by the Corresponding Angles Converse.

5∠ is congruent to the angle whose measure is given by the Alternate Exterior Angles Converse.

12. 2 180 1 180 68 112m m∠ = °− ∠ = °− ° = ° by

the Linear Pair Theorem. 3 1 68m m∠ = ∠ = ° by the Vertical Angles


Theorem.

5 1 68m m∠ = ∠ = ° by the Corresponding Angles Converse.

6 4 112m m∠ = ∠ = ° by the Alternate Interior Angles Converse.



14. 2 8 111m m∠ = ∠ = ° by the Alternate

Exterior Angles Converse. 1 180 2 180 111 69m m∠ = °− ∠ = °− ° = ° by




Angles Converse. 6 4 111m m∠ = ∠ = ° by the Alternate Interior


Theorem. 16. 1 180 4 180 34 146m m∠ = °− ∠ = °− ° = ° by







Theorem. 18. 1 7 154m m∠ = ∠ = ° by the Alternate Exterior

Angles Converse. 2 180 1 180 154 26m m∠ = °− ∠ = °− ° = ° by






Theorem.



20. Use the Linear Pair Theorem. The known angle and the angle that measures yº are supplementary.

105 18075

yy

° + ° = °=

The known angle is an alternate exterior angle to the angle that measures xº, so use the Alternate Exterior Angles Converse.

105x = 22. Use the Linear Pair Theorem. The known

angle and the angle that measures xº are supplementary.

85 18095

xx

° + ° = °=

The angle that measures xº is a corresponding angle to the angle that measures yº, so use the Corresponding Angles Converse.

95y =

24. 1 180 80 100m∠ = °− ° = ° by the Same-Side

Interior Angle Converse. 2 70m∠ = ° by the Alternate Interior Angles

Converse. 26. 1 41m∠ = ° by the Alternate Interior Angles

Converse. 2 180 1 180 41 139m m∠ = °− ∠ = °− ° = ° by

the Same-Side Interior Angle Converse. 3 2 139m m∠ = ∠ = ° by the Vertical Angles


Angles Converse. 28. 1 102m∠ = ° by the Alternate Exterior Angles

Converse. 2 102m∠ = ° by the Corresponding Angles

Converse. 3 180 40 140m∠ = °− ° = ° by the Linear Pair

Theorem. 4 3 140m m∠ = ∠ = ° by the Alternate Interior

Angles Converse. 30. 1 97m∠ = ° by the Alternate Interior Angles

Converse. 2 133m∠ = ° by the Vertical Angles

Theorem. 3 180 2 180 133 47m m∠ = °− ∠ = °− ° = ° by

the Same-Side Interior Angle Converse. 4 180 1 180 97 83m m∠ = °− ∠ = °− ° = ° by the

Same-Side Interior Angle Converse. 32. 1 115m∠ = ° by the Alternate Exterior Angles

Converse.

34. Use the Vertical Angles Theorem and the Same-Side Interior Angles Converse.

1 180 38 142m∠ = °− ° = ° because the same-side interior angle to 1∠ is a vertical angle to the given angle.

36. Use the Alternate Interior Angles Converse.

5 1105 110

5 522

xx

x

° = °

=

=

38. Use the Corresponding Angles Converse.

3 873 87

3 329

xx

x

° = °

=

=

40. Use the Alternate Exterior Angles Converse.

4 1 474 484 48

4 412

xxx

x

° − ° = °=

=

=

42. Find the measure of the angle that forms a

linear pair with the angle that measures 112º. 180 112 68° − ° = ° The angle that measures 2( 3)x + ° is an

alternate interior angle of the angle that measures 68°, so these angles are congruent and have equal measures. 2( 3) 682( 3) 68

2 23 34

31

xx

xx

+ ° = °+ =

+ ==

44. Use the Same-Side Interior Angles Converse.

5 4 1809 1809 180

9 920

x xxx

x

° + ° = °=

=

=

Substitute 20 for x to find the angle measures. 5 5(20) 100x° = ° = °

4 4(20) 80x° = ° = °



46. Use the Corresponding Angles Converse. ( 20) (250 4 )

5 20 2505 2305 230

5 546

x xx

xx

x

+ ° = − °+ =

=

=

=

Substitute 46 for x to find the angle measures. ( 20) (46 20) 66x + ° = + ° = °

(250 4 ) (250 4 46)(250 184)66

− ° = − ⋅ °= − °= °

x

48. What fact do you need to show that 1∠ and

2∠ are supplementary? It was already stated in Step 1 that || .a b Use the other fact that

was given. The answer is || .c d

50. The proof uses the facts that 3∠ and 2∠ are

supplementary and that 1∠ and 2∠ are supplementary. 1 3∠ ≅ ∠ by the Equal Supplements Theorem.

52. There are two unknown variables, so two

equations are necessary. Using the Corresponding Angles Converse, 3x y° = °. Using the Linear Pair Theorem,

180x y° + ° = °. Solve the system by

substituting 3y for x into the equation 180x y° + ° = °.

3 1804 1804 180

4 445

y yyy

y

+ ==

=

=

Substitute 45 for y into the equation 3x y=

to solve for x. 3 45 135= ⋅ =x

54. Use the Same-Side Interior Angles Converse

to solve for x. 2 ( 12) 180

3 12 1803 1923 192

3 364

x xx

xx

x

° + − ° = °− =

=

=

=

Use the Same-Side Interior Angles Converse to solve for y. 3 ( 20) 180

4 20 1804 1604 160

4 440

y yy

yy

y

° + + ° = °+ =

=

=

=

56. Correct; by the Same-Side Interior Angles

Converse, the two angles are supplementary and can be added for a sum of 180º. This equation can then be solved to find x.

58. No; the Alternate Interior Angles Converse

says that alternate interior angles are always congruent. There are no counterexamples to show that this is incorrect.

60. The information below will be used for this

proof.

If it can be proven that 1∠ and 7∠ are congruent, then the Alternate Exterior Angles Theorem is proven.

Statements Reasons

1. ||l m 1. Given

2. 1 3∠ ≅ ∠ 2. Vertical Angles Theorem

3. 3 7∠ ≅ ∠ 3. Corresponding Angles Converse

4. 1 7∠ ≅ ∠ 4. Transitive Property of Congruence

62. Sample answer: Both show congruence of

alternate angles when two parallel lines are cut by a transversal. They are different because the Alternate Interior Angles Converse shows the relationship between angles on the plane between the two parallel lines while the Alternate Exterior Angles Converse shows the relationship between angles outside that plane.



64. 1 66m∠ = ° by the Alternate Interior Angles Converse.

2 180 94 86m∠ = °− ° = ° by the Same-Side Interior Angles Converse.

66. Sometimes; two lines in intersecting planes

can intersect at any angle, not just 90º. 68. Never; two lines in parallel planes cannot

intersect, so they cannot be perpendicular.


1. Given a b and ,b c prove .a c Statements Reasons

1. a b 1. Given


3. b c 3. Given


5. 1 3∠ ≅ ∠ 5. Substitution or Transitive Property

6. a c 6. Corresponding Angles Theorem

2. Since 1 90 ,m∠ = ° then .n ⊥ If ,n ⊥ by

Theorem 3.4-3, all four of the angles formed by the intersection of n and are right angles.

49 90

41yy

° + ° = °° = °

Thus, the value of y is 41.


1. Given line and point P, how many lines can be drawn through P parallel to ? One

2. Given line and point P, how many lines

can be drawn through P perpendicular to ? One

3. The postulate that gives us the answer to



Exercise 2 above is called the Perpendicular Postulate.

5. Postulates are accepted as true and are not

proved. True

6. Theorems are accepted as true and are not proved. False

Exercise Set 3.4

2. By Theorem 3.4-3, 90.x = 4. By Theorem 3.4-3, the lines intersect to form

four right angles. The ray divides one right angle into two adjacent angles. Use the Angle Addition Postulate to solve for x.

45 9045

xx

° + ° = °=

6. By Theorem 3.4-3, the lines intersect to form


54 9036

xx

+ = °=

8. By Theorem 3.4-3, the lines intersect to form


55 9035

xx

° + ° = °=

10. 1m∠ = 90° 12. Use Theorem 3.4-3 and the Angle Addition

Postulate. 2 3m m∠ + ∠ = 90°

14. Use Theorem 3.4-3 and the Angle Addition

Postulate. If 3 35 ,m∠ = ° then 2 55 .m∠ = ° 16. Use the Vertical Angles Theorem. If

2 65 ,m∠ = ° then 5 65 .m∠ = ° 18. If the rungs of the ladder are ⊥ to one side,

then per the Corresponding Angles Theorem, the rungs are . If the sides of the ladder are

, then per the Perpendicular Transversal Theorem, the rungs of the ladder are also ⊥ to the other side.



20. If a b and ,b c then by Two Lines Parallel to a Third Line Theorem .a c If

a c and ,d c⊥ then by the Perpendicular

Transversal Theorem .a d⊥

22. Since a b⊥ and ,b c per the Perpendicular

Transversal Theorem .a c⊥ If a c⊥ and ,c d then per the Perpendicular Transversal

Theorem .a d⊥

24. Since b c and ,c d⊥ per the Perpendicular

Transversal Theorem .b d⊥ If a b⊥ and ,b d⊥ then per the Corresponding Angles

Theorem .a d

26. The triangles need to be arranged so that

opposite interior angles formed using the diagonal of the square as a transversal are congruent. Then, by the Alternate Interior Angles Theorem, the sides are parallel.

28. Since D A,⊥ A B and A C, per the

Perpendicular Transversal Theorem, D B⊥ and D C.⊥

30. Given: , , ,q r r s b q⊥ and a s⊥

Prove: a b

Statements Reasons 1. q r 1. Given

2. r s 2. Given

3. q s 3. Two Lines Parallel to a Third Line Theorem

4. b q⊥ and a s⊥ 4. Given

5. 1∠ and 3∠ are right angles

5. Theorem 3.4-3

6. 1 90m∠ = ° and 3 90m∠ = °

6. Definition of right angle


8. 1 2m m∠ = ∠ 8. Definition of ≅ 9. 2 90m∠ = ° 9. Transitive Property

of =

10. 2 390 3m m

m∠ + ∠ =°+ ∠

10. Addition Property

of =

11. 2 390 90m m∠ + ∠ =°+ °

11. Substitution

12. 2 3 180m m∠ + ∠ = °

12. Simplify.

13. a b 13. Same-Side Interior Angles Theorem

32. Given: 1 2,∠ ≅ ∠ 1∠ and 2∠ form a linear

pair Prove: m⊥

Statements Reasons 1. 1 2∠ ≅ ∠ 1. Given 2. 1 2m m∠ = ∠ 2. Definition of ≅ 3. 1 2 180m m∠ + ∠ = ° 3. Definition of a

linear pair 4. 2 1 180m∠ = ° 4. Substitution

Property of = 5. 1 90m∠ = ° 5. Division Property

of = 6. 1∠ is a right angle. 6. Definition of right

angle 7. m⊥ 7. Definition of

perpendicular lines



34. Avenue B and Main Street are perpendicular per the Perpendicular Transversal Theorem.

36–38.

Given: k and m k Prove: m

Since ,k 2 1∠ ≅ ∠ by the Corresponding

Angles Converse. Since ,m k 1∠ 3≅ ∠ (Exercise 36) for the same reason. By the Transitive Property of Congruence,

2 3.∠ ≅ ∠ By the Corresponding Angles Theorem (Exercise 38), .m

40. If ,a b then the alternate exterior angles

formed by the intersection of a transversal with a and b are congruent. Each pair of exterior angles forms a linear pair. The sum of the measures of same-side exterior angles is 180°.

( )3 2 44 1803 42 180

3 13846

xx

xx

− °+ ° = °+ =

==

Thus, a b if and only if 46.x = 42. By sight, the angle is acute.


1. Draw line n and point P not on n.

Step 1: Label points H and J on n. Use a straight

edge to draw .HP Step 2:

At P construct 1∠ congruent to .PHJ∠ Label new line l. l n

2. Draw a segment and label its length m.

Step 1:

Label segment m, .AB Step 2:

Draw point D such that D is not on .AB

Step 3:

Use a straightedge to draw ray .AD Step 4:

Construct congruent corresponding angles so

that the ray from point D is parallel to .AB Step 5:

Draw a side of length 2m. Construct point C, so that 2 .CD m=

Step 6:

Draw BC to complete the quadrilateral. The

quadrilateral is such that ,CDAB 2 ,CD m= and .AB m=

3. Use a straight edge to draw .EF

Step 1:

Construct two points on EF that are equidistant from point F. Step 2: Open the compass so that is greater than half the distance between the two points. With the compass point on the left point, draw an arc above F. Step 3: Without changing the compass setting, place the compass point on the right point. Draw an arc that intersects the arc from Step 3. Label the point of intersection G. Step 4:

Draw .FG FG EF⊥ at point F.



4. Draw CX and a point Z, not on .CX

Step 1: Open the compass to a size greater than the

distance from Z to .CX With the compass on

point Z, draw an arc that intersects CX at two points. Step 2: Place the compass point on the left hand intersection from Step 1 and make an arc. Step 3: Keep the same compass setting. With the compass point on the right hand intersection from Step 1, draw an arc that intersects the arc from Step 2. Label the intersection B. Step 4:

Draw .ZB ZB CX⊥

Exercise Set 3.5

2. Draw AB and a point J not on .AB

Step 1:

Draw .AJ Step 2: At J construct 1∠ congruent to .BAJ∠ Step 3: Draw line m through J that overlaps the new

ray from Step 2. Then, .m AB

4. Draw .AJB∆

. Step 1:

Use a straightedge to draw .AJ Step 2: At J, construct an angle congruent to .JAB∠ Step 3: Draw line m that overlaps with the new ray

formed by the angle in Step 2. .m AB

6. Draw the given triangle and point J inside.

Step 1:

Draw .AJ Step 2: Construct an angle congruent to JAB∠ at J.

Label new line m. m AB



8. Draw two segments and label their lengths a and b.

Step 1:

Draw a ray with endpoint A. The draw AB such that point B is not on the first ray. Step 2: Through point B, draw a ray parallel to the first ray. Construct congruent corresponding angles, so the rays are parallel. Step 3: Draw a side of length 2a. Construct Z, so that

2 .AZ a= Step 4: Draw a side of length b. Construct Y so that,

.BY b= Step 5:

Draw YZ to complete the quadrilateral.

10. Draw two segments and label their lengths a

and b.

Step 1:

Draw a ray with endpoint A. Then draw AB such that point B is not on the first ray. Step 2: Through point B, draw a ray parallel to the first ray. Construct congruent corresponding angles, so the rays are parallel. Step 3: Draw a side of length a. Construct Z, so that

2 .AZ a= Step 4: Draw a side of length b. Construct C so that,

.BC b=

Step 5:

Construct the perpendicular bisector of .BC Label the intersection of the bisector and

,BC Y. 1

.2

BY b=

Step 6:

Draw YZ to complete the quadrilateral.

12. Draw line with point P on .

Step 1: Place the compass point on P and construct two points on that are equidistant from point P. Step 2: Open the compass wider so that is greater

than 1

2 the distance between the two points.

With the compass point on the left hand point, draw an arc above P. Step 3: Without changing the compass setting, place the compass point on the right hand point. Draw an arc that intersects the arc from Step 2. Step 4: Use a straight edge to draw a line through the intersection and P. The new line is perpendicular to at P.



14. Draw line RS and point P not on .RS

Step 1: Open the compass to a size greater than the

distance from P to .RS With the compass point on point P, draw an arc that intersects

RS at two points. Step 2: Place the compass point on one of the intersections from Step 1 and make an arc. Step 3: Use the same compass setting. With the compass point on the other intersection from Step 1, draw an arc that intersects the arc from Step 2. Label the intersection B. Step 4:

Draw .PB PB RS⊥

16. Draw line RS and point P not on .RS

Step 1:

Open the compass to a size greater than the

distance from P to .RS With the compass point on point P, draw an arc that intersects

RS at two points. Step 2: Place the compass point on one of the intersections from Step 1 and make an arc. Step 3: Use the same compass setting. With the compass point on the other intersection from Step 1, draw an arc that intersects the arc from Step 2. Label the intersection B. Step 4:

Draw .PB PB RS⊥

18. Draw a segment with length a.

Step 1: Draw a line m and label a point A on m. Step 2: Draw a segment of length a on m. Construct B, so that .AB a= Step 3: Place the compass point on point A and construct two points on m that are equidistant from A. Step 4: Open the compass wider so that it is greater

than 1

2 the distance between the two points.

With the compass point on the left hand point, draw an arc above A. Step 5: Without changing the compass setting, place the compass point on the right hand point. Draw an arc that intersects the arc from Step 4. Step 6: Use a straightedge to draw a ray with

endpoint A. This ray is perpendicular to .AB Step 7: Repeat Steps 3–5 to construct the ray perpendicular to m at B. Step 8: Draw a segment of length a on each ray. Construct C and D so that AD a= and

.BC a= Step 9:

Draw .CD This makes a square ABCD with sides of length a.



20. Draw a ray with endpoint A.

Step 1: Draw a segment of length b on the ray. Construct B so .AB a= Step 2:

Construct ray BD perpendicular to ray AB point B. Step 3:

Draw a segment of length b on .BD Construct C so .BC b= Step 4:

Use a straightedge to construct .AC This forms a right triangle ABC∆ with legs of length a and b.

22. a. Draw DAB∠ with .AB c=

Step 1: Construct an angle at D congruent to

.DAB∠ Step 2: Open the compass to the length .AB Draw a segment of length .AB Construct point C so that .DC AB=

Step 3:

Draw segment .BC Quadrilateral ABCD

has parallel sides AB and DC with length c.

b. The other pair of sides appears to be

parallel and congruent.

c. Measure the length of AD and ,BC to see that .AD BC= Also, use a protractor to measure ADC∠ and BCD∠ to see that 180 .ADC m BCDm∠ + ∠ = ° This confirms conjecture in part b.

24. Steps i and iii use a compass. In Step i, the

compass points are located at the intersection

of the arcs with .CG In Step iii, the compass points are located at C and G.

26. Draw a segment of length b, and a line with a

point A labeled.

Step 1: Open the compass to match the endpoints of the segment of length b. Put the compass point on point A and draw an arc that intersects the line. Step 2: Using the same compass setting, put the compass point on the intersection of the line and the arc and draw another arc that intersects the line at a point other than A. Label the intersection B. Note that

2 .AB b a= = Step 3: Using a smaller compass width, draw two arcs that intersect the line at points equidistant from A. Call the points R and S. Step 4: Open the compass so that it is larger than 1

.2

RS Put the compass point on point R and

draw and arc. Then, put the compass point on point S and draw an arc that intersects the previous arc. Label the point of intersection T.



Step 5:

Use a straightedge to draw ray .AT Set the compass to the length of the given segment, as in Step 1. With the compass point on A,

draw an arc that intersects AT at a point. Label the point of intersection D. Note that

.AD b= Step 6: Repeat Steps 3 – 5 at point B, to draw

segment .BC Step 7:

Use a straightedge to draw .DC The quadrilateral is a rectangle with side lengths a

and b such that 1

.2

b a=

28. Draw a segment of length b, and a line with a point A labeled.

Step 1: Open the compass to match the endpoints of the segment of length b. Put the compass point on point A and draw an arc that intersects the line. Step 2: Using the same compass setting, put the compass point on the intersection of the line and the arc and draw another arc that intersects the line at a point other than A. Step 3: Repeat Step 2 with the new intersection point. Label the intersection .B′ Note that

3 .AB b′ = Step 4:

Construct the bisector of .AB′ Label the

intersection of the bisector and AB′ point B.

Note that 3

.2

aA bB = =

Step 5: Using a smaller compass width, draw two arcs that intersect the line at points equidistant from A. Call the points R and S.

Step 6: Open the compass so that it is larger than 1

.2

RS Put the compass point on point R and

draw and arc. Then, put the compass point on point S and draw an arc that intersects the previous arc. Label the point of intersection T. Step 7:

Use a straightedge to draw ray .AT Set the compass to the length of the given segment, as in Step 1. With the compass point on A,

draw an arc that intersects AT at a point. Label the point of intersection D. Note that

.AD b= Step 8: Repeat Steps 5–7 at point B, to draw segment

.BC Step 9:

Use a straightedge to draw .DC The quadrilateral is a rectangle with side lengths a

and b such that 2

.3

b a=

30. Draw a segment of length c and a ray with

endpoint A.

Step 1: Open the compass to the length of the segment. Put the compass point on point A and draw an arc that intersects the ray. Using the same compass setting, put the compass point on the point of intersection and draw an arc that intersects the ray at a point other than A. Label this point B. Step 2: With the same compass setting, put the compass point on point A and draw an arc

above .AB Step 3: Open the compass to length c. Put the compass point on point B and draw an arc that intersects the arc from Step 2. Label the intersection C.



Step 4:

Use a straightedge to draw AC and .BC Note that 2AC a c= = and .BC c= So,

ABC∆ has sides with the desired lengths.

32. Draw a segment of length c, a segment of

length b, and a ray with endpoint A.

Step 1: Open the compass to length b. Put the compass point on point A and draw an arc that intersects the ray. Step 2: Open the compass to length c. Put the compass point on the point of intersection and draw an arc that intersects the ray. Label this point B. Step 3: Open the compass to length b, put the compass point on point A and draw a circle. Step 4: Open the compass to length c. Put the compass point on point B and draw circle. Step 5: The circles intersect at a point on the ray. So no such triangle can be constructed.

34. Draw line w and point X not on w.

Step 1: Open the compass to a size greater than the distance from X to w. With the compass point on point X, draw an arc that intersects w at two points. Step 2: Place the compass point on one of the intersections from Step 1 and make an arc. Step 3: Use the same compass setting. With the compass point on the other intersection from Step 1, draw an arc that intersects the arc from Step 2. Label the intersection B. Step 4:

Draw .XB .XB w⊥

36. No, given any line and a point not on that

line, there is exactly one line through the point that is perpendicular to the given line.

38. 1 4 3

12 1 3

− −= =

− − −

40. 7 ( 1) 7 1 6

31 ( 1) 1 1 2

− − − − + −= = = −

− − +


1. a. For line a, use points ( )5, 7 and

( )2, 3 and the slope formula.

2 1

2 1

7 3 4

5 2 3

y ym

x x

− −= = =− −

b. For line c, use points ( )1, 7 and ( )5, 7

and the slope formula.

2 1

2 1

7 7 00

5 1 4

y ym

x x

− −= = = =− −



2. Rewrite the equation in slope-intercept form, .y mx b= +

2 3 93 2 9

23

3

x yy x

y x

− =− = − +

= −

The coefficient of x, 2

,3

is the slope, and the

y-intercept point is ( )0, 3 .−

3. To predict the price of a pass in 2025 we

need to find y when x is 25. (Since year 2000 corresponds to 0,x = year 2025 corresponds to 25.)x =

3.2 483.2(25) 4880 48128

y x= += += +=

In the year 2025, the price of an adult one-day pass to Disney World will be about $128.

4. a. Solve each equation for y. 2 3

2 31 3

2 2

x yy x

y x

− =− = − +

= −

2 32 3

x yy x

+ == − +

The slopes are 1

2 and 2,− which are not

equal. Although the slopes are not the

same their product is: 1

( 2) 1,2− = − so

they are perpendicular. b. Solve each equation for y. 4 3 2

3 4 24 2

3 3

x yy x

y x

− =− = − +

= −

8 6 66 8 6

41

3

x yy x

y x

− + = −= −

= −

The slopes are both 4

,3

so they are

parallel.

Vocabulary & Readiness Check 3.6 1. The measure of the steepness or tilt of a line

is called slope. 2. The slope of a line through two points is

measured by the ratio of vertical change to horizontal change.

3. If a linear equation is in the form ,y mx b= +

the slope of the line is m the y-intercept point is ( )0, b and the y-intercept is b.

4. The form y mx b= + is the slope-intercept

form. 5. The slope of a horizontal line is 0. 6. The slope of a vertical line is undefined. 7. Two nonvertical perpendicular lines have

slopes whose product is 1.− 8. Two nonvertical lines are parallel if they

have the same slope and different y-intercepts.

9. 7

6m = is a positive slope, so the line slants

upward. 10. 3m = − is a negative slope, so the line slants

downward. 11. 0m = has a 0 slope, so the line is horizontal. 12. m is undefined, so the line is vertical.

Exercise Set 3.6

2. Use the slope formula.

2 1

2 1

11 6 5

7 1 6

y ym

x x

− −= = =− −


2 1

2 1

4 9 5 5

6 2 4 4

y ym

x x

− − −= = = = −− −


2 1

2 1

11 7 4 4

2 3 5 5

y ym

x x

− −= = = = −− − − −


2 1

2 1

6 ( 4) 105

1 ( 3) 2

y ym

x x

− − −= = = =− − − −


2 1

2 1

5 ( 1) 6 2

6 3 9 3

y ym

x x

− − −= = = = −− − − −




2 1

2 1

0 2 2,

4 4 0

y ym

x x

− − −= = =− −

so the slope is

undefined. 14. Use the slope formula.

2 1

2 1

5 ( 5) 00

3 ( 2) 5

y ym

x x

− − − −= = = =− − −


2 1

2 1

5 2 3 3

0 5 5 5

y ym

x x

− −= = = = −− − −


2 1

2 1

6 ( 2) 41

1 3 4

y ym

x x

− − − − −= = = =− − − −

20. 2 has the greater slope because it slants

upward which means it is increasing, and 1

slants downward, which means it is decreasing.

22. 2 has the greater slope because it slants

upward, which means it is increasing, and 1

is horizontal, which means it has a slope of 0. 24. Although both slopes are positive, 1 has the

greater slope because it slants upward at a steeper increase.

26. The equation is already in slope-intercept

form. The slope is 2− and the y-intercept points is ( )0, 6 .

28. Write the equation in slope-intercept form. 5 10

5 10x y

y x− + =

= +

The slope is 5 and the y-intercept point is

( )0, 10 .

30. Write the equation in slope-intercept form. 3 4 6

3 3

4 2

x y

y x

− − =

= − −

The slope is 3

4− and the y-intercept point is

30

2, .

⎛ ⎞−⎜ ⎟⎝ ⎠

32. The equation is already in slope-intercept

form. The slope is 1

4− and the y-intercept

point is ( )0 0, .

34. D; The slope of the line is 2 and the


36. C; The slope of the line is 2− and the


38. Since both y values in the given points are

2− , the line is horizontal. So, the slope is 0. 40. Since both x values in the given points are 4,

the line is vertical. So, the slope is undefined. 42. Write the equation in slope-intercept form. 7 0

7y

y− =

=

Since both of the y values in the given points are the same, the line is horizontal. So, the slope is 0.

44. The slopes of the lines are equal and they

have different y-intercepts, so they are parallel. The slope is the steepness of the line. The y-intercept is where the line crosses through the y-axis.

46. Solve each equation for y. 2 10

2 102 10

x yy xy x

− = −− = − −

= +

2 4 24 2 2

1 1

2 2

x yy x

y x

+ == − +

= − +

The slopes are 2 and 1

,2

− which are

not equal. The product of the slopes is equal to 1,− so the two lines are perpendicular.

1

2 12

⎛ ⎞− = −⎜ ⎟⎝ ⎠



48. Solve each equation for y. 4 7

4 71 7

4 4

x yy x

y x

+ == − +

= − +

2 5 05 2

2

5

x yy x

y x

− =− = −

=

The slopes are 1

4− and

2,

5 which are

not equal. The slopes of the lines are not equal and the product is not equal to 1,− so the two lines are neither parallel nor perpendicular.

1 2

14 5⎛ ⎞− ≠ −⎜ ⎟⎝ ⎠

50. Use the slope formula for points ( )1, 0 and

( )0, 3 .

2 1

2 1

3 0 33

0 1 1

y ym

x x

− −= = = = −− − −

52. Use the slope formula for the points ( )2, 4

and ( )1, 3 .− −

2 1

2 1

1 4 51

3 2 5

y ym

x x

− − − −= = = =− − − −

54. Take the takeoff point to be ( )0, 0 and use

the definition of slope.

rise 3 0 3

run 25 0 25m

−= = =−

The slope of the climb is 3

.25

56. Take the starting point to be ( )0, 0 and use

the definition of slope.

rise 15 3

run 100 20m = = =

The slope of the road shown is 3

.20

58. a. Rewrite the equation in slope-intercept

form, .y mx b= +

266 10 27,40910 266 27,409

133 27,409

5 10

x yy x

y x

− + == +

= +

The slope of the line is 133

5 and the

y-intercept is 27 409

10

,.

b. The slope in this context means that there will be an average increase of about

26.6 nurses employed per year. c. The y-intercept represents the average

number of employed nurses there were in the year 2000.

60. a. The slope is 107.3. This is the estimated

amount that college tuition and fees are going to increase by every year.

b. The y-intercept is 1245.62. This is the yearly cost of college tuition and fees in the year 2000.

62. No, the slope of the ramp will be too steep.

The law says that the maximum slope of the

ramp is 1

0 08312

. .≈ The plan for the library

has a height of 3 ft and a length of 10 ft, so

the slope will be 3

0 310

. .= Since

0.3 0.083,> you cannot design a ramp that complies with the law.

64. A line parallel to y x= will have the same

slope, which is 1. 66. A line perpendicular to y x= will have a

slope of 1,− since the product of 1 and 1− is 1.−

68. Rewrite the equation in slope-intercept form,

.y mx b= +

3 4 104 3 10

3 5

4 2

x yy x

y x

− + == +

= +

A line parallel to 3 4 10x y− + = will have the

same slope, which is 3

4.

70. 0 3[ ( 10)]

3( 10)3 30

y xy xy x

− = − − −= − += − −



72. 9 8[ ( 4)]9 8( 4)9 8 32

8 23

y xy xy x

y x

− = − − −− = − +− = − −

= − −


1. Write the equation in slope-intercept form. If the y-intercept point is (0, 4), then the y-intercept, b, is 4. Use the given slope, m,

which is 3

.4

−

34

4

y mx b

y x

= +

= − +

2. Graph the y-intercept, 2, at (0, 2). The slope,

3,

4 is “rise over run,” so move up 3 units and

right 4 units, and then make another point. Draw the line through these points.

3. Rewrite the equation in slope-intercept form.

2 62 6

13

2

x yy x

y x

+ == − +

= − +

Graph the y-intercept, 3, at (0, 3). The slope, 1

,2

− is negative, so move down 1 unit and

right 2 units, and then make another point.

Draw the line through these points.

4. Substitute the values into point-slope form

using substituting the slope for m and the given point for 1x and 1,y and then rewrite

in slope-intercept form.

1 1( )5 4( ( 2))5 4( 2)5 4 8

4 3

y y m x xy xy xy x

y x

− = −− = − − −− = − +− = − −

= − −

5. Find the slope of the line. Use (2, 0) for

2 2( , )x y and ( 1,− 2) for 1 1( , ).x y

2 1

2 1

0 2 2

2 ( 1) 3

y ym

x x

− −= = = −− − −

Substitute the slope and one of the coordinate pairs into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )2

0 ( 2)32 4

3 3

y y m x x

y x

y x

− = −

− = − −

= − +

6. You are given two coordinate pairs,

( 2,3)− and (1, 1), substitute them for

2 2( , )x y and 1 1( , ),x y respectively to find the

slope.

2 1

2 1

3 1 2

2 1 3

y ym

x x

− −= = = −− − −

Substitute the slope and one of the coordinate pairs into the point-slope form equation.



Then rewrite in standard form.

1 1( )2

1 ( 1)32 2

13 32 5

3 32 5

3 32 3 5

y y m x x

y x

y x

y x

x y

x y

− = −

− = − −

− = − +

= − +

+ =

+ =

7. Define the variables. Define x as “years after

2000” and define y as “number of houses sold. Two pairs of data points are given, (2, 7513) and (6, 9198). Use these data points to find the slope. Substitute (6, 9198) for

2 2( , )x y and (2, 7513) for 1 1( , ).x y

2 1

2 1

9198 7513 1685421.25

6 2 4

y ym

x x

− −= = = =− −

Substitute the slope and one of the points into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )7513 421.25( 2)7513 421.25 842.5

421.25 6670.5

y y m x xy xy x

y x

− = −− = −− = −

= +

Substitute 14 for x because 2014 is 14 years after 2000, and then simplify.

421.25(14) 6670.5 12568y = + =

The amount of houses that is predicted to be sold in 2014 is 12568 houses.

8. A horizontal line contains all the points with

a particular y-value and has an equation of the form .y b= The point given here is

(6, 2),− so the equation of the horizontal line

that contains the point is 2.y = −

9. A line with undefined slope is vertical, and it

contains all the points with a particular x-value. It has an equation of the form .x c= The point given here is (6, 2),− so the

equation of the line with undefined slope that contains the point is 6.x =

10. Rewrite the given equation in slope-intercept form. 3 4 1

4 3 13 1

4 4

x yy x

y x

+ == − +

= − +

The slope is 3

.4

− Since the new line is

parallel to 3 4 1,x y+ = the lines must have

the same slope. Substitute this slope and the given point into the point-slope form equation. Then rewrite in standard form.

1 1( )3

( 3) ( 8)4

4( 3) 3( 8)4 12 3 24

4 3 12 243 4 12

y y m x x

y x

y xy x

y xx y

− = −

− − = − −

+ = − −+ = − +

+ + =+ =

11. Rewrite the given equation in slope-intercept

form. 3 4 1

4 3 13 1

4 4

x yy x

y x

+ == − +

= − +

The slope is 3

.4

− Since the new line is

perpendicular to 3 4 1,x y+ = the new line’s

slope is the negative reciprocal, or 4

.3

Substitute this slope and the given point into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )4

( 3) ( 8)34 32

33 34 41

3 3

y y m x x

y x

y x

y x

− = −

− − = −

+ = −

= −



12. The slope of the baseballs path is 1

,3

and

since the new line is parallel, its slope is the same. Substitute the slope and the given point into the point-slope form equation and then rewrite in standard form.

1 1( )1

40 ( 90)3

3 120 903 30

3 30

y y m x x

y x

y xy xx y

− = −

− = −

− = −− =− = −


1. The slope is 4.− The y-intercept point is

(0, 12).

2. The slope is 2

.3

The y-intercept point is

70, .

2⎛ ⎞−⎜ ⎟⎝ ⎠

3. The slope is 5. The y-intercept point is

(0, 0). 4. The slope is 1.− The y-intercept point is

(0, 0).

5. The slope is 1

.2

The y-intercept point is

(0, 6).

6. The slope is 2

.3

− The y-intercept point is

(0, 5). 7. Parallel; the lines have the same slope. 8. Parallel; the lines have the same slope. 9. Neither; the lines do not have the same slope,

nor are the slopes negative reciprocals of each other.

10. Neither; the slopes are reciprocals, but they

are not negative reciprocals, so the lines are not perpendicular.

Exercise Set 3.7

2. Substitute the slope and the intercept into slope-intercept form. m is the slope and b is the y-coordinate of the y-intercept point.

16

2

y mx b

y x

= +

= −

4. Substitute the slope and the intercept into

slope-intercept form. m is the slope and b is the y-coordinate of the y-intercept point.

13

5

y mx b

y x

= +

= − −

6. Substitute the slope and the intercept into

slope-intercept form. m is the slope and b is the y-coordinate of the y-intercept point.

40

54

5

y mx b

y x

y x

= +

= − +

= −

8. The equation is given in slope-intercept form,

.y mx b= + The slope is the coefficient of x

and the y-intercept is the term added to mx. Graph the y-intercept point at (0, 1) first. Then move up 2 units and right 1 unit. Make a point there, at (1, 3). Graph the line that contains both points.



10. Rewrite 3 9x y+ = in slope-intercept form.

3 9y x= − +

In slope-intercept form, ,y mx b= + the slope

is the coefficient of x and the y-intercept is the term added to mx. Graph the y-intercept at (0, 9). Then move down 3 units and right 1 unit. Make a point there, at (1, 6). Graph the line that contains both points.

12. Rewrite 2 5 16x y− + = − in slope-intercept

form. 2 16

5 5y x= −

The slope is the coefficient of x and the y-intercept is the term added to mx. The y-intercept point is difficult to graph, so find a pair of coordinates to graph that produces whole number coordinates that are easy to graph. (3, 2)− is a good choice for this.

Because the slope is 2

,5

move up 2 units and

right 5 units. Mark this point, (8, 0). Graph

the line that contains these two points.

14. Substitute the given slope and point into the

point-slope form equation. Then rewrite in slope-intercept form.

1 1( )1 4( 5)1 4 20

4 19

y y m x xy xy x

y x

− = −− = −− = −

= −

16. Substitute the given slope and point into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )( 4) 4( 2)

4 4 84 4

y y m x xy x

y xy x

− = −− − = − −

+ = − += − +



1 1( )2

4 ( ( 9))32

4 632

103

y y m x x

y x

y x

y x

− = −

− = − −

− = +

= +



1 1( )1

( 6) ( 4)51 4

65 51 26

5 5

y y m x x

y x

y x

y x

− = −

− − = − −

+ = − +

= − −

22. Find the slope. Use (7, 8) for 2 2( , )x y and

(3, 0) for 1 1( , ).x y

2 1

2 1

8 0 82

7 3 4

y ym

x x

− −= = = =− −

Substitute the slope and one of the given points into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )0 2( 3)

2 6

y y m x xy x

y x

− = −− = −

= −



24. Find the slope. Use (2, 6) for 2 2( , )x y and

(7, 4)− for 1 1( , ).x y

2 1

2 1

6 ( 4) 102

2 7 5

y ym

x x

− − −= = = = −− − −


1 1( )( 4) 2( 7)

4 2 142 10

y y m x xy x

y xy x

− = −− − = − −

+ = − += − +

26. Find the slope. Use ( 3,− 10) for 2 2( , )x y and

( 9, 2)− − for 1 1( , ).x y

2 1

2 1

10 ( 2) 122

3 ( 9) 6

y ym

x x

− − −= = = =− − − −


1 1( )10 2( ( 3))10 2 6

2 16

y y m x xy xy x

y x

− = −− = − −− = +

= +

28. Find the slope. Use (4, 8)− for 2 2( , )x y and

(8, 3)− for 1 1( , ).x y

2 1

2 1

8 ( 3) 5 5

4 8 4 4

y ym

x x

− − − − −= = = =− − −


1 1( )5

( 3) ( 8)45

3 1045

134

y y m x x

y x

y x

y x

− = −

− − = −

+ = −

= −

30. Find the slope. Use 3 3

,2 4

⎛ ⎞⎜ ⎟⎝ ⎠

for 2 2( , )x y and

1 1,

2 4⎛ ⎞−⎜ ⎟⎝ ⎠

for 1 1( , ).x y

2 1

2 1

3 114 4

13 1 12 2

y ym

x x

⎛ ⎞− −⎜ ⎟− ⎝ ⎠= = = =− −

Substitute the slope and one of the given points into the point-slope form equation.

Then rewrite in slope-intercept form.

1 1( )3 3

14 23 3

4 26 3

4 43

4

y y m x x

y x

y x

y x

y x

− = −⎛ ⎞− = −⎜ ⎟⎝ ⎠

− = −

= − +

= −

32. The y-intercept point is given at (0, 2),− and

another point is given at (2, 2). Substitute them for 2 2( , )x y and 1 1( , ),x y respectively to

find the slope.

2 1

2 1

2 22

0 2

y ym

x x

− − −= = =− −

Use the slope and the y-intercept to write the equation in slope intercept form. Then rewrite in standard form.

2 22 2

y xx y

= −− + = −

34. There are two points given, (3, 1)− and

( 4,0).− Substitute them for 2 2( , )x y and

1 1( , ),x y respectively to find the slope.

2 1

2 1

1 0 1

3 ( 4) 7

y ym

x x

− − −= = = −− − −

Substitute the slope and one of the points into the point-slope form equation. Then rewrite in standard form.

1 1( )1

0 ( ( 4))7

7 ( 4)7 47 4

y y m x x

y x

y xy x

x y

− = −

− = − − −

= − += − −

+ = −

36. A horizontal line contains one particular

y-coordinate, but every x-coordinate. Use the y-coordinate of the point to write the equation.

1y =

38. A vertical line contains one particular x-

coordinate, but every y-coordinate. Use the x-coordinate of the point to write the equation.

2x =



40. A line with undefined slope is vertical. A vertical line contains one particular x-coordinate, by every y-coordinate. Use the x-coordinate of the point to write the equation.

0x = 42. Parallel lines have the same slope. The slope

of the given line is 3. Substitute this slope and the given point into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )5 3( 1)5 3 3

3 2

y y m x xy xy x

y x

− = −− = −− = −

= +

44. Perpendicular lines have slopes that are

negative reciprocals of each other. The slope

of the given line is 2

,3

so the slope of the

new line must be 3

.2

− Substitute this slope

and the given point into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )3

8 ( ( 4))23

8 623

22

y y m x x

y x

y x

y x

− = −

− = − − −

− = − −

= − +

46. Perpendicular lines have slopes that are

negative reciprocals of each other. The slope

of the given line is 3

,2

− so the slope of the

new line must be 2

.3

Substitute this slope and

the given point into the point-slope form equation. Then rewrite in slope-intercept form.

1 1( )2

( 3) ( ( 2))32 4

33 32 5

3 3

y y m x x

y x

y x

y x

− = −

− − = − −

+ = +

= −

48. Substitute this slope and the given point into the point-slope form equation. Then rewrite in standard form.

1 1( )2 3( ( 4))2 3 12

3 143 14

y y m x xy xy x

y xx y

− = −− = − −− = +

= +− + =

50. Find the slope. Substitute (8, 6) for 2 2( , )x y

and (2, 9) for 1 1( , ).x y

2 1

2 1

6 9 3 1

8 2 6 2

y ym

x x

− −= = = − = −− −

Substitute this slope and one of the given points into the point-slope form equation. Then rewrite in standard form.

1 1( )1

6 ( 8)2

2 12 82 20

y y m x x

y x

y xx y

− = −

− = − −

− = − ++ =

52. Substitute the slope and the y-intercept into

the slope-intercept form equation. Then rewrite in standard form.

24

92

49

36 9 2

y mx b

y x

y x

x y

= +

= − +

+ =

+ =

54. A horizontal line contains one particular y-

coordinate, but every x-coordinate. Use the y-coordinate of the point to write the equation.

0y =

56. Parallel lines have the same slope. Rewrite

the given equation in slope-intercept form to find the slope. 6 2 5

2 6 55

32

x yy x

y x

+ == − +

= − +

The slope is 3.− Substitute this slope and the given point into the point-slope form equation. Then rewrite in standard form.

1 1( )( 3) 3( 8)

3 3 243 21

y y m x xy x

y xx y

− = −− − = − −

+ = − ++ =



58. a. Two coordinate pairs are given. Written in the form (years past 2009, value of computer), they are (2, 2000) and

(4, 800). Find the slope by substituting the points for 1 1( , )x y and 2 2( , ),x y

respectively.

2 1

2 1

800 2000

4 21200

2600

y ym

x x

−=

−−=−

= −

= −

Now use the slope and one of the coordinate pairs to write an equation in point-slope form. Then rewrite in slope-intercept form.

1 1( )800 600( 4)800 600 2400

600 3200

y y m x xy xy x

y x

− = −− = − −− = − +

= − +

b. Substitute the number of years after 2009, which is 5, for x. Then simplify.

600(5) 32003000 3200

200

y = − += − +=

The computer is estimated to be worth $200 in 2014.

60. a. Two coordinate pairs are given. Written in the form (years past 2000, value of building), they are (7, 165,000) and

(12, 180,000). Find the slope by substituting the points for 1 1( , )x y and

2 2( , ),x y respectively.

2 1

2 1

180,000 165,000

12 715,000

53000

y ym

x x

−=

−−=−

=

=


1 1( )165,000 3000( 7)165,000 3000 21,000

3000 144,000

y y m x xy xy x

y x

− = −− = −− = −

= +


3000(20) 144,00060000 144,000204,000

y = += +=

The value of the building is estimated to be $204,000 in 2020.

62. a. There are two coordinate pairs given. In

the form (years past 2004, thousands of systems analysts), they are (0, 487) and (10, 640). Find the slope by substituting the points for 1 1( , )x y and 2 2( , ),x y

respectively.

2 1

2 1

640 487

10 0153

1015.3

y ym

x x

−=

−−=−

=

=


1 1( )487 15.3( 0)487 15.3

15.3 487

y y m x xy xy x

y x

− = −− = −− =

= +


15.3(14) 487214.2 487701.2

y = += +=

The number of systems analysts in 2018 is estimated to be 701,200.

64. True; two vertical lines will have the same

slope, and two lines with the same slope are parallel.

66. Find the midpoint of the segment using the

Midpoint Formula.

1 2 1 2, 2 2

6 8 3 1,

2 214 4

,2 2

( 7, 2)

x x y yM

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠− − − −⎛ ⎞= ⎜ ⎟

⎝ ⎠− −⎛ ⎞= ⎜ ⎟

⎝ ⎠= − −



Find the slope of the segment.

2 1

2 1

1 ( 3) 21

8 ( 6) 2

y ym

x x

− − − −= = = = −− − − − −

The line and segment are perpendicular, so the slope of the perpendicular bisector is the negative reciprocal of the segment. Find the negative reciprocal of 1,− which is 1. Using the slope of 1 and the midpoint of the segment, write an equation for the perpendicular bisector in point-slope form. Then rewrite in standard form.

1 1( )( 1) 1( ( 8))

1 87

7

y y m x xy x

y xy x

x y

− = −− − = − −

+ = += +

− + =

68. Find the midpoint of the segment using the

Midpoint Formula.

1 2 1 2,2 2

5 7 8 2,

2 212 10

,2 2

(6,5)

x x y yM

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

Find the slope of the segment.

2 1

2 1

2 8 63

7 5 2

y ym

x x

− −= = = − = −− −

The line and segment are perpendicular, so the slope of the perpendicular bisector is the negative reciprocal of the segment. Find the

negative reciprocal of 3,− which is 1

.3

Using

the slope and the midpoint of the segment, write an equation for the perpendicular bisector in point-slope form. Then rewrite in standard form.

1 1( )1

8 ( 5)3

3 24 53 193 19

y y m x x

y x

y xy x

x y

− = −

− = −

− = −= +

− + =

70. Use the Distance Formula. 2 2

2 1 2 1

2 2

2 2

( ) ( )

( 6 ( 8)) ( 3 ( 1))

2 ( 2)

4 4

8

2 2

d x x y y= − + −

= − − − + − − −

= + −= +==

72. Use the Distance Formula.

2 22 1 2 1

2 2

2 2

( ) ( )

(5 7) (8 2)

( 2) 6

4 36

40

2 10

d x x y y= − + −

= − + −

= − += +==

Chapter 3 Vocabulary Check

1. Planes that do not intersect are called parallel

planes. 2. Coplanar lines that do not intersect are called

parallel lines. 3. A line that intersects two or more coplanar

lines at different points is called a traversal. 4. Skew lines are not coplanar; they are not

parallel and do not intersect. 5. The measure of the steepness or tilt of a line

is called slope. 6. The slope of a line through two points is

measured by the ratio of vertical change to horizontal change.

7. If a linear equation is in the form ,y mx b= +

the slope of the line is m, the y-intercept point is (0, b), and the y-intercept is b.

8. The form = +y mx b is called the slope-

intercept form. 9. The slope of a horizontal line is 0.

10. The slope of a vertical line is undefined. 11. Two non-vertical perpendicular lines have

slopes whose product is 1.−



12. Two non-vertical lines are parallel if they have the same slope and different y-intercepts or x-intercepts.

13. Angles 3, 4, 5, and 6 are called interior

angles. 14. Angles 1, 2, 7, and 8 are called exterior

angles. 15. Angles 2 and 6 are called corresponding

angles. 16. Angles 4 and 5 are called alternate interior

angles. 17. Angles 1 and 8 are called alternate exterior

angles. 18. Angles 3 and 5 are called same-side interior

angles. 19. Angles 2 and 3 are called vertical angles. 20. Angles 1 and 4 are called vertical angles. 21. Given line and point P, how many lines

can be drawn through P parallel to ? Exactly one.

22. Given line and point P, how many lines

can be drawn through P perpendicular to ? Exactly one.



24. The postulate that gives the answer to

Exercise 22 above is called the Perpendicular Postulate.

Chapter 3 Review

1. 2 and 7;∠ ∠ lines a and b with traversal d

3 and 6;∠ ∠ lines c and d with traversal e 3 and 8;∠ ∠ lines b and e with traversal c 2. 2 and 6;∠ ∠ lines a and e with traversal d 5 and 8;∠ ∠ lines a and b with traversal c

3. 1 and 4;∠ ∠ lines c and d with traversal b 1 and 5;∠ ∠ lines a and b with traversal c 2 and 4;∠ ∠ lines a and b with traversal d

2 and 5;∠ ∠ lines c and d with traversal a 4. 1 and 7;∠ ∠ lines c and d with traversal b 5. 1∠ and 2∠ are corresponding angles. 6. 1∠ and 2∠ are alternate interior angles. 7. Use the Alternate Interior Angles Theorem

and solve for x

(3 5) 653 5 65

3 6020

+ ° = °+ =

==

xx

xx

8. Use the Vertical Angles Theorem and

Same-Side Interior Angles Theorem.

(2 10) 130 1802 10 50

2 4020

xx

xx

+ °+ ° = °+ =

==

9. Given that 1 9,∠ ≅ ∠ the corresponding

angles of lines r and p made by traversal are congruent. So, r p by the

Corresponding Angles Theorem. 10. Given 3 6 180∠ + ∠ = °m m is always true,

there is not enough information to make an assertion about any lines.

11. Given 2 3 180 ,m m∠ + ∠ = ° the same-side

interior angles of lines and m with traversal r are supplementary, thus parallel by the Same-Side Interior Angles Theorem.

12. Given 5 11,∠ ≅ ∠ the alternate interior angles

of lines r and p made by traversal m are

congruent. So, r p by the Alternate Interior

Angles Theorem. 13. Given ,a b it follows that 120 1° = ∠m by

the Corresponding Angles Converse. Since 1∠ and 2∠ are vertical angles, we have

2 1 120∠ = ∠ = °m m by the Vertical Angles Theorem.



14. Given ,p q 2 105∠ = °m by the Alternate

Interior Angles Converse. Given 1∠ and 2∠ are supplementary, the sum of the

measure of their angles is 180 .°

1 2 1801 105 180

1 75

m mm

m

∠ + ∠ = °∠ + ° = °

∠ = °

15. By the Corresponding Angles Converse,

118.x = 16. The angles that measure y° and 25° can be

added by the Angle Addition Postulate. By the Same-Side Interior Converse, the sum of the measures of the angles that measure ( 25)y + ° and x° is 180 .°

( 25) 180

25 118 18037

+ °+ ° = °+ + =

=

y xy

y

17. If and ,c db b⊥ ⊥ then c d because if two

lines are perpendicular to the same line, then they are parallel to each other by the Two Lines Perpendicular to a Third Line Theorem.

18. If ,c d then .a c⊥ Because it is given that

and ,d da c⊥ the result follows from the Perpendicular Transversal Theorem.

19. Draw line m and point Q not on m.

Step 1: Put the compass point at Q and draw an arc that intersects line m at two points, call them E and F. Step 2: Open the compass to more than half of .EF Put the compass point on point F and draw an arc. Using the same compass setting, put the compass point on point E and draw an arc that intersects the previous arc at a point H. Step 3: Use a straightedge to draw a line connecting

Q and H. Line QH is perpendicular to line m

and passes through point Q.

20. Use the segments below, and draw a horizontal line and label a point on the line C.

Step 1: Open the compass to the endpoints of the segment of length b. Put the compass point on point C and draw an arc intersecting the line, label it D. So, .CD b= Step 2: Using a small compass setting, draw arcs on either side of C that intersect the line. Label the intersections E and F. Step 3: Open the compass and put the compass point on point E. Draw an arc above the line. Using the same compass setting, put the compass point on point F and draw an arc intersecting the previous arc. Label the intersection O. Draw the ray from point C through point O. The ray is perpendicular to .CD Step 4: Repeat Steps 2 and 3 for point D to draw ray

.DT The ray is perpendicular to .CD Step 5: Open the compass, so the points of the compass are on the endpoints of the segment of length a. Using this compass setting, put the compass point on point C and draw an arc

that intersects ray .CO Label the point of intersection A. So, .CA a= Step 6: Repeat Step 5 for point D. Label the point of intersection B.



Step 7: Use a straightedge to draw segment .AB The quadrilateral ABDC is a rectangle with side lengths a and b.

21. Draw a horizontal line and label a point on

the line C.

Step 1: Open the compass so that the points of the compass are on the endpoints of the segment of length b. Put the point of the compass on C and draw an arc that intersects the line. Using the same compass settings, put the point of the compass on the point where the arc and the line intersect and draw an arc that intersects the line at a point other than C. Label the point D. Note that 2 .CD b= Step 2: Repeat Steps 1–7 from Exercise 20 to construct the rectangle ABDC with side lengths a and 2b.

22. Draw .CAP∠

Step 1: Open the compass so that it is less than .CA Put the point of the compass on point A and draw an arc that intersects both sides of

.CAP∠ Label the points of intersection E and F. Step 2: Using the same compass setting, put the compass point on point C and draw an arc

that intersects .AC Label the point G.

Step 3: Open the compass to the length .EF Put the compass point on point G and draw an arc that intersects the previous arc. Step 4: Use a straightedge to construct a ray with its endpoint at point C that passes through the intersection of the arcs in the previous step. Step 5: Open the compass so that the points are on the endpoint of the segment of length a. Put the compass point on point A and draw an arc

that intersects .AP Using the same compass setting, put the compass point on the intersection of the ray and the arc, and draw an arc that intersects the ray at a point other than A. Label the point B. Step 6: Using the same compass settings, repeat Step 5 for point C. Label the intersection D. Step 7: Use a straightedge to draw .BD The quadrilateral ABDC has one pair of opposite sides, each with length 2 .a

23. 1 2

1 2

2 3 51.

6 1 5

y y

xm

x

− − −= = = = −−

−−

The slope is 1.−

24. 1 2

1 2

5 2 7

7 ( 7) 0

y ym

x x

− − −= = = −− − − −

.

The slope is undefined since the line passing through the two given points is vertical.

25. The equation is given in the form y mx b= +

where m is the slope and b is the y-intercept. So, the slope is 2 and the y-intercept point is (0, 1).−

26. Write the equation in slope-intercept form by

solving for y.

3 2( 5)3 2 10

2 7

− = − +− = − −

= − −

y xy x

y x

The slope is 2− and the y-intercept point is (0, 7).−



27. Given the slope, 1

,2

m = − and the

y-intercept, 12,b = use the slope-intercept form = +y mx b to write the equation

112.

2y x= − +

28. First find the slope:

2 1

2 1

2 ( 2) 44

4 3 1xm

y y

x

− −−

== =−

=−

Choose a point and apply the point-slope formula, then write the equation in slope-intercept form.

1 1( )2 4( 4)2 4 16

4 14

y y m x xy xy x

y x

− = −− = −− = −

= −

29. Compare the slopes of the two lines.

2 1

2 1

11 ( 4) 155

2 ( 1) 3ABm

y y

x x

− −= = =− −

−=

−

2 1

2 1

10 1 93

4 1 3CD

y y

x xm

−=

−−= = =−

Since the slopes are not equal, nor is their product 1,− they are neither parallel nor perpendicular.


2 1

2 1

2 8 10 10

1 2 3 3AB

y ym

x x

−= − − −= =

−−=

− −

2 1

2 1

3 7 10 10

0 3 3 3CD

y ym

x x

−= − − −= =

−=

− −

Since the slopes are equal, the two lines are parallel.


2 1

2 1

3 2 1

3 0 3AB

y y

xm

x

−=

−−= = −

− −

2 1

2 1

3 ( 6) 93

1 ( 2) 3CD

y ym

x x

−= − −=

−= =

− −

Since the product of the slopes is 1,− the lines are perpendicular.


2 1

2 1

8 3 51

4 ( 1) 5AB

y y

x xm

−=−

= =− −−

=

2 1

2 1

8 0 81

2 ( 6) 8CD

y y

x xm

−=−

= =− −−

=

Since the slopes are the same, the two lines are parallel.

33. Since the line must be parallel to 8 1,y x= −

the slope must be 8. Use the point-slope formula and the given point.

1 1

2 8( (( )

6))2 8( 6)

my yx

y x

x xy−− = − −−

=

+

−

=

34. The slope of the line must be the opposite

reciprocal of 1

.6

The slope is, therefore, 6.−

Use the point-slope formula and the given point.

1 1

( 3) 6( 3)(

( )

3 6 3)

y yy x

x

m

y

x x−− − = − −

+ = − −

= −

35. Answers may vary. 1 and 3;∠ ∠ 2 and 4;∠ ∠

6 and 8;∠ ∠ and 5 and 7∠ ∠ are corresponding angles.

36. 8∠ measures 110° by the Vertical Angles

Theorem. 6∠ corresponds to the angle labeled 110 ,° and 3∠ corresponds to 8,∠ so both will measure 110° by the Corresponding Angles Converse.

37. Use the Alternate Interior Angles Theorem,

and solve for x.

(2 ) 10653

xx° = °=

38. By the Corresponding Angles Converse,

.m It is given that ,n so it follows that m n by transitivity. It is given that

and .a a m⊥ ⊥ ⊥a n by the Perpendicular Transversal Theorem.

39. Since Morris Ave intersects both 1st and 3rd

streets at right angles, the Corresponding Angles Theorem gives that 1st and 3rd streets are parallel. Since 3rd Street and 5th Street are parallel, 1st Street is parallel to 5th Street by the Transitive Property.



40. Use the slope-intercept formula ,y mx b= +

where m is the slope and b is the y-intercept, to obtain 5 6.y x= − +

41. Use the point-slope formula.

1 1

8 3( (( )

2))8 3( 2)

my yx

y x

x xy−− = − −−

=

+

−

=

42. Use the point slope formula.

1 1

( 9) 3( 1)9 3

)

)

(

( 1

y yy x

m x x

y x

−− −

+ =

== −

−

−

43. The slope of the line must be the opposite

reciprocal to the given slope 2. The slope is,

therefore, 1

.2

− Use the point slope formula.

1 1

1( 3) ( 1)

21

3 ( 1)2

( )my y

y x

y x

x x−

− − = − −

+ = − −

= −

Chapter 3 Test

1. AB and DH do not intersect and are not

coplanar, so they are skew lines.

2. BC and AD are coplanar and do not intersect, so they are parallel lines.

3. DA and HD are incident at ,D so they are neither parallel nor skew lines.

4. Use the Corresponding Angles Converse to

find 1 65 .m∠ = ° Use the Vertical Angles Theorem or The Alternate Interior Angles Theorem to find 2 65 .m∠ = °

5. Use the Alternate Angles Theorem to obtain

1 85 .m∠ = ° By the Same-Side Interior Angles Theorem, 2 70 180 .m∠ + ° = ° Thus,

2 110 .m∠ = ° 6. Use the Corresponding Angles Converse to

find 1 85 .m∠ = ° By the Same-Side Interior Angles Converse, 1 2 180 ,m m∠ + ∠ = ° so

2 180 1 180 85 95 .m m∠ = °− ∠ = ° − ° = ° So 2 95 .m∠ = °

7. 2∠ and the angle labeled 70° form a linear pair, so 2 180 70 110 ,m∠ = °− ° = ° 1∠ and the angle labeled 70° are corresponding angles with respect to the parallel lines, so

1 70m∠ = ° by the Corresponding Angles Converse.

8. m if and only if the angles labeled

(14 5)x − ° and 13x° have equal measure, by

the Alternate Interior Angles Theorem and its converse. Set the two to be equal and solve for .x

(14 5) 13

5

x

x

x ° = °=

−

9. Use the Alternate Exterior Angles Theorem

with the angle labeled (2 30)x − ° and the

sum of the angles labeled an ,45 d x° ° which may be added by the Angle Addition Postulate.

( 45(2 3 )0)

2 75

75

x

x x

x

x

° = + °= +=

−

10. The sum of the angles labeled x° and 25°

make a vertical angle with the labeled right angle. Use the Vertical Angles Theorem and the Angle Addition Postulate to solve for .x

25 90

65x

x° + ° = °=

11.

Statements Reasons

1. m 1. Given


3. 2 4∠ ≅ ∠ 3. Given

4. 1 4∠ ≅ ∠ 4. Transitive Property

5. n p 5. Corresponding Angles Theorem

12. Draw line m and point T not on m.



Step 1: Put the compass point at Q and draw an arc that intersects line m at two points, call them E and F. Step 2: Open the compass to more than half of .EF Put the compass point on point F and draw an arc. Using the same compass setting, put the compass point on point E and draw an arc that intersects the previous arc at a point H. Step 3: Use a straightedge to draw a line connecting

T and H. Line TH is perpendicular to line m and passes through point T.

13. Draw .ABC∠

Step 1: Put the compass point on point B and draw and arc that intersects both sides of the angle. Step 2: With the same compass setting, put the compass point on point A and draw an arc

that intersects .BA Step 3: Open the compass to the length between the two intersections in Step 1. Put the point of the compass on the intersection of the arc and

BA from Step 2 and draw an arc that intersects the previous arc. Step 4: Use a straightedge to draw line m through point A and the intersection from Step 3. By

construction, A ABC∠ ≅∠ so .m BC

14. Rewrite the given equation in slope-intercept form, .y mx b= +

2 3 6

3 2 6

22

3

x y

y

y x

x

− = −− = −

= +

−

The slope is 2

3 and the y-intercept is 2. Plot

the y-intercept point (0, 2) and the point up

two and to the right three, (4,3). Draw a

straight line through the points to graph the given equation.

15. The line has a slope of zero, so it is

horizontal, with all points having a y-coordinate of 3.−

16. Use the slope formula with the points given.

2 1

2 1

( 8) 10 18 3

5 ( 7) 12 2

y y

x xm

− − − −= == = −− − −

17. To find the slope and y-intercept, write the

equation in slope-intercept form ,y mx b= +

where m is the slope and b is the y-intercept.

3 12 8

12 3 8

1 2

4 3

x y

y x

y x

+ == − +

+= −

So, the slope is 1

4− and the y-intercept point

is 2

0, .3

⎛ ⎞⎜ ⎟⎝ ⎠



18. A horizontal line has slope 0, so the horizontal line through the point (2, 8)− is

given by 8.y = − This is because a

horizontal line must pass through all values of ,x but only one value of .y

19. Use the point-slope formula with the point

and the slope that are given, and write the equation in standard form.

1 1

( 1) 3( 4)

1 3 12

3 1

(

1

)y y

y x

y x

x y

m x x−− − = − −

+ = − ++ =

= −

20. Use the two points given to find the slope of

the line.

2 1

2 1

3 ( 2) 1

6 4 2m

y y

x x

− − − −= = −− −

=

Choose a point and apply the point-slope formula with the point and the slope

1,

2m = − and write the equation in standard

form.

1 1

( 2) ( 4

(

)

)

1

2

21

0

21

2

2

m x xy y

y

x

x

y x

y

= −−

− − = − −

− +

+

+

=

=

21. First, use the given line to determine the

necessary slope. The slope of the given line is 3, so the slope of the perpendicular line must

be the opposite reciprocal 1

.3

− Use the

point-slope formula, and write the equation in standard form.

1 1( )

12 ( ( 1))

31 1

23 3

1 12

3 31

3

5

3

m x x

y x

y

y y

x

x y

x y

− = −

− = − − −

− = − −

+ = − +

+ = −

22. Use the line given to determine the slope of

the line. The slope of the line given is 1

,2

−

so the slope of a line parallel to it must also

have a slope of 1

.2

− Now use the point-slope

formula, and write the equation in slope-intercept form.

1 1( )

1( 2) ( 3)

21 3

22 21 1

2 2

m x x

y x

y x

y x

y y = −

− − = − −

+ = − +

= − −

−

23. First find the slope of the two lines. For 1,L

write the equation in slope-intercept form ,y mx b= + where m is the slope and b is

the y-intercept.

2 5 8

5 2

8

8

2

5 5

x y

y x

y x

− =− = − +

−=

So, 1

2.

5Lm = Use the slope formula and the

two points given to find the slope of 2.L

2

2 1

2 1

1 4 5

1 1 2L

y y

x xm

m

−−− −= =− −

=

Since 1 2

,L Lm m≠ 1 2and LL are not parallel.

Since 1 2· 1,L Lm m ≠ − 1 2 and LL are not

perpendicular.

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Chapter 3 ISM