hscc geom wsk 03 - schoolwires · 2015. 9. 29. · chapter 3 2. a. ⃖ab ##⃗ and ⃖bc##⃗ are...

Copyright © Big Ideas Learning, LLC Geometry 71All rights reserved. Worked-Out Solutions

Chapter 3

Chapter 3 Maintaining Mathematical Profi ciency (p. 123)

1. m = 2 − (−1) — −1 − 3 = 3 — −4

slope = − 3 — 4

2. m = −1 − 2 — −3 − (−2) = −3 — −3 + 2

= −3 — −1 = 3

slope = 3

3. m = −2 − (−2) — 1 − (−3)

= −2 + 2 — 1 + 3

= 0 — 4 = 0

slope = 0

4. y = mx + b 1 = −3(6) + b 1 = −18 + b 19 = b The equation is y = −3x + 19.

5. y = mx + b 8 = −2(−3) + b 8 = 6 + b 2 = b The equation is y = −2x + 2.

6. y = mx + b 5 = 4(−1) + b 5 = −4 + b 9 = b The equation is y = 4x + 9.

7. y = mx + b −4 = 1 — 2 (2) + b −4 = 1 + b −5 = b The equation is y = 1 — 2 x − 5.

8. y = mx + b −5 = − 1 — 4 (−8) + b −5 = 2 + b −7 = b The equation is y = − 1 — 4 x − 7.

9. y = mx + b 9 = 2 — 3 (0) + b 9 = 0 + b 9 = b The equation is y = 2 — 3 x + 9.

10. When calculating the slope of a horizontal line, the vertical change is zero. This is the numerator of the fraction, and zero divided by any number is zero. When calculating the slope of a vertical line, the horizontal change is zero. This is the denominator of the fraction, and any number divided by zero is undefi ned.

Chapter 3 Mathematical Practices (p. 124) 1. These lines are perpendicular. They have slopes m1 = −

1 — 2

and m2 − 2.

6

−4

4

−6

y = −12x + 1

y = 2x − 4

2. These lines are coincident, because their equations are equivalent.

6

−4

4

−6

y = −12x + 1

3. These lines are parallel. Their slopes are equal, m1 = − 1 — 2

and m2 = − 1 — 2 .

6

−4

4

−6

y = −12x + 1

y = −12x − 1

4. These lines are neither parallel nor perpendicular because their slopes are m1 = −

1 — 2 and m2 = 1. They intersect

at (−2, 2).

6

−4

4

−6

y = −12x + 1

y = x + 4

3.1 Explorations (p. 125) 1. a. Parallel lines have no common points.

b. Intersecting lines have one point in common.

c. Coincident lines have infi nitely many points in common.

72 Geometry Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 3

2. a. ⃖ ##⃗ AB and ⃖ ##⃗ BC are intersecting lines. They intersect at point B.

b. ⃖ ##⃗ AD and ⃖ ##⃗ BC are parallel lines. They are coplanar and will never intersect.

c. ⃖ ##⃗ EI and ⃖ ##⃗ IH are coincident lines. Points E, I, and H are collinear.

d. ⃖ ##⃗ BF and ⃖ ##⃗ EH are skew lines. They are not coplanar and will never intersect.

e. ⃖ ##⃗ EF and ⃖ ##⃗ CG are skew lines. They are not coplanar and will never intersect.

f. ⃖ ##⃗ AD and ⃖ ##⃗ GH are parallel lines. They both lie on plane ABG, which is not drawn, and they will never intersect.

3. a. Vertical angles: ∠1, ∠3; ∠2, ∠4; ∠5, ∠7; ∠6, ∠8Two pairs of opposite rays are formed by each of these pairs of angles.

b. Linear pairs: ∠1, ∠2; ∠1, ∠4; ∠2, ∠3; ∠4, ∠3; ∠5, ∠8; ∠5, ∠6; ∠6, ∠7; ∠7, ∠8One pair of opposite rays is formed by each of these pairs of angles.

4. Two lines are parallel if they are coplanar and do not intersect. Two lines intersect if they are coplanar and have exactly one point in common. Coincidental lines are coplanar and share all the same points because the equations of the lines are the same. Skew lines are lines that do not intersect and are not coplanar.

5. Sample answer: ⃖ ##⃗ EC and ⃖ ##⃗ BD are skew lines because they are not coplanar and they do not intersect. ⃖ ##⃗ DH and ⃖ ##⃗ CG are parallel because they are coplanar and will never intersect.

⃖ ##⃗ AF and ⃖ ##⃗ FH are intersecting lines because they intersect at point F.

3.1 Monitoring Progress (pp. 126–128) 1. The only line skew to ⃖ ##⃗ EH that contains F is ⃖ ##⃗ CF .

2. yes; Because ####⃗ DM is perpendicular to — BF at M, no other line could also be perpendicular to the same line through the same point, according to the Perpendicular Postulate (Post. 3.2).

3. ∠1 and ∠5 are corresponding angles.

4. ∠2 and ∠7 are alternate exterior angles.

5. ∠4 and ∠5 are alternate interior angles.

3.1 Exercises (pp. 129–130)

Vocabulary and Core Concept Check 1. Two lines that do not intersect and are not parallel are

skew lines.

2. ∠2 and ∠3 do not belong because they are vertical angles formed by one pair of intersecting lines and one point of intersection. The other three pairs of angles are formed by two lines cut by a transversal.

Monitoring Progress and Modeling with Mathematics 3. The line containing point B parallel to ⃖ ##⃗ CD is ⃖ ##⃗ AB .

4. The line containing point B that appears to be perpendicular to ⃖ ##⃗ CD is ⃖ ##⃗ BC .

5. The line containing point B that is skew to ⃖ ##⃗ CD is ⃖ ##⃗ BF .

6. The plane containing point B parallel to plane CDH is plane ABF.

7. Parallel lines: ⃖ ###⃗ MK and ⃖ ##⃗ LS

8. Perpendicular lines: ⃖ ##⃗ NP and ⃖ ##⃗ PQ

9. no; ⃖ ##⃗ NP intersects ⃖ ###⃗ KM

10. no; by the Perpendicular Postulate (Post. 3.2), only one line can be perpendicular to ⃖ ##⃗ NP at point P. Because ⃖ ##⃗ PQ is marked as perpendicular to ⃖ ##⃗ NP , ⃖ ##⃗ PR cannot be perpendicular to ⃖ ##⃗ NP .

11. Corresponding angles: ∠1, ∠5; ∠2, ∠6; ∠3, ∠7; ∠4, ∠8

12. Alternate interior angles: ∠3, ∠6; ∠4, ∠5

13. Alternate exterior angles; ∠2, ∠7; ∠1, ∠8

14. Consecutive interior angles: ∠3, ∠5; ∠4, ∠6

15. ∠1 and ∠5 are corresponding angles.

16. ∠11 and ∠3 are consecutive interior angles.

17. ∠6 and ∠13 are consecutive interior angles.

18. ∠2 and ∠11 are alternate interior angles.

19. Lines that do not intersect could be skew lines. Correct statement: If two coplanar lines do not intersect, then they are parallel.

20. There are an infi nite number of lines through a given point that can intersect with the line, but only one is perpendicular. Correct statement: If there is a line and point not on the line, then there is exactly one line through the point that is perpendicular to the given line.

21. a. true; The fl oor is level with the horizontal just like the ground.

b. false; The lines intersect the plane of the ground, so they intersect certain lines of that plane.

c. true; The balusters appear to be vertical, and the fl oor of the treehouse is horizontal. So, they are perpendicular.


Chapter 3

22. no; All three lines could intersect at the same point or two skew lines intersected by a third.

23. yes; If the two lines cut by the transversal are parallel and the transversal is perpendicular to both lines, then all eight angles are right angles.

24. a. The lines parallel to ⃖ ##⃗ NQ are ⃖ ##⃗ MP , ⃖ ##⃗ LS , and ⃖ ##⃗ KR . b. The lines that intersect ⃖ ##⃗ NQ are ⃖ ###⃗ NM , ⃖ ##⃗ NK , ⃖ ##⃗ QP , and ⃖ ##⃗ QR . c. The lines that are skew to ⃖ ##⃗ NQ are ⃖ ##⃗ PS , ⃖ ##⃗ ML , ⃖ ##⃗ KL , and ⃖ ##⃗ RS . d. yes; If lines do not intersect, then they are either parallel

or skew depending on whether they are coplanar.

25. ∠HFB and ∠GJH are corresponding angles to ∠BCG.

26. ∠AJH is a consecutive interior angle to ∠BCG.

27. ∠HJC and ∠DFC are alternate interior angles to ∠FCJ.

28. ∠HJG is an alternate exterior angle to ∠FCA.

29. no; They can both be in a plane that is slanted with respect to the horizontal.

Maintaining Mathematical Profi ciency 30. m∠1 = m∠3 = 76°, because ∠1 and ∠3 are vertical angles

and vertical angles are congruent.m∠2 = 180° − 76° = 104°, because ∠1 and ∠2 are supplementary angles and their sum is 180°.m∠2 = m∠4 = 104°, because ∠2 and ∠4 are vertical angles and vertical angles are congruent.

31. m∠1 = 180° − 159° = 21°, because ∠1 and ∠2 are supplementary angles and their sum is 180°.m∠1 = m∠3 = 21°, because ∠1 and ∠3 are vertical angles and vertical angles are congruent.m∠2 = m∠4 = 159°, because ∠2 and ∠4 are vertical angles and vertical angles are congruent.

3.2 Explorations (p. 131) 1. m∠1 = m∠3 = m∠5 = m∠7, m∠2 = m∠4 = m∠6 = m∠8,

and any odd-numbered angle is supplementary to any even-numbered angle.

2. a. Corresponding angles are congruent when they are formed by two parallel lines and a transversal.

b. Alternate interior angles are congruent when they are formed by two parallel lines and a transversal.

c. Alternate exterior angles are congruent when they are formed by two parallel lines and a transversal.

d. Consecutive interior angles are supplementary when they are formed by two parallel lines and a transversal.

3. When two parallel lines are cut by a transversal, the pairs of angles that are congruent are alternate interior angles, alternate exterior angles, and corresponding angles.

4. m∠2 = 100°, m∠3 = 80°, m∠4 = 100°, m∠5 = 80°, m∠6 = 100°, m∠7 = 80°, m∠8 = 100°

3.2 Monitoring Progress (pp. 133–134) 1. m∠4 = 105° by the Vertical Angles Congruence Theorem

(Thm. 2.6).m∠5 = 105° by the Corresponding Angles Theorem (Thm. 3.1).m∠8 = 105° by the Alternate Exterior Angles Theorem (Thm. 3.3).

2. ∠3 and ∠7 are corresponding angles, so m∠3 = m∠7. ∠7 and ∠8 are supplementary angles.

m∠3 + m∠8 = 180° 68° + (2x + 4)° = 180° 2x + 72 = 180 2x = 108 x = 54

3. yes; The congruence of ∠3 and ∠2 is not dependent on the congruence of ∠1 and ∠3, so the order does not matter.

4. 41°; Because the Sun’s rays are parallel, ∠1 and ∠2 are alternate interior angles. By the Alternate Interior Angles Theorem (Thm. 3.2), ∠1 ≅ ∠2. So, by the defi nition of congruent angles, m∠1 = m∠2 = 41°.


Vocabulary and Core Concept Check 1. Both theorems refer to two pairs of congruent angles that

are formed when two parallel lines are cut by a transversal, and the angles that are congruent are on opposite sides of the transversal. However with the Alternate Interior Angles Theorem (Thm. 3.2), the congruent angles lie between the parallel lines, and with the Alternate Exterior Angles Theorem (Thm. 3.3), the congruent angles lie outside the parallel lines.

2. m∠2 and m∠3 is the pair that does not belong. These are consecutive interior angles, which are supplementary. The other three are pairs of congruent angles.


Chapter 3

Monitoring Progress and Modeling with Mathematics 3. m∠1 = 117° by the Vertical Angles Theorem (Thm. 2.6).

m∠2 = 117° by the Alternate Exterior Angles Theorem (Thm. 3.3).

4. m∠1 = 150° by the Corresponding Angles Theorem (Thm. 3.1).m∠2 = 150° by the Alternate Exterior Angles Theorem (Thm. 3.3).

5. m∠1 = 122° by the Alternate Interior Angles Theorem (Thm. 3.2).122° + m∠2 = 180°

m∠2 = 180° − 122° = 58° by the Consecutive Interior Angles Theorem (Thm. 3.4)

6. m∠1 = 140° by the Alternate Interior Angles Theorem (Thm. 3.2).m∠2 = 40° by the Linear Pair Postulate (Post. 2.8).

7. Alternate interior angles are congruent.

128° = 2x°

128 — 2 = 2x —

2

64 = x

8. Consecutive interior angles are supplementary.

72° + (7x + 24)° = 180° 96 + 7x = 180 96 − 96 + 7x = 180 − 96 7x = 84

7x — 7 = 84 —

7

x = 12

9. m∠5 = 65° because alternate interior angles are congruent.m∠5 + (11x − 17)° = 180° by the defi nition of supplementary angles.

65° + (11x − 17)° = 180° 11x + 48 = 180 11x + 48 − 48 = 180 − 48 11x = 132

11x — 11

= 132 — 11

x = 12

10. m∠4 + 118° = 180° by the defi nition of supplementary angles.So, m∠4 = 180° − 118° = 62°.

Alternate interior angles are congruent.

62 = 8x + 6 56 = 8x

56 — 8 = 8x —

8

x = 7

11. m∠1 = 100°, m∠2 = 80°, m∠3 = 100°; Because the 80° angle is a consecutive interior angle with both ∠1 and ∠3, they are supplementary by the Consecutive Interior Angles Theorem (Thm. 3.4). Because ∠1 and ∠2 are consecutive interior angles, they are supplementary by the Consecutive Interior Angles Theorem (Thm. 3.4).

12. m∠1 = 47°, m∠2 = 133°, m∠3 = 47°; Because ∠1 is consecutive interior angles with the angle that is a vertical angle with the 133° angle, they are supplementary by the Consecutive Interior Angles Theorem (Thm. 3.4). The vertical angle is also 133° by the Vertical Angels Congruence Theorem (Thm. 2.6). Because the 133° angle and ∠2 are alternate interior angles, they are congruent by the Alternate Interior Angles Theorem (Thm. 3.2). Because the 133° angle and ∠3 are consecutive interior angles, they are supplementary by the Consecutive Interior Angles Theorem (Thm. 3.4).

13. In order to use the Corresponding Angles Theorem (Thm. 3.1), the angles need to be formed by two parallel lines cut by a transversal, but none of the lines in this diagram appear to be parallel; ∠9 and ∠10 are corresponding angles.

14. a. When — AD & — BC , with — DB as the transversal, then ∠ADB and ∠CBD are alternate interior angles, and they are congruent (Thm. 3.2). With — AC as the transversal, ∠BCA and ∠DAC are alternate interior angles, and they are congruent (Thm. 3.2).

b. Two pairs of supplementary angles when — AB & — DC are ∠BAD and ∠CDA and ∠ABC and ∠DCB by the Consecutive Interior Angles Theorem (Thm. 3.4).

15. Given p & q, and t as the transversal. Prove Alternate exterior angles are congruent.

1

3

p

t

q2

STATEMENTS REASONS

1. p & q, and t as the transversal.

1. Given

2. ∠1 ≅ ∠3 2. Corresponding Angles Theorem (Thm. 3.1)

3. ∠3 ≅ ∠2 3. Vertical Angles Congruence Theorem (Thm. 2.6)

4. ∠1 ≅ ∠2 4. Transitive Property of Congruence


Chapter 3

16. Given p & q, and t as the transversal. Prove Consecutive interior angles are supplementary.

1

2

p

t

q3

STATEMENTS REASONS

1. p & q, and t as the transversal.

1. Given

2. ∠1 ≅ ∠3 2. Alternate Interior Angles Theorem (Thm. 3.2)

3. m∠1 = m∠3 3. Defi nition of congruent angles

4. m∠2 + m∠3 = 180° 4. Linear Pair Postulate (Post. 2.8)

5. m∠2 + m∠1 = 180° 5. Substitution Property of Equality

6. ∠1 and ∠2 are supplementary angles.

6. Defi nition of supplementary angles

17. Because the trees form parallel lines, and the rope is a transversal, ∠2 and the 76° are consecutive interior angles. So, they are supplementary by the Consecutive Interior Angles Theorem (Thm. 3.4).

76° + m∠2 = 180° m∠2 = 180° − 76° m∠2 = 104° 18. a. ∠1 and ∠2 are alternate interior angles,

so m∠1 = m∠2 = 70°. ∠1 and ∠3 are consecutive interior angles. 70° + m∠3 = 180° m∠3 = 180° − 70° = 110° b. ∠1 and ∠2 are congruent by the Alternate Interior Angles

Theorem (Thm. 3.2). ∠1 and ∠3 are supplementary by the Consecutive Interior Angels Theorem (Thm. 3.4). By substitution, ∠2 and ∠3 are supplementary. So, ∠ABC is a straight angle.

c. yes; m∠2 will be 60° and m∠3 will be 120°. The opening of the box will be more steep because ∠1 is smaller.

19. yes; If the transversal is perpendicular to the parallel lines, the angles formed at the intersection are all right angles (90°).

20. no; It is impossible to have parallel lines in spherical geometry. Because all lines are circles with the same diameter, any two lines will always intersect in two points.

21. 5x° + (14x − 10)° = 180° 2y° = 5x° 19x + 10 = 180 2y = 5(10) 19x = 190 2y = 50

19 — 19 x = 190

— 19 2 — 2 y =

50 — 2

x = 10 y = 25

22. 2y° + (2x + 12)° = 180° 4x° + (y + 6)° = 180° 2y + 2x = 168 4x + y = 174

2 — 2 y + 2 — 2 x =

168 — 2 y = −4x + 174

y + x = 84 y = −x + 84

−x + 84 = −4x + 174 3x + 84 = 174 3x = 90 x = 30 y = −30 + 84 = 54

23. no; In order to make the shot, you must hit the cue ball so that m∠1 = 65°. the angle that is complementary to ∠1 must have a measure of 25° because this angle is alternate interior angles with the angle formed by the path of the cue ball and the vertical line drawn.

24. 60°; ∠1 ≅ ∠5 by the Corresponding Angles Theorem (Thm. 3.1), ∠2 ≅ ∠4 by the Alternate Interior Angles Theorem (Thm. 3.2), ∠2 ≅ ∠3 by the defi nition of angle bisector, and ∠4 ≅ ∠5 is given. So, by the Transitive Property of Congruence, all fi ve of the angles labeled must be congruent to each other. From the diagram, m∠1 + m∠2 + m∠3 = 180°, and because they all have the same measure, it must be that they each have a measure of

180° — 3 = 60°.

Maintaining Mathematical Profi ciency 25. If two angles are congruent, then they are vertical angles.

(false)

26. If you see a tiger, then you went to the zoo. (false)

27. If two angles are supplementary, then they form a linear pair. (false)

28. If we go to the park, then it is warm outside. (false)


Chapter 3

3.3 Explorations (p. 137) 1. a. If two lines are cut by a transversal so that the

corresponding angles are congruent, then the lines are parallel. The converse is true.

b. If two lines are cut by a transversal so that the alternate interior angles are congruent, then the lines are parallel. The converse is true.

c. If two lines are cut by a transversal so that the alternate exterior angles are congruent, then the lines are parallel. The converse is true.

d. If two lines are cut by a transversal so that the consecutive interior angles are supplementary, then the lines are parallel. The converse is true.

2. The converse is true for all four of the theorems involving parallel lines and transversals.

3. If you assume the converse of the Corresponding Angles Theorem (Thm. 3.1), then you can use it to prove the converse of the other three theorems.

3.3 Monitoring Progress (pp. 138–141) 1. yes; The angle that is corresponding with the 75° angle also

forms a linear pair with the 105° angle. So, it must be 180° − 105° = 75° by the Linear Pair Postulate (Post. 2.8).Because the corresponding angles have the same measure, they are congruent by defi nition. So, m & n by the Corresponding Angles Converse (Thm. 3.5).

2. The hypothesis and conclusion of the Corresponding Angles Converse (Thm. 3.5) are the reverse of the Corresponding Angles Theorem (Thm. 3.1).

3. Given ∠1 ≅ ∠8 Prove j & k

4. It is given that ∠4 ≅ ∠5. By the Vertical Angles Congruence Theorem (Thm. 2.6), ∠1 ≅ ∠4. Then by the Transitive Property of Congruence (Thm. 2.2), ∠1 ≅ ∠5. So, by the Corresponding Angles Converse (Thm. 3.5), g & h.

5. Using the Transitive Property of Parallel Lines (Thm. 3.9) over and over again, you can show that the ground is parallel to the step above it and the one above that, and so on, until you have stated that the line formed by the ground is parallel to the line formed by the top step.

6. m∠8 = 65°; By the Transitive Property of Parallel Lines (Thm. 3.9), p & r. By the Corresponding Angles Theorem (Thm. 3.1), the angle at the intersection of line r and line s that is corresponding with the 115° angle would also have a measure of 115°. This angle also forms a linear pair with ∠8. So, by the Linear Pair Postulate (Post. 2.8), m∠8 = 180° − 115° = 65°.


Vocabulary and Core Concept Check 1. corresponding angles, alternate interior angles, and alternate

exterior angles

2. Two lines cut by a transversal are parallel if and only if the corresponding angles are congruent.

Two lines cut by a transversal are parallel if and only if the alternate interior angles are congruent.

Two lines cut by a transversal are parallel if and only if the alternate exterior angles are congruent.

Two lines cut by a transversal are parallel if and only if the consecutive interior angles are supplementary.

Monitoring Progress and Modeling with Mathematics 3. Lines m and n are parallel when the marked corresponding

angles are congruent.

120° = 3x°

120 — 3 = 3x —

3

x = 40

4. Lines m and n are parallel when the marked corresponding angles are congruent.

135° = (2x + 15)° 120 = 2x

120 — 2 = 2x —

2

x = 60

5. Lines m and n are parallel when the marked consecutive interior angles are supplementary.

180° = 150° + (3x − 15)° 180 = 135 + 3x

45 = 3x

45 — 3 = 3x —

3

x = 15

6. Lines m and n are parallel when the marked alternate exterior angles are congruent.

x° = (180 − x)° 2x = 180

2x — 2 = 180 —

2

x = 90


Chapter 3

7. Lines m and n are parallel when the marked consecutive interior angles are supplementary.

x° + 2x° = 180° 3x = 180

3x — 3 = 180 —

3

x = 60

8. Lines m and n are parallel when the marked alternate interior angles are congruent.

3x° = (2x + 20)° x = 20

9. Let A and B be two points on line m. Draw ⃖ ##⃗ AP and construct an angle ∠1 on n at P so that ∠PAB and ∠1 are corresponding angles.

P

A B

1

m

n

10. Let A and B be two points on line m. Draw ⃖ ##⃗ AP and construct an angle ∠1 on n at P so that ∠PAB and ∠1 are corresponding angles.

PA

B

n

m1

11. Given ∠1 ≅ ∠8

k

j

1

2

8

Prove j & k

STATEMENTS REASONS

1. ∠1 ≅ ∠8 1. Given



4. j & k 4. Corresponding Angles Theorem (Thm. 3.1)

12. Given ∠3 and ∠5 are supplementary.

k

j3 25

Prove j & k

STATEMENTS REASONS

1. ∠3 and ∠5 are supplementary.

1. Given


2. Linear Pair Postulate (Post. 2.8)

3. m∠3 + m∠5 = 180°, m∠2 + m∠3 = 180°


4. m∠3 + m∠5 = m∠2 + m∠3

4. Transitive Property of Equality

5. m∠2 = m∠5 5. Subtraction Property of Equality

6. ∠2 ≅ ∠5 6. Defi nition of congruent angles

7. j & k 7. Corresponding Angles Converse (Thm. 3.5)

13. yes; Alternate Interior Angles Converse (Thm. 3.6)

14. yes; Alternate Exterior Angles Converse (Thm. 3.7)

15. no

16. yes; Corresponding Angles Converse (Thm. 3.5)

17. no

18. yes; Alternate Exterior Angles Converse (Thm. 3.7)

19. This diagram shows that vertical angles are always congruent. Lines a and b are not parallel unless x = y, and you cannot assume that they are equal.

20. It would be true that a & b if you knew that ∠1 and ∠2 were supplementary, but you cannot assume that they are supplementary unless it is stated or the diagram is marked as such. You can say that ∠1 and ∠2 are consecutive interior angles.

21. yes; m∠DEB = 180° − 123° = 57° by the Linear Pair Postulate (Post. 2.8). So, by defi nition, a pair of corresponding angles are congruent, which means that

⃖ ##⃗ AC & ⃖ ##⃗ DF by the Corresponding Angles Converse (Thm. 3.5).

22. yes; m∠BEF = 180° − 37° = 143° by the Linear Pair Postulate (Post. 2.8). So, by defi nition, a pair of corresponding angles are congruent, which means that

⃖ ##⃗ AC & ⃖ ##⃗ DF by the Corresponding Angles Converse (Thm. 3.5).

23. no; The marked angles are vertical angles. You do not know anything about the angles formed by the intersection of ⃖ ##⃗ DF and ⃖ ##⃗ BE .


Chapter 3

24. yes; m∠EBC = 115° by the Vertical Angles Congruence Theorem (Thm. 2.6). Because m∠EBC + m∠FEB = 115° + 65° = 180°, ∠EBC and ∠FEB are supplementary by defi nition, which means that ⃖ ##⃗ AC & ⃖ ##⃗ DF by the Consecutive Interior Angles Converse (Thm. 3.8).

25. Yes, all streets are parallel to each other. E 20th Ave. is parallel to E 19th Ave. by the Corresponding

Angles Converse (Thm. 3.5). E 19th Ave. is parallel to E. 18th Ave. by the Alternate Exterior Angles Converse (Thm. 3.7). E 18th Ave. is parallel to E 17th Ave. by the Alternate Interior Angles Converse (Thm. 3.6). So, they are all parallel to each other by the Transitive Property of Parallel Lines (Thm. 3.9).

26. Because each rung of the ladder is parallel to the one directly above, using the Transitive Property of Parallel Lines (Thm. 3.9), the top rung is parallel to the bottom rung.

27. The two angles marked as 108° are corresponding angles. Because they have the same measure, they are congruent to each other. So, m & n by the Corresponding Angles Converse (Thm. 3.5).

28. ###⃗ EA & ###⃗ HC by the Corresponding Angles Converse (Thm. 3.5). ∠AEH ≅ ∠CHG by defi nition because m∠AEH = 62° + 58° = 120° and m∠CHG = 59° + 61° = 120°. However, ###⃗ EB is not parallel to ###⃗ HD because corresponding angles ∠BEH and ∠DHG do not have the same measure and are therefore not congruent.

29. A, B, C, D; The Corresponding Angles Converse (Thm. 3.5) can be used because the angle marked at the intersection of line m and the transversal is vertical angles with, and therefore congruent to, an angle that is corresponding with the other marked angle. The Alternate Interior Angles Converse (Thm. 3.6) can be used because the angles that are marked as congruent are alternate interior angles. The Alternate Exterior Angles Converse (Thm. 3.7) can be used because the angles that are vertical with, and therefore congruent to, the marked angles are alternate exterior angles. The Consecutive Interior Angles Converse (Thm. 3.8) can be used because each of the marked angles forms a linear pair with, and is therefore supplementary to, an angle that is a consecutive interior angles with the other marked angle.

30. m∠1 = 32°; The 32° angle that is marked is corresponding with ∠2. So m∠2 = 32° by the Corresponding Angles Theorem (Thm. 3.1). Considering the line formed by the top of the step and the line formed by the fl oor, ∠1 and ∠2 are alternate interior angles. So, if ∠1 ≅ ∠2, then the top of the step will be parallel to the fl oor by the Alternate Interior angles Converse (Thm. 3.6).

31. Two angles must be given. Sample answer:

∠2 ≅ ∠7 or ∠4 ≅ ∠5 by the Alternate Interior Angles Converse (Thm. 3.6)

∠1 ≅ ∠8 or ∠3 ≅ ∠6 by the Alternate Exterior Angles Converse (Thm. 3.7)

∠2 ≅ ∠5 or ∠4 ≅ ∠7 by the Consecutive Interior Angle Converse (Thm. 3.8)

∠1 ≅ ∠5, ∠2 ≅ ∠6, ∠3 ≅ ∠7, ∠4 ≅ ∠8 by the Corresponding Angles Converse (Thm. 3.5)

32. Sample answer:

1

3

r

t

s

2

In this diagram, angles from only one intersection are marked as being congruent. In order to prove that the two lines are parallel, you need to know something about at least one angle formed by each of the intersections that the transversal makes with the two other lines. For instance, if you knew something about the measure of ∠3, you would be able to determine whether line r is parallel to line s.

33. Given m∠1 = 115°, m∠2 = 65°

m12

n

Prove m & n

STATEMENTS REASONS

1. m∠1 = 115°, m∠2 = 65° 1. Given

2. m∠1 + m∠2 = m∠1 + m∠2

2. Refl exive Property of Equality

3. m∠1 + m∠2 = 115° + 65° 3. Substitution Property of Equality

4. m∠1 + m∠2 = 180° 4. Simplify.5. ∠1 and ∠2

are supplementary.5. Defi nition of

supplementary angles

6. m & n 6. Consecutive Interior Angles Converse (Thm. 3.8)


Chapter 3

34. Given ∠1 and ∠3 are supplementary.

m1

23

n

Prove m & n

STATEMENTS REASONS


1. Given

2. m∠1 and m∠3 = 180°




5. m∠2 + m∠3= 180°

5. Substitution Property of Equality



7. m & n 7. Consecutive Interior Angles Converse (Thm. 3.8)

35. Given ∠1 ≅ ∠2, ∠3 ≅ ∠4 A

B

C

DE

12

3

4

Prove — AB & — CD

STATEMENTS REASONS

1. ∠1 ≅ ∠2, ∠3 ≅ ∠4 1. Given




5. — AB & — CD 5. Alternate Interior Angles Converse (Thm. 3.6)

36. Given a & b, ∠2 ≅ ∠3

1 2a

c d

b3 4

Prove c & d

STATEMENTS REASONS

1. a & b, ∠2 ≅ ∠3 1. Given

2. ∠3 ≅ ∠1 2. Alternate Interior Angles Theorem (Thm. 3.2)


4. c & d 4. Corresponding Angles Converse (Thm. 3.5)

37. no; Based on the diagram, — AB & — DC by the Alternate Interior Angles Converse (Thm. 3.6), but you cannot be sure that

⃖ ##⃗ AD & ⃖ ##⃗ BC .

38. no; In order to conclude that r & s, you would need to show that ∠1 ≅ ∠3. In order to conclude that p & q, you would need to show that either ∠1 ≅ ∠2 or ∠3 ≅ ∠4.

39. a. p q r b. Given p & q, q & r Prove p & r

c. p

1 2 3

q r

STATEMENTS REASONS

1. p & q, q & r 1. Given

2. ∠1 ≅ ∠2, ∠2 ≅ ∠3

2. Corresponding Angles Theorem (Thm. 3.1)


4. p & r 4. Corresponding Angles Converse (Thm. 3.5)

40. a. Use the Corresponding Angles Converse (Thm. 3.5).

(2x + 2)° = (x + 56) ° x + 2 = 56 x = 54 If x = 54, p & q. b. Use the Vertical Angles Congruence Theorem (Thm. 2.6)

and the Consecutive Interior Angles Converse (Thm. 3.8).

(y + 7)° + (3y − 17) ° = 180° 4y − 10 = 180 4y = 190 4 — 4 y =

190 — 4

y = 47.5 If y = 47.5, r & s. c. no; If x = 54, then (x + 56)° = 110°. If y = 47.5, then

(y + 7)° = 54.5°. Because these two angles form a linear pair, their sum should be 180°, but 110° + 54.5° = 164.5°. So, both pairs of lines cannot be parallel at the same time.


Chapter 3

Maintaining Mathematical Profi ciency 41. d = √

—— (−2 − 1)2 + (9 − 3)2

= √——

(−3)2 + (6)2 = √

— 9 + 36 = √

— 45 ≈ 6.71

42. d = √———

(8 − (−3))2 + (−6 − 7)2 = √

—— (8 + 3)2 + (−13)2

= √——

(11)2 + (−13)2 = √

— 121 + 169 = √

— 290 ≈ 17.03

43. d = √——

(0 − 5)2 + (8 − (−4))2 = √

—— (−5)2 + (8 + 4)2

= √——

(−5)2 + (12)2 = √

— 25 + 144 = √

— 169 = 13

44. d = √——

(9 − 13)2 + (−4 − 1)2 = √

—— (−4)2 + (−5)2

= √—

16 + 25 = √—

41 ≈ 6.40

3.1–3.3 What Did You Learn? (p. 145) 1.

A B

C

J

H

G

2. For part (a), I started by writing the equation (2x + 2)° = (x + 56)°, because the angles represented by these two expressions are corresponding angles with respect to lines p and q. So, in order for lines p and q to be parallel by the Corresponding Angles Theorem (Thm. 3.1), the expressions must be equal to each other. For part (b), I started by writing the equation, (y + 7)° + (3y − 17) ° = 180°. In order for lines r and s to be parallel, the angles represented by these two expressions must be supplementary because each one forms a linear pair with one of the consecutive interior angles formed by lines r and s and transversal q.

3.1–3.3 Quiz (p. 146) 1. ⃖ ##⃗ GH is parallel to ⃖ ##⃗ EF . 2. ⃖ ##⃗ FG is perpendicular to ⃖ ##⃗ EF .

3. ⃖ ##⃗ GC is skew to ⃖ ##⃗ EF .

4. Plane GCB is parallel to plane ADF. Any three of the four points G, C, B, and F can be used to form the parallel plane.

5. ∠3 and ∠5, and ∠4 and ∠6 are consecutive interior angles.

6. ∠3 and ∠6, and ∠4 and ∠5 are alternate interior angles.

7. ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, and ∠4 and ∠8 are corresponding angles.

8. ∠1 and ∠8, and ∠2 and ∠7 are alternate exterior angles.

9. By the Linear Pair Postulate (Post. 2.3):

m∠1 + 138° = 180° m∠1 = 42° By the Alternate Exterior Angles Theorem (Thm. 3.3): m∠1 = m∠2 = 42° 10. m∠1 = 123° by the Corresponding Angles Theorem

(Thm. 3.1).

m∠2 = 123° by the Vertical Angle Congruence Theorem (Thm. 2.6).

11. By the Linear Pair Postulate (Post. 2.8):

m∠1 + 57° = 180° m∠1 = 123° m∠2 = 57° by the Alternate Interior Angles Theorem

(Thm. 3.2).

12. yes; Consecutive Interior Angles Converse (Thm. 3.8) (69° + 111° = 180°)

13. no

14. yes; Transitive Property of Parallel Lines (Thm. 3.9)

15. a. All of the bars are parallel to each other by the Transitive Property of Parallel Lines (Thm. 3.9).

b. ∠1 corresponds to ∠2 by the Corresponding Angles Theorem (Thm. 3.1). So, m∠1 = m∠2 = 58°.

16. a. Sample answer: q & p and m & k b. Sample answer: n ⊥ m and n ⊥ k c. Sample answer: Lines k and q are skew, and linesℓand m

are skew.

d. Because m & k, ∠1 ≅ ∠2 by the Alternate Exterior Angles Theorem (Thm. 3.3).

3.4 Explorations (p. 147) 1. a. — AB ⊥ — CD ; — AB is parallel to the horizontal edge of the

paper because points A, O, and B are all the same distance from the edge. Similarly, — CD is parallel to the vertical edge of the paper because points C, O, and D are the same distance from the edge. The horizontal and vertical edges form right angles in the corners. So, lines parallel to them will also be perpendicular.

b. — AO ≅ — OB ; Point O must be the midpoint of — AB because the paper was folded in half. So, — AO and — OB are congruent by defi nition of midpoints.

2. a. Check students’ work.

b. They are all right angles.


Chapter 3

3. a. Check students’ work.

b. Check students’ work.

c. Check students’ work; — CD is perpendicular to — AB , and point O is the midpoint of — AB . Point C is the same distance from A as it is from B, and D is the same distance from A as it is from B. So, the segment connecting C and D contains all the points that are equidistant from points A and B.

4. Sample answer: If you have a segment, and you fold it in half so that both halves match, the fold will be perpendicular to the segment. When lines are perpendicular, all four angles are right angles. If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular. In a plane, if a transversal is perpendicular to one of the two parallel lines, then it is perpendicular to the other line. Finally, in a plane, if two lines are perpendicular to the same line, then they are parallel to each other.

5. If AB = 4 units, AO = 2 units and OB = 2 units.

3.4 Monitoring Progress (pp. 148–151) 1. EG = √

——— (1 − (−4))2 + (2 − (−3))2

= √——

(1 + 4)2 + (2 + 3)2 = √

— (5)2 + (5)2

= √—

(25 + 25) = √—

50 ≈ 7.1 The distance from E to ⃖ ##⃗ EF is about 7.1 units.

2. Given h & k, j ⊥ h h

j

k

1 2

43

5 687

Prove j ⊥ k

STATEMENTS REASONS

1. h & k, j ⊥ h 1. Given

2. m∠2 = 90° 2. Defi nition of perpendicular lines

3. ∠2 ≅ ∠7 3. Alternate Exterior Angles Theorem (Thm. 3.3)


5. m∠7 = 90° 5. Transitive Property of Equality

6. j ⊥ k 6. Defi nition of perpendicular lines

3. yes; Because a ⊥ d and b ⊥ d, you can conclude that b & a by the Lines Perpendicular to a Transversal Theorem (Thm. 3.12).

4. yes; Because b ⊥ d and c & d, you can conclude that b ⊥ c by the Perpendicular Transversal Theorem (Thm. 3.11).


Vocabulary and Core Concept Check 1. The perpendicular bisector of a segment is the line that

passes through the midpoint of the segment at a right angle.

2. “Find XZ” is different, which asks for the distance between two points. The other three ask for the length of a perpendicular segment from the point X to lineℓ. — XZ is not a perpendicular segment.

XZ = √——

(−3 − 4)2 + (3 − 4)2 = √

—— (− 7)2 + (−1)2

= √—

50 ≈ 7.1 units

XY = √——

(−3 − 3)2 + (3 − 1)2 = √

—— (− 6)2 + (2)2

= √—

40 ≈ 6.3 units

Monitoring Progress and Modeling with Mathematics 3. AY = √

—— (3 − 0)2 + (0 − 1)2

= √——

(3)2 + (−1)2 = √

— 9 + 1 = √

— 10 ≈ 3.16

The distance from point A to ⃖ ##⃗ XZ is about 3.16 units.

4. AZ = √——

(3 − 4)2 + [3 − (−1)]2 = √

—— (−1)2 + (3 + 1)2

= √—

12 + 42 = √—

1 + 16 = √—

17 ≈ 4.12 The distance from point A to ⃖ ##⃗ XZ is about 4.12 units.

5. Using P as the center, draw two arcs intersecting with line m. Label the intersections as points X and Y. Using X and Y as centers and an appropriate radius, draw arcs that intersect. Label the intersection as Z. Draw ⃖ ##⃗ PZ .

P

m

X Y

Z

6. Using P as the center, draw two arcs intersecting with line m. Label the intersections as points X and Y. Using X and Y as centers and an appropriate radius, draw arcs that intersect. Label the intersection as Z. Draw ⃖ ##⃗ PZ .

m

XY

Z

P


Chapter 3

7. Using P as the center and any radius, draw arcs intersecting m and label those intersections as X and Y. Using X as the center, open the compass so that it is greater than half of XP and draw an arc. Using Y as the center and retaining the same compass setting, draw an arc that intersects with the fi rst. Label the point of intersection as Z. Draw ⃖ ##⃗ PZ .

m

X

Y

Z

P

8. Using P as the center and any radius, draw arcs intersecting m and label those intersections as X and Y. Using X as the center, open the compass so that it is greater than half of XP and draw an arc. Using Y as the center and retaining the same compass setting, draw an arc that intersects with the fi rst. Label the point of intersection as Z. Draw ⃖ ##⃗ PZ .

P

mX Y

Z

9. Using a compass setting greater than half of AB, draw two arcs using A and B as the centers. Connect the points of intersections of the arcs with a straight line.

A

B

10. Using a compass setting greater than half of AB, draw two arcs using A and B as the centers. Connect the points of intersections of the arcs with a straight line.

A

B

11. In order to claim parallel lines by the Lines Perpendicular to a Transversal Theorem (Thm. 3.12), both lines must be marked as perpendicular to the transversal. The correct statement is: Lines x and z are perpendicular.

12. The distance form point C to ⃖ ##⃗ AB is not 12 centimeters, it is 8 centimeters. The segment from C to — AB must be perpendicular to — AB in order to equal the distance from C to — AB .

13. Given ∠1 ≅ ∠2

h

g

1 2

Prove g ⊥ h

STATEMENTS REASONS

1. ∠1 ≅ ∠2 1. Given

2. m∠1 = m∠2 2. Defi nition of congruence

3. m∠1 + m∠2 = 180° 3. Linear Pair Postulate (Post. 2.8)

4. m∠2 + m∠2 = 180° 4. Substitution Property of Equality

5. 2(m∠2) = 180° 5. Distributive Property

6. m∠2 = 90° 6. Division Property of Equality


8. g ⊥ h 8. Defi nition of perpendicular lines

14. Given m ⊥ p and n ⊥ p

p

m

1

n

2

Prove m & n

STATEMENTS REASONS

1. m ⊥ p 1. Given2. ∠1 is a right angle. 2. Defi nition of perpendicular

lines

3. n ⊥ p 3. Given

4. ∠2 is a right angle. 4. Defi nition of perpendicular lines

5. ∠1 ≅ ∠2 5. Right Angles Congruence Theorem (Thm. 2.3)

6. m & n 6. Corresponding Angles Converse (Thm. 3.5)


Chapter 3

15. Given a ⊥ b

a

b

1 243

Prove ∠1, ∠2, ∠3, and ∠4 are right angles.

STATEMENTS REASONS

1. a ⊥ b 1. Given 2. ∠1 is a right angle. 2. Defi nition of perpendicular

lines

3. m∠1 = 90° 3. Defi nition of right angle

4. ∠1 ≅ ∠4 4. Vertical Angle Congruence Theorem (Thm. 2.6)


6. ∠1 and ∠2 are a linear pair.

6. Defi nition of linear pair


7. Linear Pair Postulate (Post. 2.8)

8. m∠1 + m∠2= 180°


9. 90° + m∠2 = 180° 9. Substitution Property of Equality

10. m∠2 = 90° 10. Subtraction Property of Equality

11. ∠2 ≅ ∠3 11. Vertical Angle Congruence Theorem (Thm. 2.6)


13. ∠1, ∠2, ∠3, and ∠4 are right angles.

13. Defi nition of right angle

16. Given ###⃗ BA ⊥ ###⃗ BC A

B

12

C

Prove ∠1 and ∠2 are complementary.

STATEMENTS REASONS

1. ###⃗ BA ⊥ ###⃗ BC 1. Given2. ∠ABC is a right angle. 2. Defi nition of

perpendicular lines

3. m∠ABC = 90° 3. Defi nition of right angle

4. m∠ABC = m∠1 + m∠2 4. Angle Addition Postulate (Post. 1.4)

5. 90° = m∠1 + m∠2 5. Transitive Property of Equality

6. ∠1 and ∠2 are complementary.

6. Defi nition of complementary angles

17. none; The only thing that can be concluded in this diagram is that v ⊥ y. In order to say that lines are parallel, you need to know something about both of the intersections between the transversal and the two lines.

18. b & c; Because a ⊥ b and a ⊥ c, lines b and c are parallel by the Lines Perpendicular to a Transversal Theorem (Thm. 3.12).

19. m & n; Because m ⊥ q and n ⊥ q, lines m and n are parallel by the Lines Perpendicular to a Transversal Theorem (Thm. 3.12). The other lines may or may not be parallel.

20. none; The only things that can be concluded in this diagram are that a ⊥ d and b ⊥ c. In order to say that lines are parallel, you need to know something about both of the intersections between the transversal and the two lines.

21. n & p; Because k ⊥ n and k ⊥ p, lines n and p are parallel by the Lines Perpendicular to a Transversal Theorem (Thm. 3.12).

22. y & z and w & x; Because w ⊥ y and w ⊥ z, lines y and z are parallel by the Lines Perpendicular to a Transversal Theorem (Thm 3.12). Because w ⊥ z and x ⊥ z, lines w and x are parallel by the Lines Perpendicular to a Transversal Theorem (Thm. 3.12).

23. m∠1 = 90°, m∠2 = 60°, m∠3 = 30°, m∠4 = 20°, m∠5 = 90°; m∠1 = 90°, because it is marked as a right angle.

m∠2 = 90° − 30° = 60°, because it is complementary to the 30° angle.

m∠3 = 30°, because it is vertical angles with, and therefore congruent to, the 30° angle.

m∠4 = 90° − (30° + 40°) = 20°, because it is forms a right angle together with ∠3 and the 40° angle.

m∠5 = 90°, because it is vertical angles with, and therefore congruent to, ∠1.

24. no; The shortest distance from a point on one line to the other line will be different for different points on the line unless the lines are parallel.

25. For a ⊥ b: For b & c: (9x + 18)° = [5(x + 7) + 15]° 9x + 18 = 5x + 35 + 15 9x + 18 = 5x + 50 9x + 18 − 18 = 5x + 50 − 18 9x = 5x + 32 9x − 5x = 5x − 5x + 32 4x = 32 x = 8

(9x + 18)° = 90° 9x + 18 − 18 = 90 − 18 9x = 72 x = 8

So, x = 8 when a ⊥ b and b & c.


Chapter 3

26. point C; Because — AC appears to be perpendicular to the water’s edge, it would represent the shortest distance from point A to the line formed by the opposite edge of the stream.

27. In A, C, D, and E, — AC & — BD and — AC ⊥ — BD . 28. There are eight right angles. Because two lines always

intersect in two points, and each intersection creates four right angles, there will be eight right angles formed by two perpendicular lines.

29.

A B

30. The line segments that are perpendicular to the crosswalk require less paint, because they represent the shortest distance from one side of the crosswalk to the other.

31. Because a ⊥ c, b ⊥ c, and d & c, then by the Perpendicular Transversal Theorem (Thm. 3.11), a ⊥ d and b ⊥ d. There are four right angles and opposite sides are equal. So, the shape is a rectangle.

32. Using x = 0, the y-intercept of y = 3 — 2 x + 4 is 4. Find the equation of the line perpendicular to −3x + 2y = −1 through the point (0, 4) on y = 3 — 2 x + 4.

Rewrite the equation in the form of y = mx + b. −3x + 2y = −1 2y = 3x − 1

y = 3 — 2 x − 1 —

2

The slope is 3 — 2 , so, the slope of the perpendicular line is − 2 — 3 .

y = − 2 — 3 x + b

4 = − 2 — 3 ⋅ 0 + b

4 = b The equation of the line is y = − 2 — 3 x + 4.

Find the intersection of the lines y = 3 — 2 x − 1 — 2 and

y = − 2 — 3 x + 4.

3 — 2 x − 1 —

2 = − 2 —

3 x + 4

6 ⋅ 3 — 2 x − 1 — 2 ⋅ 6 = − 2 —

3 x ⋅ 6 + 4 ⋅ 6

9x − 3 = −4x + 24

13x = 27

x = 27 — 13

y = − 2 — 3 x + 4

y = − 2 — 3 ⋅ 27 — 13 + 4 ⋅ 13 — 13

y = −18 — 13

+ 52 — 13

y = 34 — 13

The point of intersection is ( 27 — 13 , 34 — 13 ) .

distance = √——

( 27 — 13 − 0 ) 2 + ( 34 — 13 − 4 )

2

= √—— ( 27 — 13 ) 2 + ( 34 — 13 − 4 ⋅ 13 — 13 ) 2 = √—— ( 27 — 13 ) 2 + ( 34 — 13 − 52 — 13 ) 2 = √—— ( 729 — 169 ) + ( −18 — 13 ) 2 = √

——

( 729 — 169 ) + ( 324 — 169 ) = √—

1053 — 169

≈ 2.5 units


Chapter 3

33. Find the length of the segment that is perpendicular to the plane and that has one endpoint on the given point and one endpoint on the plane; You can fi nd the distance from a line to a plane only if the line is parallel to the plane. Then you can pick any point on the line and fi nd the distance from that point to the plane. If a line is not parallel to a plane, then the distance from the line to the plane is not defi ned because it would be different for each point on the line.

Maintaining Mathematical Profi ciency

34. 6 + 4 — 8 − 3

= 10 — 5 = 2

35. 3 − 5 — 4 − 1

= − 2 — 3

36. 8 + 3 — 7 + 2

= 11 — 9

37. 13 − 4 — 2 + 1

= 9 — 3 = 3

38. slope = 3; y-intercept = 9

39. slope = − 1 — 2 ; y-intercept = 7

40. slope = 1 — 6 ; y-intercept = −8

41. slope = −8; y-intercept = −6

3.5 Explorations (p. 155) 1. a. y = 3 — 2 x + b 2 = 3 — 2 ⋅ 0 + b 2 = b The line y = 3 — 2 x + 2 is parallel to the line y =

3 — 2 x − 1.

The slopes are equal.

6

−4

−6

4

y = x + 232

y = x − 132

b. m = 3 — 2 , so the perpendicular line will have a slope of m = − 2 — 3 .

y = − 2 — 3 x + b 1 = 3 — 2 ⋅ 0 + b 1 = b The line y = − 2 — 3 x + 1 is perpendicular to the line

y = 3 — 2 x − 1. The slopes are opposite reciprocals and have a product of −1.

6

−4

−6

4y = − x + 123

y = x − 132

c. y = 1 — 2 x + b

−2 = 1 — 2 ⋅ (2) + b −2 = 1 + b −3 = b The line y = 1 — 2 x − 3 is parallel to the line y =

1 — 2 x + 2.


9

−6

−9

6

y = x + 212

y = x − 312

d. m = −2, so the perpendicular line will have a slope of m = 1 — 2 .

y = −2x + b

−3 = −2 ⋅ (2) + b −3 = −4 + b 1 = b The line y = −2x + 1 is perpendicular to the line

y = 1 — 2 x + 2. The slopes are opposite reciprocals and have a product of −1.

9

−6

−9

6

y = x + 212

y = −2x + 1

e. y = −2x + b

−2 = −2 ⋅ (0) + b −2 = b The line y = −2x − 2 is parallel to the line y = −2x + 2.


6

−4

−6

4

y = −2x + 2

y = −2x − 2


Chapter 3

f. m = −2, so the perpendicular line will have a slope of m = 1 — 2 .

y = 1 — 2 x + b

0 = 1 — 2 ⋅ (4) + b 0 = 2 + b −2 = b The line y = 1 — 2 x − 2 is perpendicular to the line

y = 2x + 2. The slopes are opposite reciprocals and have a product of −1.

6

−4

−6

4

y = −2x + 2

y = x − 212

2. a. For the blue line, use the points (0, 3) and ( − 3 — 2 , 0 ) to write the equation.

slope = 0 − 3 — − 3 — 2 − 0

= −3 — − 3 — 2

= −3 ⋅ ( − 2 — 3 ) = 2 The y-intercept is 3.

y = mx + b y = 2x + 3 For the red line, use the points (1, 0) and (0, −2) to write

the equation.

slope = −2 − 0 — 0 − 1

= −2 — −1 = 2

The y-intercept is −2. y = mx + b

6

−4

−6

4

y = 2x − 2

y = 2x + 3 y = 2x − 2

b. For the blue line, use the points (0, −3) and ( − 3 — 2 , 0 ) to write the equation.

slope = 0 − (−3)

— − 3 — 2 − 0

= 3 — − 3 — 2

= 3 ⋅ ( − 2 — 3 ) = −2 The y-intercept is −3. y = mx + b y = −2x − 3 For the red line, use the points (−4, 0) and (0, 2) to write

the equation.

slope = 2 − 0 — 0 − (−4)

= 2 — 4 = 1 —

2

The y-intercept is 2.

y = mx + b

4

−4

−8

4

y = −2x − 3

y = x + 212

y = 1 — 2 x + 2

3. For a line parallel to a given line, the slopes will be the same. For a line perpendicular to the given line, the slopes will be opposite reciprocals. Find the y-intercept of a line by substituting the slope and the given point into the slope-intercept form of a line, y = mx + b, and solving for b. Once you know the slope and y-intercept of a line, you can get the equation of the line by substituting them into y = mx + b.

4. The slope is 3.

y = 3x + b −2 = 3 ⋅ (1) + b −2 = 3 + b −5 = b The line y = 3x − 5 is parallel to the line y = 3x + 2

through the point (1, −2). The slope is 3. The slope of the perpendicular line is − 1 — 3 . y = − 1 — 3 x + b

−2 = − 1 — 3 ⋅ (1) + b −2 = − 1 — 3 + b −6 = −1 + 3b −5 = −3b 5 — 3 = b

The line y = − 1 — 3 x − 5 — 3 is perpendicular to the line y = 3x + 2

through the point (1, −2).

3.5 Monitoring Progress (pp. 156–159) 1. In order to divide the segment in the ratio 4 to 1, partition the

segment into 4 + 1, or 5 congruent parts. P is 4 — 5 of the way from point A to point B.

slope of — AB = 4 − 3 — 8 − 1

= 1 — 7

To fi nd the coordinates of P, add 4 — 5 (or 0.8) of the run to the x-coordinate and add 4 — 5 (or 0.8) of the rise to the y-coordinate.

run = 7 ⋅ 0.8 = 5.6 x = 1 + 5.6 = 6.6 rise = 1 ⋅ 0.8 = 0.8 y = 3 + 0.8 = 3.8 So, the point is P(6.6, 3.8).

2. In order to divide the segment in the ratio 3 to 7, partition the segment into 3 + 7, or 10 congruent parts. P is 3 — 10 of the way from point A to point B.

slope of — AB = 5 − 1 — 4 + 2

= 4 — 6


run = 6 ⋅ 0.3 = 1.8 x = −2 + 1.8 = −0.2 rise = 4 ⋅ 0.3 = 1.2 y = 1 + 1.2 = 2.2 So, the point is P(−0.2, 2.2).


Chapter 3

3. The slope of line a is 4 — 2 = 2.

The slope of line b is 4 — 2 = 2.

The slope of line c is 3 — 1 = 3.

The slope of line d is − 1 − 0 — 0 + 3

= − 1 — 3 .

Lines a and b have equal slopes. Therefore, a & b. The product of the slopes of the lines c and d is −1.

So, c ⊥ d. 4. a. slope of the parallel line = 3 y = 3x + b 5 = 3(1) + b 5 = 3 + b 2 = b Because m = 3 and b = 2, an equation of the line is

y = 3x + 2. b. slope of perpendicular line: 3 ⋅ m = −1 m = − 1 — 3 y = − 1 — 3 x + b 5 = − 1 — 3 (1) + b 5 = − 1 — 3 + b 16 — 3 = b Because m = − 1 — 3 and b =

16 — 3 , an equation of the line is

y = − 1 — 3 x + 16

— 3 .

5. The graph of x = 4 is a vertical line and the graph of y = 2 is a horizontal line. So, they are perpendicular by Theorem 3.14.

6. The slope of y = x + 4 is 1, so the line perpendicular to y = x + 4 will have a slope of −1.

y = −x + b 4 = −6 + b 10 = b The line perpendicular to y = x + 4 is y = −x + 10. Find the point of intersection.

x + 4 = −x + 10 Equation 1 2x + 4 = 10 Equation 2

x + 4 = −x + 10 2x + 4 = 10 2x = 6 2 — 2 x =

6 — 2

x = 3

y = 3 + 4 = 7 So, the perpendicular lines intersect at (3, 7).

Find the distance from (6, 4) to (3, 7).


(3 − 6)2 + (7 − 4)2 = √——

(−3)2 + (3)2 = √

— 9 + 9 = √

— 18 ≈ 4.24 units

7. The slope of y = −2x is −2, so the line perpendicular to y = −2x will have a slope of 1 — 2 .

y = 1 — 2 x + b

6 = 1 — 2 ⋅ (−1) + b

6 = − 1 — 2 + b

12 = −1 + 2b 13 = 2b

13 — 2 = b

The line perpendicular to y = 2x is y = 1 — 2 x + 13

— 2 .

Find the point of intersection.

y = 2x Equation 1

y = 1 — 2 x + 13 —

2 Equation 2

−2x = 1 — 2 x + 13 —

2

−4x = x + 13 −5x = 13

−5 — 5 x = 13 —

5

x = − 13 — 5

y = −2 ( − 13 — 5 ) = 26 — 5 So, the perpendicular lines intersect at ( − 13 — 5 , 26 — 5 ) .

Find the distance from (−1, 6) to ( − 13 — 5 , 26 — 5 ) .

distance = √———

( − 13 — 5 − (−1) ) 2 + ( 26 — 5 − 6 )

2

= √——— ( −13 + 5 — 5 ) 2 + ( 26 − 30 — 5 ) 2 = √—— ( −8 — 5 ) 2 + ( −4 — 5 ) 2 = √

—

64 — 25

+ 16 — 25

= √—

80 — 25

≈ 1.79 units


Vocabulary and Core Concept Check 1. A directed line segment AB is a segment that represents

moving from point A to B.

2. Two lines are perpendicular if the product of their slopes equals −1.


Chapter 3

Monitoring Progress and Modeling with Mathematics 3. In order to divide the segment in the ratio 1 to 4, partition the


slope of — AB = −2 − 0 — 3 − 8

= −2 — −5 = 2 —

5


run = −5 ⋅ 0.2 = −1 x = 8 − 1 = 7 rise = −2 ⋅ 0.2 = −0.4 y = 0 − 0.4 = −0.4 So, the point is P(7, −0.4).


slope of — AB = 1 − (−4) — 6 − (−2)

= 1 + 4 — 6 + 2

= 5 — 8


run = 8 ⋅ 0.6 = 4.8 x = −2 + 4.8 = 2.8 rise = 5 ⋅ 0.6 = 3 y = −4 + 3 = −1 So, the point is P(2.8, −1).


slope of — AB = −3 − 6 — −2 − 1 = −9 — −3

= 3

To fi nd the coordinates of P, add 5 — 6 of the run to the x-coordinate and add 5 — 6 of the rise to the y-coordinate.

run = − 3 ⋅ 5 — 6 = −2.5 x = 1 − 2.5 = −1.5 rise = −9 ⋅ 5 — 6 = −7.5 y = 6 − 7.5 = −1.5 So, the point is P(−1.5, −1.5).

6. In order to divide the segment in the ratio 2 to 6, partition the segment into 2 + 6, or 8 congruent parts. P is 2 — 8 =

1 — 4 of the

way from point A to point B.

slope of — AB = −4 − 2 — 5 − (−3)

= −6 — 8 = − 3 —

4


run = 8 ⋅ 0.25 = 2 x = −3 + 2 = −1 rise = −6 ⋅ 0.25 = −1.5 y = 2 − 1.5 = 0.5 So, the point is P(−1, 0.5).

7. The slope of line a is 6 − 4 — 5 + 1

= 2 — 6 = 1 —

3 .

The slope of line b is 2 − 1 — 3 + 1

= 1 — 4 .

The slope of line c is 0 + 2 — 3 + 3

= 2 — 6 = 1 —

3 .

The slope of line d is 6 + 2 — 1 − 3

= 8 — −2 = −4.

Lines a and c have equal slopes. Therefore, a & c.

The product of the slopes of the lines b and d is −1. Therefore, b ⊥ d.

8. The slope of line a is 2 − 3 — 2 − (−2)

= −1 — 4 = − 1 —

4 .

The slope of line b is −2 − 0 — 3 + 3

= −2 — 6 = − 1 —

3 .

The slope of line c is 4 + 2 — 2 − 0

= 6 — 2 = 3.

The slope of line d is 6 − 0 — 0 − (−2)

= 6 — 2 = 3.

Lines c and d have equal slopes. Therefore, c & d.

The product of the slopes of the lines b and c is −1 and the product of the slopes of the lines b and d is −1. Therefore, b ⊥ c and b ⊥ d.

9. Line 1 has a slope of 4 − 0 — 7 − 1

= 6 — 4 = 2 —

3 .

Line 2 has a slope of 6 − 0 — 3 − 7

= 6 — −4 = − 3 —

2 .

The product of the slopes is 2 — 3 ⋅ ( − 3 — 2 ) = −1.

Therefore, the two lines are perpendicular by the Slopes of Perpendicular Lines Theorem (Thm. 3.14).

10. Line 1 has a slope of −2 − 1 — −7 − (−3) = −3 — −4

= 3 — 4 .

Line 2 has a slope of 4 − (−1) — 8 − 2

= 5 — 6 .

The slopes are not equal nor do their products equal −1. So, the lines are neither perpendicular nor parallel.

11. Line 1 has a slope of 7 − 3 — −5 − (− 9) = 4 —

4 = 1.

Line 2 has a slope of 2 − 6 —— −7 − (−11) = −4 —

4 = −1.

The product of the slopes is 1 ⋅ (−1) = −1. Therefore, the two lines are perpendicular by the Slopes of

Perpendicular Lines Theorem (Thm. 3.14).

12. Line 1 has a slope of 9 − 5 — −8 − 10 = 4 — −18

= − 2 — 9 .

Line 2 has a slope of −6 − (−4) — 11 − 2

= −2 — 9 = − 2 —

9 .

The slopes are equal. Therefore, the two lines are parallel by the Slopes of Parallel Lines Theorem (Thm. 3.13).


Chapter 3

13. slope of the parallel line = −2 y = −2x + b −1 = −2 ⋅ 0 + b −1 = 0 + b −1 = b Because m = −2 and y

x

23

1

−2

31−2−1

y = −2x + 3

y = −2x − 1

b = −1, an equation of the line is y = −2x − 1.

14. slope of the parallel line = 1 — 5

y = 1 — 5 x + b

8 = 1 — 5 ⋅ 3 + b

40 = 3 + 5b 37 = 5b

37 — 5 = 5b —

5

37 — 5 = b

Because m = 1 — 5 and b = 37

— 5 , y

x

23456

8

−2

4 5321−2−1

y = 15x + 357

y = 15x + 45

an equation of the line is

y = 1 — 5 x + 37

— 5 .

15. The slope is undefi ned, because x = −5 is a vertical line. So, the line parallel to x = −5 must be vertical and go through the x-axis. Therefore, the equation is x = −2.

y

x

456

23

1

−2

−1−3−4−6

x = −5

x = −2

16. −x + 2y = 12 2y = x + 12 y = 1 — 2 x + 6 slope of the parallel line = 1 — 2

y = 1 — 2 x + b

0 = 1 — 2 ⋅ 4 + b 0 = 2 + b −2 = b Because m = 1 — 2 and b = −2, y

x

45

7

23

1

−3

421−2−1−3−4

y = 12x + 6

y = 12x − 2

an equation of the line is y = 1 — 2 x − 2.

17. The slope of the line is −9. The slope of the perpendicular line is

− 9 ⋅ m = −1 m = 1 — 9 .

y = 1 — 9 x + b 0 = 1 — 9 ⋅ 0 + b b = 0 Because m = 1 — 9 and b = 0, an equation of the line is

y = 1 — 9 x. y

x

678

−2

4 5321−1

y = 19x

y = 9x −− 1

18. The slope of y = −3 is 0, so the line perpendicular to y = −3 has an undefi ned slope. Therefore, the equation is x = 4.

y

x

4

23

1

−4−5−6

−2

5321−2−1−3−4

x = 4

y = −3


Chapter 3

19. y − 4 = −2(x + 3) y − 4 = −2x − 6 y = −2x − 6 + 4 y = −2x − 2 The slope of the line is −2. The slope of the perpendicular line is

−2 ⋅ m = −1 m = 1 — 2 .

y = 1 — 2 x + b 3 = 1 — 2 ⋅ 2 + b 3 = 1 + b 2 = b Because m = 1 — 2 and b = 2, an equation of the line is

y = 1 — 2 x + 2. y

x

45

3

1

−3−4−5

−2

4 5321−2−1−3−4

y = 12x + 2

y = −2x − 2

20. 3x − 5y = 6 −5y = −3x + 6

−5 — −5 y = −3 — −5

x + 6 — −5

y = 3 — 5 x − 6 —

5

The slope of the line is 3 — 5 .

The slope of the perpendicular line is

3 — 5 ⋅ m = −1

m = − 5 — 3 .

y = − 5 — 3 x + b

3y = −5x + 3b 3(0) = −5 ⋅ (−8) + 3b 0 = 40 + 3b

− 40 — 3 = b

Because m = − 5 — 3 and b = − 40

— 3 , y

x

2

−6−8

−14−16

−4

8642−4−2

y = 35x − 65

y = −53x − 430

an equation of the line is y = − 5 — 3 x −

40 — 3 .

21. The slope of y = 3x is 3, so the line perpendicular to y = 3x will have a slope of − 1 — 3 .

y = − 1 — 3 x + b 7 = − 1 — 3 ⋅ (−1) + b 7 = 1 — 3 + b 21 = 1 + 3b 20 = 3b 20 — 3 = b The line perpendicular to y = 3x is y = − 1 — 3 x +

20 — 3 .


y = 3x Equation 1 y = − 1 — 3 x +

20 — 3 Equation 2

3x = − 1 — 3 x + 20

— 3

9x = −x + 20 10x = 20 10 — 10 x =

20 — 10

x = 2 y = 3 ⋅ 2 = 6 So, the perpendicular lines intersect at (2, 6).

Find the distance from (−1, 7) to (2, 6). distance = √

—— (2 − (−1))2 + (6 − 7)2 = √—

32 + (−1)2 = √

— 9 + 1 = √

— 10 ≈ 3.16 units

22. The slope of y = x − 6 is 1, so the line perpendicular to y = x − 6 will have a slope of −1.

y = −1x + b −3 = −1 ⋅ (−9) + b −3 = 9 + b −12 = b The line perpendicular to y = x − 6 is y = −x − 12. Find the point of intersection.

y = x − 6 Equation 1 y = −x − 12 Equation 2

x − 6 = −x − 12

2x = −6

2 — 2 x = −6 —

2

x = −3

y = −3 − 6 = −9

So, the perpendicular lines intersect at (−3, −9).

Find the distance from (−9, −3) to (−3, −9).


(−3 − (−9))2 + (−9 − (−3))2 = √

— 62 + (−6)2 = √

— 36 + 36 = √

— 72 ≈ 8.49 units


Chapter 3

23. 5x + 2y = 4 2y = −5x + 4

2y — 2 = −5x —

2 + 4 —

2

y = −5 — 2 x − 2

The slope of 5x + 2y = 4 is − 5 — 2 , so the line perpendicular to 5x + 2y = 4 will have a slope of 2 — 5 .

y = 2 — 5 x + b

−21 = 2 — 5 ⋅ 15 + b

−21 = 6 + b −27 = b The line perpendicular to 5x + 2y = 4 is y = 2 — 5 x − 27.


y = − 5 — 2 x + 2 Equation 1

y = 2 — 5 x − 27 Equation 2

− 5 — 2 x + 2 = 2 —

5 x − 27

− 5 — 2 x = 2 —

5 x − 29

− 25 — 10

x − 4 — 10

x = −29

− 29 — 10

x = −29

− 10 — 29

⋅ ( − 29 — 10 x ) = −29 ⋅ ( − 10 — 29 ) x = 10

y = − 5 — 2 ⋅ 10 + 2

y = −25 + 2

y = −23

So, the perpendicular lines intersect at (10, − 23).

Find the distance from (15, −21) to (10, −23).


(10 − 15)2 + (−23 − (−21))2

= √——

(−5)2 + (−2)2

= √—

25 + 4 = √—

29 ≈ 5.39 units

24. −x + 2y = 14

2y = x + 14

2y — 2 = x —

2 + 14 —

2

y = 1 — 2 x + 7

The slope of −x + 2y = 14 is 1 — 2 , so the line perpendicular to −x + 2y = 14 will have a slope of −2.

y = −2x + b

5 = −2 ⋅ ( − 1 — 4 ) + b 5 = 1 —

2 + b

10 = 1 + 2b

9 = 2b

b = 9 — 2

The line perpendicular to −x + 2y = 14 is y = −2x + 9 — 2 .


y = 1 — 2 x + 7 Equation 1

y = −2x + 9 — 2 Equation 2

1 — 2 x + 7 = −2x + 9 —

2

x + 14 = −4x + 9 5x = −5 x = −1

y = 1 — 2 ⋅ (−1) + 7

y = − 1 — 2 + 14 —

2

y = 13 — 2

So, the perpendicular lines intersect at ( −1, 13 — 2 ) . Find the distance from ( − 1 — 4 , 5 ) to ( −1, 13 — 2 ) .

distance = √——— ( −1 − ( − 1 — 4 ) ) 2 + ( 13 — 2 − 5 ) 2 = √——— ( −4 — 4 + 1 — 4 ) 2 + ( 13 — 2 − 10 — 2 ) 2 = √—— ( −3 — 4 ) 2 + ( 3 — 2 ) 2 = √

—

9 — 16

+ 9 — 4

= √— 9 — 16 + 36 — 16 = √—

45 — 16

≈ 1.68 units

25. Because the slopes are opposites but not reciprocals, their product does not equal −1. Lines 1 and 2 are neither parallel nor perpendicular.


Chapter 3

26. Parallel lines have the same slope, not the same y-intercept.

y = 2x + 1, (3, 4)

4 = 2(3) + b −2 = b The line y = 2x − 2 is parallel to the line y = 2x + 1.

27. midpoint = ( −4 + 4 — 2 , 3 − 1 — 2 ) = (0, 1) slope = −1 − 3 —

4 − (−4) = −4 —

8 = − 1 —

2

The slope of the perpendicular line is 2.

y = 2x + b 1 = 2 ⋅ 0 + b 1 = b

Because m = 2 and b = 1, the equation of the perpendicular bisector of — PQ is y = 2x + 1.

28. midpoint = ( −5 + 3 — 2 , −5 + 3 — 2 ) = (−1, −1) slope = 3 − (−5) —

3 − (−5) = 8 —

8 = 1

The slope of the perpendicular line is −1. y = −1x + b −1 = −1 ⋅ (−1) + b −1 = 1 + b −2 = b

Because m = −1 and b = −2, the equation of the perpendicular bisector of — PQ is y = −x − 2.

29. midpoint = ( 0 + 6 — 2 , 2 − 2 — 2 ) = (3, 0) slope = −2 − 2 —

6 − 0 = −4 —

6 = − 2 —

3

The slope of the perpendicular line is 3 — 2 .

y = 3 — 2 x + b

0 = 3 — 2 ⋅ 3 + b

0 = 9 — 2 + b

− 9 — 2

= b

Because m = 3 — 2 and b = − 9 — 2 , the equation of the

perpendicular bisector of — PQ is y = 3 — 2 x − 9 — 2 .

30. midpoint = ( −7 + 1 — 2 , 0 + 8 — 2 ) = ( −6 — 2 , 8 — 2 ) = (−3, 4) slope = 8 − 0 —

1 − (−7) = 8 —

8 = 1

The slope of the perpendicular line is −1. y = −1x + b 4 = −1 ⋅ (−3) + b 4 = 3 + b 1 = b Because m = − 1 and b = 1, the equation of the perpendicular bisector of — PQ is y = −x + 1. 31. In order to divide the segment in the ratio 1 to 4, partition the


slope = 2 − (−2) — 5 − (−4)

= 4 — 9


run = 9 ⋅ 0.2 = 1.8, x = −4 + 1.8 = −2.2 rise = 4 ⋅ 0.2 = 0.8, y = −2 + 0.8 = −1.2 The point on the graph that represents the school is

(−2.2, −1.2)

32. slope of — QR = 6 − 4 — 2 − 6

= 2 — −4 = − 1 —

2

slope of — RS = 4 − 1 — 6 − 5

= 3 — 1 = 3

slope of — ST = 1 − 3 — 5 − 1

= −2 — 4 = − 1 —

2

slope of — QT = 6 − 3 — 2 − 1

= 3 — 1 = 3

Quadrilateral QRST is a parallelogram. — QT & — RS because they have the same slope (m = 3), and — ST & — QR because they have the same slope ( m = − 1 — 2 ) .

33. slope of — LM = 8 − 6 — 5 − 0

= 2 — 5

slope of — MN = −1 − 8 — 4 − 5

= −9 — −1 = 9

slope of — LN = 6 + 1 — 0 − 4

= 7 — −4

Triangle LMN is not a right triangle because the slopes of

the sides are 2 — 5 , − 7 —

4 , and 9. No combination of the products

of two slopes equal −1. So, none of the segments are perpendicular.

34. Train tracks: y = 2x, V(−2, 3) y = 2x + b 3 = 2 ⋅ (−2) + b 3 = −4 + b 7 = b Because m = 2 and b = 7, the equation of the line that

represents the new road is y = 2x + 7.


Chapter 3

35. Train tracks: y = − 2 — 3 , P(2, 2) The slope of the perpendicular line is 3 — 2 .

y = 3 — 2 x + b

2 = 3 — 2 ⋅ 2 + b 2 = 3 + b −1 = b Because m = 3 — 2 and b = −1, the equation of the line that

represents the new road is y = 3 — 2 x − 1.

36. The distance between the gazebo and the nature trail is 42.4 feet.

Nature trail: y = 1 — 3 x − 4, gazebo: (−6, 4) The slope of the perpendicular line is −3. y = −3x + b 4 = −3 ⋅ (−6) + b 4 = 18 + b −14 = b The line perpendicular to y = 1 — 3 x − 4 is y = − 3x − 14. Find the point of intersection.

y = 1 — 3 x − 4 Equation 1 y = −3x − 14 Equation 1

1 — 3 x − 4 = −3x − 14 3 ⋅ 1 — 3 x − 3 ⋅ 4 = 3 ⋅ (−3x) − 3 ⋅ 14 x − 12 = −9x − 42 10x − 12 = −42 10x = −30 x = −3 y = 1 — 3 ⋅ (−3) − 4 = −1 − 4 = −5 So, the perpendicular lines intersect at (−3, −5). Find the distance from (−6, 4) and (−3, −5). distance = √

——— [−6 − (−3)]2 + [4 − (−5)]2

= √——

(−6 + 3)2 + (4 + 5)2

= √——

(−3)2 + (9)2

= √—

9 + 81 = √—

90 ≈ 9.49 units

Because each unit in the coordinate plane corresponds to 10 feet, the distance between the gazebo and the nature trail is about 10 ⋅ 9.49 = 94.9 feet.

37. The slope of a line perpendicular toℓmust be the opposite reciprocal of the slope of lineℓ. Then the slope must be negative, and have an absolute value greater than 1. So, an inequality that represents the slope m of a line perpendicular toℓis m < − 1.

38. The angles of the quadrilateral are all right angles because the sides are all formed by horizontal or vertical lines. Also, the length of each side is n. So, JKLM is a square.

39. XY is the same as YX. If the ratio XP to PY is 3 to 5, then the ratio YP to PX is 5 to 3. This is the same point P in both cases.

40. yes; If two lines have the same y-intercept, then they intersect in that point. But parallel lines do not intersect.

41. a. Substitute 4x + 9 for y in the second equation. 4x − (4x + 9) = 1 4x − 4x − 9 = 1 −9 = 1 −9 ≠ 1 Because there is no solution, the lines do not intersect and

are, therefore, parallel.

b. Solve the second equation for y.

2x − y = 18 −y = −2x + 18 y = 2x − 18 Substitute the result for y in the fi rst equation.

3(2x − 18) + 4x = 16 6x − 54 + 4x = 16 10x − 54 = 16 10x = 70 x = 7 y = 2x − 18 y = 2 ⋅ 7 − 18 y = 14 − 18 y = −4 The lines intersect at one point (7, −4). c. Substitute −5x + 6 for y in the second equation. 10x + 2(−5x + 6) = 12 10x − 10x + 12 = 12 12 = 12 Because the statement 12 = 12 is always true, there are

infi nitely many solutions, and the lines are the same.


Chapter 3

42. Given ax + by = 0, (x0, y0) ax + by = 0 by = −ax y = − a —

b x

The slope is − a — b .

The perpendicular line has slope b — a .

Let c be the y-intercept.

y = b — a x + c

y0 = b — a x0 + c

y0 − b — a x0 = c

ay0 − bx0 — a = c

y = b — a x +

ay0 − bx0 — a

Find the point of intersection of the line ax + by = 0 and the perpendicular line.

y = b — a x +

ay0 − bx0 — a

− a — b x = b —

a x +

ay0 − bx0 — a

− ax — b = bx —

a +

ay0 − bx0 — a

− ax — b − bx —

a =

ay0 − bx0 — a

− ( ax — b + bx — a ) = ay0 − bx0

— a

− ( a2x + b2x — ab ) = ay0 − bx0 — a ( − ( a2 + b2 — ab ) ) x = ay0 − bx0 — a x = ( ay0 − bx0 — a ) ( − ab — a2 + b2 ) x = ( ay0 − bx0 — 1 ) ( − b — a2 + b2 ) x = ( − b(ay0 − bx0) —— a2 + b2 ) y = − a —

b x

y = − a — b ( − b(ay0 − bx0) —— a2 + b2 )

y = a — 1 ( (ay0 − bx0) — a2 + b2 )

y = ( a(ay0 − bx0) —— a2 + b2 ) Point of intersection: ( − b(ay0 − bx0) —— a2 + b2 , a(ay0 − bx0) —— a2 + b2 )

Find the distance between (x0, y0) and

( − b(ay0 − bx0) —— a2 + b2 , a(ay0 − bx0) —— a2 + b2 ) .distance = √————— ( x0 − ( − b(ay0 − bx0) —— a2 + b2 ) ) 2 + ( y0 − a(ay0 − bx0) —— a2 + b2 ) 2 = √——————

( x0(a2 + b2) + b(ay0 − bx0) ——— a2 + b2 ) 2 + ( y0(a2 + b2) − a(ay0 − bx0)

——— a2 + b2

) 2 = √—————— ( x0a2 + x0b2 + aby0 − b2x0 ——— a2 + b2 ) 2 + ( y0a2 + y0b2 − a2y0 + abx0 ——— a2 + b2 ) 2 = √——— ( x0a2 + aby0 — a2 + b2 ) 2 + ( y0b2 + abx0 — a2 + b2 ) 2 = √————— ( (x0a2 + aby0)(x0a2 + aby0) ——— (a2 + b2)2 ) + ( (y0b2 + abx0)(y0b2 + abx0) ——— (a2 + b2)2 ) = √——————

( a2(x02a2 + 2x0aby0 + b2y02) ——— (a2 + b2)2 ) + ( b2(y02b2 + 2abx0y0 + a2x02) ——— (a2 + b2)2 ) = √—————— ( a2(a2x02 + 2abx0y0 + b2y02) ——— (a2 + b2)2 ) + ( b2(a2x02 + 2abx0y0 + b2y02) ——— (a2 + b2)2 ) = √——— (a2 + b2)(a2x02 + 2abx0y0 + b2y02) ———

(a2 + b2)2

= √——— (a2x02 + 2abx0y0 + b2y02) ——— (a2 + b2)

= √—— ((ax0 + by0)2) —— (a2 + b2)

= ∣ ax0 + by0 ∣

— √—

a2 + b2

So, the distance between the line ax + by = 0 and the

perpendicular line y = b — a x +

ay0 − bx0 — a is

∣ ax0 + by0 ∣ —

√—

a2 + b2 .

Sample answer: Use the line and point from Monitoring Progress Exercise 7 on page 159.

∣ (2)(−1) + (1)(6) ∣

—— √—

22 + 12 =

∣ −2 + 6 ∣ —

√—

5 =

∣ 4 ∣ —

√—

5 = 4 —

√—

5 ≈ 1.79 units

43. slope = −2 − k — −7 − (−1) = −2 − k — −7 + 1

= −2 − k — −6 = −(2 + k) — −6

= (2 + k) — 6

The slope of the parallel line is 1.

1 = 2 − k — 6

6 = 2 + k 4 = k


Chapter 3

44. slope = 0 − 2 — 7 − k

= −2 — 7 − k

The slope of the given line is 1. The slope of the perpendicular line is −1.

−1 = −2 — 7 − k

−1(7 − k) = −2 −7 + k = −2 k = 5

45. Using points A(3, 2) and B(6, 8), fi nd the coordinates of point P that lies beyond point B along ###⃗ AB so that the ratio of

AB to BP is 3 to 2. In order to keep the ratio, AB — BP

= 3 — 2 ,

solve this ratio for BP to get BP = 2 — 3 AB. Next, fi nd the rise

and run from point A to point B. Leave the slope in terms of

rise and run and do not simplify. mAB = 8 − 2 — 6 − 3

= 6 — 3 = rise —

run .

Add 2 — 3 of the run to the x-coordinate of B, which is

2 — 3 ⋅ 3 + 6 = 8. Add 2 — 3 of the rise to the y-coordinate of B,

which is 2 — 3 ⋅ 6 + 8 = 12. So, the coordinates of P are (8, 12).

46. The slope of the perpendicular line is − 1 — 2 . y = − 1 — 2 x + b Use the y-intercept of y = 2x + 5, (0, 5). y = − 1 — 2 x + b 5 = − 1 — 2 ⋅ 0 + b 5 = b y = − 1 — 2 x + 5 Find the intersection of the perpendicular lines y = − 1 — 2 x + 5

and y = 2x. 2x = − 1 — 2 x + 5 4x = −x + 10 5x = 10 x = 2 y = 2 ⋅ 2 = 4 Find the distance between (0, 5) and (2, 4).


(2 − 0)2 + (4 − 5)2 = √

— 22 + (−1)2 = √

— 4 + 1 = √

— 5 ≈ 2.24 units

47. If lines x and y are perpendicular to line z, then by the Slopes of Perpendicular Lines Theorem (Thm. 3.14), mx ⋅ mz = −1 and my ⋅ mz = − 1. By the Transitive Property of Equality, mx ⋅ mz = my ⋅ mz, and by the Division Property of Equality mx = my. Therefore, by the Slopes of Parallel Lines Theorem (Thm. 3.13), x & y.

48. If x & y and y & z, then by the Slopes of Parallel Lines Theorem (Thm. 3.13), mx = my and my = mz. Therefore, by the Transitive Property of Equality, mx = mz. So, by the Slopes of Parallel Lines Theorem, (Thm. 3.13), x & z.

49. If lines x and y are vertical lines and they are cut by any horizontal transversal, z, then x ⊥ z and y ⊥ z by Theorem 3.14. Therefore, x & y by the Lines Perpendicular to Transversal Theorem (Thm. 3.12).

50. If lines x and y are horizontal, then by defi nition mx = 0 and my = 0. So, by the Transitive Property of Equality, mx = my. Therefore, by the Slopes of Parallel Lines Theorem (Thm. 3.13), x & y.

51. By defi nition, the x-axis is perpendicular to the y-axis. Let m be a horizontal line, and let n be a vertical line. Because any two horizontal lines are parallel, m is parallel to the x-axis. Because any two vertical lines are parallel, n is parallel to the y-axis. By the Perpendicular Transversal Theorem, (Thm 3.11), n is perpendicular to the x-axis. Then, by the Perpendicular Transversal Theorem (Thm. 3.11), n is perpendicular to m.

Maintaining Mathematical Profi ciency 52. y

x

4567

321

321−1

A(3, 6)

53. y

x

321

−3−4−5

−2

321−2−1−3

B(0, −4)

54. y

x

4321

−2

4 5321−1

C(0, 5)

55. y

x

21

−3

21−2−1−3

D(−1, −2)

56. x −2 −1 0

y = x + 9 −2 + 9 = 7 −1 + 9 = 8 0 + 9 = 9 x 1 2

y = x + 9 1 + 9 = 10 2 + 9 = 11

57. x −2 −1 0

y = x − 3 — 4 −2 − 3 — 4 = −

11 — 4 −1 −

3 — 4 = −

7 — 4 0 −

3 — 4 = −

3 — 4

x 1 2

y = x − 3 — 4 1 − 3 — 4 =

1 — 4 2 −

3 — 4 =

5 — 4


Chapter 3

3.4 – 3.5 What Did You Learn? (p. 163) 1. You can fi nd the distance between two lines only if the two

lines are parallel. If you choose a point on one line and fi nd the distance from that point to the other line, the answer will always be the same when the lines are parallel. But if the lines are not parallel, the answer will be different for every point on the line.

2. After drawing the perpendicular lines going through each endpoint of the given segment, you could pick an arbitrary point on one of the perpendicular lines. Then set the compass to the distance from this point to the corresponding endpoint, and use the same compass sett

hscc geom wsk 03 - schoolwires · 2015. 9. 29. · chapter 3 2. a. ⃖ab ##⃗ and ⃖bc##⃗ are...

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