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Chapter 3
Chapter 3 Maintaining Mathematical Proficiency (p. 123)
3
2 − (−1)
−1 − 3
−4
3
slope = −—
4
−3
−3
−1 − 2
2. m = — = — = — = 3
−3 − (−2) −3 + 2 −1
1. m = — = —
slope = 3
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 undefined.
Chapter 3 Mathematical Practices (p. 124)
1
−2 − (−2)
1 − (−3)
−2 + 2
1+3
0
4
3. m = — = — = — = 0
1. These lines are perpendicular. They have slopes m1 = −—2
and m2 = 2.
4
slope = 0
y = 2x − 4
4. y = mx + b
−6
6
1 = −3(6) + b
1
1 = −18 + b
−4
19 = b
The equation is y = −3x + 19.
2. These lines are coincident, because their equations are
equivalent.
4
5. y = mx + b
8 = −2(−3) + b
−6
8=6+b
6
2=b
1
−4
The equation is y = −2x + 2.
5 = 4(−1) + b
1
4
5 = −4 + b
1
y = −2 x + 1
9=b
The equation is y = 4x + 9.
−6
6
1
y = −2 x − 1
y = mx + b
−4 =
1
—2 (2)
−4
+b
4. These lines are neither parallel nor perpendicular because
1
their slopes are m1 = −—2 and m2 = 1. They intersect
−4 = 1 + b
at (−2, 2).
−5 = b
4
The equation is y = —12 x − 5.
y=x+4
8. y = mx + b
−5 =
1
−—4(−8)
y = −2 x + 1
3. These lines are parallel. Their slopes are equal, m1 = −—2
1
and m2 = −—2.
6. y = mx + b
7.
y = −2 x + 1
−6
+b
6
1
−5 = 2 + b
−4
−7 = b
1
The equation is y = −—4 x − 7.
9. y = mx + b
9 = —23(0) + b
9=0+b
y = −2 x + 1
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 infinitely many points in common.
9=b
The equation is y = —23 x + 9.
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71
Chapter 3
2. a. ⃖⃗
AB and ⃖⃗
BC are intersecting lines. They intersect at
point B.
⃖⃗ and ⃖⃗
b. AD
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
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.
never intersect.
4. The line containing point B that appears to be perpendicular
never intersect.
5. The line containing point B that is skew to ⃖⃗
CD is ⃖⃗
BF.
ABG, which is not drawn, and they will never intersect.
6. The plane containing point B parallel to plane CDH is
e. ⃖⃗
EF and ⃖⃗
CG are skew lines. They are not coplanar and will
⃖⃗ and ⃖⃗
f. AD
GH are parallel lines. They both lie on plane
3. a. Vertical angles: ∠1, ∠3; ∠2, ∠4; ∠5, ∠7; ∠6, ∠8
Two 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, ∠8
One 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.
to ⃖⃗
CD is ⃖⃗
BC.
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 infinite 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.
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.
21. a. true; The floor 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 floor of
the treehouse is horizontal. So, they are perpendicular.
72
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Chapter 3
22. no; All three lines could intersect at the same point or two
skew lines intersected by a third.
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.
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 Proficiency
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. 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 definition of
congruent angles, m∠1 = m∠2 = 41°.
3.2 Exercises (pp. 135–136)
Vocabulary and Core Concept Check
1. Both theorems refer to two pairs of congruent angles that
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.
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.
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73
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 2x
—=—
2
2
64 = x
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
8. Consecutive interior angles are supplementary.
72° + (7x + 24)° = 180°
96 + 7x = 180
96 − 96 + 7x = 180 − 96
7x = 84
7x 84
—=—
7
7
x = 12
and ∠ CBD are alternate interior angles, and they are
— as the transversal, ∠ BCA
congruent (Thm. 3.2). With AC
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.
t
9. m∠5 = 65° because alternate interior angles are congruent.
1
p
m∠5 + (11x − 17)° = 180° by the definition of
supplementary angles.
65° + (11x − 17)° = 180°
3
2
q
11x + 48 = 180
11x + 48 − 48 = 180 − 48
11x = 132
11x 132
—=—
11
11
x = 12
10. m∠4 + 118° = 180° by the definition of supplementary
angles.
So, m∠4 = 180° − 118° = 62°.
Alternate interior angles are congruent.
62 = 8x + 6
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
56 = 8x
56 8x
—=—
8
8
x=7
74
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Chapter 3
16. Given p q, and t as the transversal.
21. 5x° + (14x − 10)° = 180°
Prove Consecutive interior angles are supplementary.
19x − 10 = 180
19x = 190
t
p
1
3
19
x
—
19
=
190
—
19
x = 10
2
q
22. 2y° + (2x + 12)° = 180°
2y + 2x = 168
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. Definition of congruent
angles
4. m∠2 + m∠3 = 180°
4. Linear Pair Postulate
(Post. 2.8)
2
—2 y
+
2
—2 x
=
2y° = 5x°
2y = 5(10)
2y = 50
50
2
—2 y = —
2
y = 25
4x° + (y + 6)° = 180°
4x + y = 174
168
—
2
y = −4x + 174
y + x = 84
y = −x + 84
5. m∠2 + m∠1 = 180°
5. Substitution Property
of Equality
6. ∠1 and ∠2 are
supplementary angles.
6. Definition 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.
−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 definition of angle
bisector, and ∠ 4 ≅ ∠ 5 is given. So, by the Transitive
Property of Congruence, all five 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°
— = 60°.
3
Maintaining Mathematical Proficiency
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)
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.
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Geometry
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75
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
3.3 Exercises (pp. 142–144)
Vocabulary and Core Concept Check
1. corresponding angles, alternate interior angles, and alternate
exterior angles
interior angles are congruent, then the lines are parallel.
The converse is true.
2. Two lines cut by a transversal are parallel if and only if the
c. If two lines are cut by a transversal so that the alternate
Two lines cut by a transversal are parallel if and only if the
alternate interior angles are congruent.
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.
corresponding 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
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 definition. 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°.
76
Geometry
Worked-Out Solutions
angles are congruent.
120° = 3x°
120 3x
—=—
3
3
x = 40
4. Lines m and n are parallel when the marked corresponding
angles are congruent.
135° = (2x + 15)°
120 = 2x
120 2x
—=—
2
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 3x
3
3
x = 15
—=—
6. Lines m and n are parallel when the marked alternate exterior
angles are congruent.
x° = (180 − x)°
2x = 180
2x 180
—=—
2
2
x = 90
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Chapter 3
7. Lines m and n are parallel when the marked consecutive
interior angles are supplementary.
Prove j k
x° + 2x° = 180°
3x = 180
3x 180
—=—
3
3
x = 60
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
m
A
3 2
5
j
k
8. Lines m and n are parallel when the marked alternate interior
n
12. Given ∠ 3 and ∠ 5 are supplementary.
1
STATEMENTS
REASONS
1. ∠ 3 and ∠ 5 are
supplementary.
1. Given
2. ∠ 2 and ∠ 3 are
supplementary.
2. Linear Pair Postulate
(Post. 2.8)
3. m∠ 3 + m∠ 5 = 180°,
m∠ 2 + m∠ 3 = 180°
3. Definition of
supplementary angles
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. Definition of congruent
angles
6. ∠ 2 ≅ ∠ 5
7. j k
B
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.
n
13. yes; Alternate Interior Angles Converse (Thm. 3.6)
14. yes; Alternate Exterior Angles Converse (Thm. 3.7)
15. no
m
P
1
7. Corresponding Angles
Converse (Thm. 3.5)
16. yes; Corresponding Angles Converse (Thm. 3.5)
A
17. no
B
18. yes; Alternate Exterior Angles Converse (Thm. 3.7)
11. Given ∠ 1 ≅ ∠ 8
1
Prove j k
2
8
j
k
STATEMENTS
REASONS
1. ∠ 1 ≅ ∠ 8
1. Given
2. ∠ 1 ≅ ∠ 2
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. ∠ 8 ≅ ∠ 2
3. Transitive Property of Congruence
(Thm. 2.2)
4. j k
4. Corresponding Angles Converse
(Thm. 3.5)
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 definition, 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 definition, 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.
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Geometry
Worked-Out Solutions
77
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 the Consecutive
by definition, which means that ⃖⃗
AC DF
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).
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:
t
1
3
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 definition
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.
r
2
26. Because each rung of the ladder is parallel to the one directly
s
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°
Prove m n
1
2
m
n
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 floor, ∠ 1 and ∠ 2 are
alternate interior angles. So, if ∠ 1 ≅ ∠ 2, then the top of
the step will be parallel to the floor by the Alternate Interior
angles Converse (Thm. 3.6).
78
Geometry
Worked-Out Solutions
STATEMENTS
REASONS
1. m∠ 1 = 115°, m∠ 2 = 65°
1. Given
2. m∠ 1 + m∠ 2 =
m∠ 1 + m∠ 2
2. Reflexive 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. Definition of
supplementary angles
6. m n
6. Consecutive Interior
Angles Converse
(Thm. 3.8)
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Chapter 3
——
34. Given ∠ 1 and ∠ 3 are supplementary.
37. no; Based on the diagram, AB DC by the Alternate Interior
1
Prove m n
m
2
3
n
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.
STATEMENTS
REASONS
1. ∠ 1 and ∠ 3 are
supplementary.
1. Given
2. m∠ 1 and
m∠ 3 = 180°
2. Definition of supplementary
angles
3. ∠ 1 ≅ ∠ 2
3. Vertical Angles Congruence
Theorem (Thm. 2.6)
4. m∠ 1 = m∠ 2
5. m∠ 2 + m∠ 3
= 180°
4. Definition of congruent
angles
5. Substitution Property of
Equality
6. ∠ 2 and ∠ 3 are
supplementary.
6. Definition of supplementary
angles
7. m n
7. Consecutive Interior Angles
Converse (Thm. 3.8)
p
39. a.
35. Given ∠ 1 ≅ ∠ 2, ∠ 3 ≅ ∠ 4
A
1
— CD
—
Prove AB
c.
2 E
3
4
C
STATEMENTS
REASONS
1. ∠ 1 ≅ ∠ 2, ∠ 3 ≅ ∠ 4
1. Given
2. ∠ 2 ≅ ∠ 3
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. ∠ 1 ≅ ∠ 3
3. Transitive Property of
Congruence
4. ∠ 1 ≅ ∠ 4
4. Transitive Property of
Congruence
a b, ∠ 2 ≅ ∠ 3
Prove c d
c
3
q
2
r
3
STATEMENTS
REASONS
1. p q, q r
1. Given
2. ∠ 1 ≅ ∠ 2,
∠2 ≅ ∠3
2. Corresponding Angles Theorem
(Thm. 3.1)
3. ∠ 1 ≅ ∠ 3
3. Transitive Property of
Congruence
4. p r
4. Corresponding Angles Converse
(Thm. 3.5)
40. a. Use the Corresponding Angles Converse (Thm. 3.5).
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°
d
1
p
p q, q r
Prove p r
b. Given
r
(2x + 2)° = (x + 56) °
5. Alternate Interior Angles
Converse (Thm. 3.6)
36. Given
q
1
D
B
— CD
—
5. AB
Angles Converse (Thm. 3.6), but you cannot be sure that
⃖⃗
AD ⃖⃗
BC.
2
4
a
4y − 10 = 180
b
4y = 190
4
190
—4 y = —
4
STATEMENTS
REASONS
1. a b, ∠ 2 ≅ ∠ 3
1. Given
2. ∠ 3 ≅ ∠ 1
2. Alternate Interior Angles
Theorem (Thm. 3.2)
3. ∠ 1 ≅ ∠ 2
3. Transitive Property of
Congruence
4. c d
4. Corresponding Angles
Converse (Thm. 3.5)
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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.
Geometry
Worked-Out Solutions
79
Chapter 3
Maintaining Mathematical Proficiency
——
41. d = √ (−2 −
+ (9 −
1)2
——
9. By the Linear Pair Postulate (Post. 2.3):
m∠ 1 + 138° = 180°
3)2
= √(−3)2 + (6)2
m∠ 1 = 42°
—
—
= √9 + 36 = √45 ≈ 6.71
By the Alternate Exterior Angles Theorem (Thm. 3.3):
———
42. d = √ (8 − (−3))2 + (−6 − 7)2
——
= √ (8 + 3)2 + (−13)2
—
10. m∠ 1 = 123° by the Corresponding Angles Theorem
(Thm. 3.1).
——
= √(11)2 + (−13)2
m∠ 1 = m∠ 2 = 42°
—
= √ 121 + 169 = √ 290 ≈ 17.03
m∠ 2 = 123° by the Vertical Angle Congruence Theorem
(Thm. 2.6).
——
43. d = √ (0 − 5)2 + (8 − (−4))2
——
= √(−5)2 + (8 + 4)2
m∠ 1 + 57° = 180°
——
= √(−5)2 + (12)2
—
m∠ 1 = 123°
—
= √25 + 144 = √ 169 = 13
——
44. d = √ (9 − 13)2 + (−4 − 1)2
——
= √(−4)2 + (−5)2
—
3.1–3.3 What Did You Learn? (p. 145)
(69° + 111° = 180°)
13. no
14. yes; Transitive Property of Parallel Lines (Thm. 3.9)
G
J
15. a. All of the bars are parallel to each other by the Transitive
H
Property of Parallel Lines (Thm. 3.9).
C
A
m∠ 2 = 57° by the Alternate Interior Angles Theorem
(Thm. 3.2).
12. yes; Consecutive Interior Angles Converse (Thm. 3.8)
—
= √16 + 25 = √41 ≈ 6.40
1.
11. By the Linear Pair Postulate (Post. 2.8):
b. ∠ 1 corresponds to ∠ 2 by the Corresponding Angles
B
Theorem (Thm. 3.1). So, m∠ 1 = m∠ 2 = 58°.
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)
⃖⃗ is parallel to ⃖⃗
1. GH
EF.
2. ⃖⃗
FG is perpendicular to ⃖⃗
EF.
⃖⃗ is skew to ⃖⃗
3. GC
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.
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
— is parallel to the vertical
from the edge. Similarly, CD
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 definition of midpoints.
2. a. Check students’ work.
b. They are all right 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.
80
Geometry
Worked-Out Solutions
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Chapter 3
3.4 Exercises (pp. 152–154)
3. a. Check students’ work.
b. Check students’ work.
—
—
—
point O is the midpoint of AB . Point C is the same
c. Check students’ work; CD is perpendicular to AB , and
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.
between two points. The other three ask for the length of a
— is not a
perpendicular segment from the point X to lineℓ. XZ
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
—
= √(1 + 4)2 + (2 + 3)2
—
The distance from point A to ⃖⃗
XZ is about 3.16 units.
—
——
4. AZ = √ (3 − 4)2 + [3 − (−1)]2
—
= √(25 + 25) = √ 50 ≈ 7.1
The distance from E to ⃖⃗
EF is about 7.1 units.
Prove j ⊥ k
2. “Find XZ” is different, which asks for the distance
= √ 9 + 1 = √10 ≈ 3.16
——
2. Given h k, j ⊥ h
passes through the midpoint of the segment at a right angle.
= √ (3)2 + (−1)2
———
1. EG = √ (1 − (−4))2 + (2 − (−3))2
—
1. The perpendicular bisector of a segment is the line that
——
3.4 Monitoring Progress (pp. 148–151)
= √(5)2 + (5)2
Vocabulary and Core Concept Check
j
1 2
3 4
5 6
7 8
h
——
= √ (−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
k
STATEMENTS
REASONS
1. h k, j ⊥ h
1. Given
2. m∠ 2 = 90°
2. Definition of perpendicular
lines
3. ∠ 2 ≅ ∠ 7
3. Alternate Exterior Angles
Theorem (Thm. 3.3)
4. m∠ 2 = m∠ 7
4. Definition of congruent
angles
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
5. m∠ 7 = 90°
6. j ⊥ k
5. Transitive Property of
Equality
6. Definition of perpendicular
lines
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 .
P
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).
Y
X
m
Z
4. yes; Because b ⊥ d and c d, you can conclude that b ⊥ c
by the Perpendicular Transversal Theorem (Thm. 3.11).
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Geometry
Worked-Out Solutions
81
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 first. Label the point of intersection as Z.
Draw ⃖⃗
PZ .
12. The distance form point C to ⃖⃗
AB is not 12 centimeters,
— must be
it is 8 centimeters. The segment from C to AB
—
perpendicular to AB in order to equal the distance from C
—.
to AB
13. Given ∠ 1 ≅ ∠ 2
g
Prove g ⊥ h
1
Z
2
h
X
P
Y
m
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 first. Label the point of intersection as Z.
Draw ⃖⃗
PZ .
m
X
STATEMENTS
REASONS
1. ∠ 1 ≅ ∠ 2
1. Given
2. m∠ 1 = m∠ 2
2. Definition 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
7. m∠ 1 = 90°
7. Transitive Property of
Equality
8. g ⊥ h
8. Definition of perpendicular
lines
Y
P
Z
14. Given m ⊥ p and n ⊥ p
m
n
Prove m n
9. Using a compass setting greater than half of AB, draw two
1
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.
B
p
STATEMENTS
REASONS
1. m ⊥ p
1. Given
2. ∠ 1 is a right angle.
2. Definition of perpendicular
lines
3. n ⊥ p
3. Given
4. ∠ 2 is a right angle.
4. Definition of perpendicular
lines
5. ∠ 1 ≅ ∠ 2
5. Right Angles Congruence
Theorem (Thm. 2.3)
6. Corresponding Angles
Converse (Thm. 3.5)
6. m n
A
2
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.
82
Geometry
Worked-Out Solutions
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Chapter 3
15. Given a ⊥ b
17. none; The only thing that can be concluded in this diagram is
b
Prove ∠ 1, ∠ 2, ∠ 3, and ∠ 4
are right angles.
1
2
3 4
a
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
STATEMENTS
Lines Perpendicular to a Transversal Theorem (Thm. 3.12).
REASONS
19. m n; Because m ⊥ q and n ⊥ q, lines m and n are parallel by
1. a ⊥ b
1. Given
2. ∠ 1 is a right angle.
2. Definition of perpendicular
lines
3. Definition of right angle
3. m∠ 1 = 90°
4. ∠ 1 ≅ ∠ 4
4. Vertical Angle Congruence
Theorem (Thm. 2.6)
5. m∠ 4 = 90°
5. Transitive Property of
Equality
6. ∠ 1 and ∠ 2 are a
linear pair.
6. Definition of linear pair
7. ∠ 1 and ∠ 2 are
supplementary.
7. Linear Pair Postulate
(Post. 2.8)
8. m∠ 1 + m∠ 2
= 180°
8. Definition of
supplementary angles
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)
12. m∠ 3 = 90°
12. Transitive Property of
Equality
13. ∠ 1, ∠ 2, ∠ 3, and
∠ 4 are right angles.
13. Definition of right angle
16. Given ⃗
BA ⊥ ⃗
BC
Prove ∠ 1 and ∠ 2 are
A
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.
other line will be different for different points on the line
unless the lines are parallel.
2
C
STATEMENTS
REASONS
BA ⊥ ⃗
BC
1. ⃗
1. Given
2. ∠ ABC is a right angle.
2. Definition of
perpendicular lines
3. Definition 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. Definition of
complementary angles
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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.
24. no; The shortest distance from a point on one line to the
1
B
6. ∠ 1 and ∠ 2 are
complementary.
20. none; The only things that can be concluded in this diagram
m∠ 5 = 90°, because it is vertical angles with, and therefore
congruent to, ∠ 1.
complementary.
3. m∠ ABC = 90°
the Lines Perpendicular to a Transversal Theorem
(Thm. 3.12). The other lines may or may not be parallel.
For b c:
25. For a ⊥ b:
(9x + 18)° = [5(x + 7) + 15]°
(9x + 18)° = 90°
9x + 18 − 18 = 90 − 18
9x + 18 = 5x + 35 + 15
9x = 72
9x + 18 = 5x + 50
x=8
9x + 18 − 18 = 5x + 50 − 18
9x = 5x + 32
9x − 5x = 5x − 5x + 32
4x = 32
So, x = 8 when
a ⊥ b and b c.
x=8
Geometry
Worked-Out Solutions
83
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 .
3
32. Using x = 0, the y-intercept of y = —2 x + 4 is 4. Find the
equation of the line perpendicular to −3x + 2y = −1
through the point (0, 4) on y = —32 x + 4.
Rewrite the equation in the form of y = mx + b.
−3x + 2y = −1
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.
2y = 3x − 1
1
3
y = —x − —
2
2
2
The slope is —32 , so, the slope of the perpendicular line is −—3 .
2
y = −—x + b
3
2
4 = −— 0 + b
3
4=b
⋅
2
The equation of the line is y = −—3 x + 4.
29.
Find the intersection of the lines y = —32 x − —12 and
2
y = −—3x + 4.
2
3
2
1
3
6 —x − — 6 = −—x 6 + 4 6
3
2
2
3
2
1
2
—x − — = −—x + 4
A
B
⋅
⋅
⋅
⋅
9x − 3 = −4x + 24
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.
13x = 27
27
x=—
13
2
y = −—x + 4
3
2 27 4 13
y = −— — + —
3 13
13
−18 52
y=—+—
13
13
34
y=—
13
⋅
⋅
(
)
27 34
The point of intersection is —
,— .
13 13
——
27
34
√( 13 − 0 ) + ( 13 − 4 )
27
34 4 13
= √( ) + ( − ⋅ )
13
13
13
27
34 52
= √( ) + ( − )
13
13 13
729
−18
= √(
+
169 ) ( 13 )
324
1053
= (
+
=√
≈ 2.5 units
√ 729
169 ) ( 169 )
169
distance =
2
—
2
—
——
—
2
—
—
2
——
—
2
—
—
2
——
—
—
——
—
84
Geometry
Worked-Out Solutions
—
2
—
—
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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 find 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 find 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 defined because it
would be different for each point on the line.
c.
y = —12 x + b
⋅
−2 = —12 (2) + b
−2 = 1 + b
−3 = b
The line y = —12 x − 3 is parallel to the line y = —12x + 2.
The slopes are equal.
6
Maintaining Mathematical Proficiency
6 + 4 10
34. — = — = 2
5
8−3
2
3−5
35. — = −—
3
4−1
8 + 3 11
9
7+2
13 − 4 9
37. — = — = 3
3
2+1
36. — = —
1
y = 2x + 2
−9
9
1
y = 2x − 3
−6
d. m = −2, so the perpendicular line will have a slope of
m = —12 .
y = −2x + b
⋅
−3 = −2 (2) + b
−3 = −4 + b
38. slope = 3; y-intercept = 9
1=b
1
39. slope = −—2 ; y-intercept = 7
The line y = −2x + 1 is perpendicular to the line
y = —12 x + 2. The slopes are opposite reciprocals and have a
product of −1.
1
40. slope = —6 ; y-intercept = −8
41. slope = −8; y-intercept = −6
6
1
y = 2x + 2
3.5 Explorations (p. 155)
−9
3
1. a. y = —2 x + b
9
y = −2x + 1
⋅
2 = —32 0 + b
2=b
−6
The line y = —32x + 2 is parallel to the line y = —32 x − 1.
The slopes are equal.
e.
y = −2x + b
3
The line y = −2x − 2 is parallel to the line y = −2x + 2.
The slopes are equal.
y = 2x + 2
−6
⋅
−2 = −2 (0) + b
−2 = b
4
6
3
4
y = 2x − 1
y = −2x + 2
−4
3
b. m = —2 , so the perpendicular line will have a slope of
2
m = −—3 .
2
y = −—3x + b
−6
6
y = −2x − 2
−4
⋅
1 = —32 0 + b
1=b
2
The line y = −—3x + 1 is perpendicular to the line
y = —32x − 1. The slopes are opposite reciprocals and
have a product of −1.
2
y = −3x + 1
4
−6
6
3
y = 2x − 1
−4
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Geometry
Worked-Out Solutions
85
Chapter 3
f. m = −2, so the perpendicular line will have a slope of
m = —12.
y = —12 x
0 = —12
+b
⋅ (4) + b
0=2+b
−2 = b
The line y = —12x − 2 is perpendicular to the line
y = 2x + 2. The slopes are opposite reciprocals and have
a product of −1.
−5 = b
6
The line y = 3x − 5 is parallel to the line y = 3x + 2
through the point (1, −2).
−4
(
3
)
2. a. For the blue line, use the points (0, 3) and −—2 , 0 to write
the equation.
0−3
−3
slope = — = — = −3
3
3
−—2 − 0 −—2
The y-intercept is 3.
⋅(
)
2
−— = 2
3
y=
−2 =
1
−—3x + b
1
−—3 (1) +
1
−—3 + b
⋅
b
−6 = −1 + 3b
−5 = −3b
y = 2x + 3
For the red line, use the points (1, 0) and (0, −2) to write
the equation.
−2 − 0 −2
slope = — = — = 2
0−1
−1
The y-intercept is −2.
y = mx + b
3.5 Monitoring Progress (pp. 156–159)
4−3 1
—=—
=—
slope of AB
8−1 7
y = 2x − 2
−4
b. For the blue line, use the points (0, −3) and
⋅(
1
The line y = −—3x − —53 is perpendicular to the line y = 3x + 2
through the point (1, −2).
from point A to point B.
6
write the equation.
0 − (−3)
3
slope = — = — = 3
3
3
−—2 − 0
−—2
The y-intercept is −3.
5
—3 = b
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 —45 of the way
4
y = 2x + 3
−6
(
3
−—2,
)
0 to
To find the coordinates of P, add —45 (or 0.8) of the run to the
x-coordinate and add —45 (or 0.8) of the rise to the y-coordinate.
⋅
⋅
run = 7 0.8 = 5.6
rise = 1 0.8 = 0.8
)
2
−— = −2
3
x = 1 + 5.6 = 6.6
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
3
segment into 3 + 7, or 10 congruent parts. P is —
of the way
10
y = mx + b
y = −2x − 3
from point A to point B.
For the red line, use the points (−4, 0) and (0, 2) to write
the equation.
2 1
2−0
slope = — = — = —
0 − (−4) 4 2
The y-intercept is 2.
4
y = −2x − 3
−8
4
Geometry
Worked-Out Solutions
5−1 4
—=—
=—
slope of AB
4+2 6
3
To find the coordinates of P, add —
(or 0.3) of the run
10
3
to the x-coordinate and add —
(or
0.3)
of the rise to the
10
y-coordinate.
⋅
⋅
run = 6 0.3 = 1.8
rise = 4 0.3 = 1.2
x = −2 + 1.8 = −0.2
y = 1 + 1.2 = 2.2
So, the point is P(−0.2, 2.2).
1
y = 2x + 2
86
1
The slope is 3. The slope of the perpendicular line is −—3 .
−2 =
y = mx + b
1
y = —x + 2
2
y = 3x + b
⋅
1
y = 2x − 2
y = mx + b
4. The slope is 3.
−2 = 3 + b
y = −2x + 2
y = 2x − 2
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.
−2 = 3 (1) + b
4
−6
3. For a line parallel to a given line, the slopes will be the
−4
Copyright © Big Ideas Learning, LLC
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Chapter 3
4
2
4
The slope of line b is — = 2.
2
3
The slope of line c is — = 3.
1
1
−1 − 0
The slope of line d is — = −—.
3
0+3
3. The slope of line a is — = 2.
7. The slope of y = −2x is −2, so the line perpendicular to
y = −2x will have a slope of —12 .
⋅
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
5 = 3(1) + b
Find the point of intersection.
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
1
m = −—3
5=
5=
16
—
3
1
−—3 x + b
1
−—3(1) + b
1
−—3 + b
=b
1
16
Because m = −—3 and b = —
, an equation of the line is
3
1
16
y = −—3 x + —
.
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 = 2x
Equation 1
13
1
y = —x + — Equation 2
2
2
13
1
−2x = —x + —
2
2
−4x = x + 13
−5x = 13
13
−5
—x = —
5
5
13
x = −—
5
13
26
y = −2 −— = —
5
5
( )
(
(
)
13 26
Find the distance from (−1, 6) to −—
,— .
5 5
———
13
√( − 5 − (−1) ) + ( 265 − 6 )
−13 + 5
= √(
) + ( 26 −5 30 )
5
−8
−4
= √(
+(
5 )
5 )
distance =
2
—
—
4 = −6 + b
2
—
2
—
2
——
—
10 = b
The line perpendicular to y = x + 4 is y = −x + 10.
Find the point of intersection.
x + 4 = −x + 10
2x + 4 = 10
)
13 26
So, the perpendicular lines intersect at −—
,— .
5 5
———
y = −x + b
Equation 1
Equation 2
2
—
—
√
2
—
√
64 16
80
= — + — = — ≈ 1.79 units
25 25
25
3.5 Exercises (pp. 160–162)
Vocabulary and Core Concept Check
x + 4 = −x + 10
1. A directed line segment AB is a segment that represents
2x + 4 = 10
moving from point A to B.
2x = 6
2
13 = 2b
13
—=b
2
13
The line perpendicular to y = 2x is y = —12 x + —
.
2
y = 3x + b
y=
1
y = —x + b
2
1
6 = — (−1) + b
2
1
6 = −— + b
2
12 = −1 + 2b
2. Two lines are perpendicular if the product of their slopes
6
—2 x = —2
equals −1.
x=3
y=3+4=7
So, the perpendicular lines intersect at (3, 7).
Find the distance from (6, 4) to (3, 7).
——
——
distance = √(3 − 6)2 + (7 − 4)2 = √ (−3)2 + (3)2
—
—
= √ 9 + 9 = √18 ≈ 4.24 units
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Geometry
Worked-Out Solutions
87
Chapter 3
Monitoring Progress and Modeling with Mathematics
3. In order to divide the segment in the ratio 1 to 4, partition the
segment into 1 + 4, or 5 congruent parts. P is —15 of the way
from point A to point B.
−2 − 0 −2 2
—=—
=—=—
slope of AB
3−8
−5 5
To find the coordinates of P, add —15 (or 0.2) of the run to the
x-coordinate and add —15 (or 0.2) of the rise to the y-coordinate.
⋅
⋅
run = −5 0.2 = −1
rise = −2 0.2 = −0.4
x=8−1=7
y = 0 − 0.4 = −0.4
So, the point is P(7, −0.4).
4. In order to divide the segment in the ratio 3 to 2, partition the
segment into 3 + 2, or 5 congruent parts. P is —35 of the way
from point A to point B.
1 − (−4) 1 + 4 5
—=—
=—=—
slope of AB
6 − (−2) 6 + 2 8
To find the coordinates of P, add —35 (or 0.6) of the run to the
x-coordinate and add —35 (or 0.6) of the rise to the y-coordinate.
⋅
⋅
run = 8 0.6 = 4.8
rise = 5 0.6 = 3
x = −2 + 4.8 = 2.8
y = −4 + 3 = −1
So, the point is P(2.8, −1).
5. In order to divide the segment in the ratio 5 to 1, partition the
segment into 5 + 1, or 6 congruent parts. P is —56 of the way
from point A to point B.
−3 − 6 −9
—=—
=—=3
slope of AB
−2 − 1 −3
To find the coordinates of P, add —56 of the run to the
x-coordinate and add —56 of the rise to the y-coordinate.
5
x = 1 − 2.5 = −1.5
run = − 3 — = −2.5
6
5
y = 6 − 7.5 = −1.5
rise = −9 — = −7.5
6
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 —28 = —14 of the
way from point A to point B.
−6
3
−4 − 2
—=—
= — = −—
slope of AB
5 − (−3)
8
4
To find the coordinates of P, add —14 (or 0.25) of the run
to the x-coordinate and add —14 (or 0.25) of the rise to the
y-coordinate.
⋅
run = 8 0.25 = 2
rise = −6 0.25 = −1.5
⋅
x = −3 + 2 = −1
y = 2 − 1.5 = 0.5
So, the point is P(−1, 0.5).
6−4 2 1
5+1 6 3
2−1 1
The slope of line b is — = —.
3+1 4
0+2 2 1
The slope of line c is — = — = —.
3+3 6 3
8
6+2
The slope of line d is — = — = −4.
1 − 3 −2
7. The slope of line a is — = — = —.
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.
1
−1
2−3
4
2 − (−2)
4
1
−2 − 0 −2
The slope of line b is — = — = −—.
3
3+3
6
4+2 6
The slope of line c is — = — = 3.
2−0 2
6
6−0
The slope of line d is — = — = 3.
0 − (−2) 2
8. The slope of line a is — = — = −—.
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.
4−0 6 2
7−1 4 3
3
6
6−0
Line 2 has a slope of — = — = −—.
2
3 − 7 −4
3
2
The product of the slopes is — −— = −1.
2
3
9. Line 1 has a slope of — = — = —.
⋅(
)
Therefore, the two lines are perpendicular by the Slopes of
Perpendicular Lines Theorem (Thm. 3.14).
−3
−2 − 1
−7 − (−3) −4
4 − (−1) 5
Line 2 has a slope of — = —.
8−2
6
3
4
10. Line 1 has a slope of — = — = —.
The slopes are not equal nor do their products equal −1. So,
the lines are neither perpendicular nor parallel.
4
7−3
−5 − (− 9) 4
−4
2−6
Line 2 has a slope of —— = — = −1.
−7 − (−11)
4
11. Line 1 has a slope of — = — = 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).
2
4
9−5
9
−8 − 10 −18
2
−6 − (−4) −2
Line 2 has a slope of — = — = −—.
9
11 − 2
9
12. Line 1 has a slope of — = — = −—.
The slopes are equal. Therefore, the two lines are parallel by
the Slopes of Parallel Lines Theorem (Thm. 3.13).
88
Geometry
Worked-Out Solutions
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Chapter 3
13. slope of the parallel line = −2
16. −x + 2y = 12
y = −2x + b
2y = x + 12
−1 = −2 0 + b
y = —12x + 6
⋅
slope of the parallel line = —12
−1 = 0 + b
−1 = b
y = —12x + b
Because m = −2 and
b = −1, an equation of the
line is y = −2x − 1.
3
y = −2x + 3
2
1
−2−1
1
3 x
y = −2x − 1 −2
1
5
14. slope of the parallel line = —
1
y = —x + b
5
1
8=— 3+b
5
40 = 3 + 5b
⋅
0 = —12 4 + b
y
0=2+b
−2 = b
Because m = —12 and b = −2,
an equation of the line
is y = —12x − 2.
7
5
4
3
−4 −3 −2−1
−3
37 = 5b
+
4 x
1
y = 2x − 2
The slope of the perpendicular line is
⋅
− 9 m = −1
8
6
5
4
3
2
37
.
—
5
−2−1
y
1
y = 5x +
y=
1
5x
+
37
5
4
5
m = —19 .
y = —19 x + b
⋅
0 = —19 0 + b
b=0
Because m = —19 and b = 0, an equation of the line is
y = —19 x.
1 2 3 4 5 x
8
7
6
−2
15. The slope is undefined, because x = −5 is a vertical line. So,
the line parallel to x = −5 must be vertical and go through
(−2, 6). Therefore, the equation is x = −2.
y
y = −9x − 1
1
y = 9x
x = −2
−1
6
1 2
17. The slope of the line is −9.
37 5b
5
5
37
=
b
—
5
37
Because m = —15 and b = —
,
5
an equation of the line is
—=—
y=
1
y = 2x + 6
2
1
⋅
1
—5 x
y
y
1 2 3 4 5 x
−2
5
4
3
18. The slope of y = −3 is 0, so the line perpendicular to
y = −3 has an undefined slope. Therefore, the equation is
x = 4.
2
1
−6
x = −5
−4 −3
−1
x
4
3
−2
2
1
−4 −3 −2−1
y
x=4
1 2 3
5 x
−2
y = −3
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−4
−5
−6
Geometry
Worked-Out Solutions
89
Chapter 3
19. y − 4 = −2(x + 3)
21. The slope of y = 3x is 3, so the line perpendicular to y = 3x
1
will have a slope of −—3 .
y − 4 = −2x − 6
1
y = −—3 x + b
y = −2x − 6 + 4
1
7 = —13 + b
The slope of the line is −2.
The slope of the perpendicular line is
21 = 1 + 3b
−2 m = −1
20 = 3b
⋅
m=
20
=b
—
3
1
—2 .
1
20
The line perpendicular to y = 3x is y = −—3 x + —
.
3
y = —12x + b
⋅
Find the point of intersection.
3 = —12 2 + b
y = 3x
3=1+b
y=
2=b
5
4
3
−2
−3
−4
−5
+
Equation 2
20
—
3
10x = 20
10
20
x=—
—
10
10
x=2
⋅
1
−4 −3 −2−1
+
1
−—3 x
20
—
3
9x = −x + 20
y
1
Equation 1
1
−—3x
3x =
Because m = —12 and b = 2, an equation of the line is
y = —12x + 2.
y=3 2=6
y = 2x + 2
So, the perpendicular lines intersect at (2, 6).
1 2 3 4 5 x
Find the distance from (−1, 7) to (2, 6).
y = −2x − 2
——
—
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
20. 3x − 5y = 6
y = x − 6 will have a slope of −1.
−5y = −3x + 6
6
−3
−5
—y = —x + —
−5
−5
−5
6
3
y = —x − —
5
5
The slope of the line is —35 .
y = −1x + b
⋅
−3 = −1 (−9) + b
−3 = 9 + b
−12 = b
The line perpendicular to y = x − 6 is y = −x − 12.
The slope of the perpendicular line is
3
— m = −1
5
5
m = −—.
3
5
y = −—x + b
3
3y = −5x + 3b
Find the point of intersection.
⋅
y=x−6
Equation 1
y = −x − 12
Equation 2
x − 6 = −x − 12
2x = −6
−6
2
—x = —
2
2
x = −3
⋅
3(0) = −5 (−8) + 3b
0 = 40 + 3b
40
−— = b
3
y = −3 − 6 = −9
40
Because m = −—53 and b = −—
3,
an equation of the line is
5
40
y = −—3x − —
.
3
2
−4−2
So, the perpendicular lines intersect at (−3, −9).
y
2 4 6 8 x
3
6
5
Geometry
Worked-Out Solutions
———
—
−6
−8
—
—
= √62 + (−6)2 = √36 + 36 = √ 72 ≈ 8.49 units
5
−14
−16
Find the distance from (−9, −3) to (−3, −9).
distance = √ (−3 − (−9))2 + (−9 − (−3))2
−4
y = 5x −
90
⋅
7 = −—3 (−1) + b
y = −2x − 2
y = −3 x −
40
3
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Chapter 3
23. 5x + 2y = 4
24. −x + 2y = 14
2y = −5x + 4
2y = x + 14
2y −5x 4
—=—+—
2
2
2
−5
y = —x − 2
2
5
The slope of 5x + 2y = 4 is −—2, so the line perpendicular to
5x + 2y = 4 will have a slope of —25.
2
y = —x + b
5
2
−21 = — 15 + b
5
−21 = 6 + b
⋅
−27 = b
The line perpendicular to 5x + 2y = 4 is y = —25 x − 27.
y = −2x + b
5 = −2
⋅ ( −—4 ) + b
1
1
5=—+b
2
10 = 1 + 2b
9
b=—
2
5
y = −—x + 2 Equation 1
2
2
y = —x − 27 Equation 2
5
The line perpendicular to −x + 2y = 14 is y = −2x + —92 .
5
2
−—x + 2 = —x − 27
2
5
5
2
−—x = —x − 29
2
5
25
4
−—x − —x = −29
10
10
29
−—x = −29
10
10
29
10
−— −—x = −29 −—
29
10
29
x = 10
⋅(
)
The slope of −x + 2y = 14 is —12 , so the line perpendicular to
−x + 2y = 14 will have a slope of −2.
9 = 2b
Find the point of intersection.
⋅(
x 14
2
2
1
y = —x + 7
2
2y
2
—=—+—
Find the point of intersection.
1
y = —x + 7
2
Equation 1
9
Equation 2
y = −2x + —
2
9
1
—x + 7 = −2x + —
2
2
x + 14 = −4x + 9
)
5x = −5
x = −1
⋅
y = −23
1
y = — (−1) + 7
2
1 14
y = −— + —
2
2
13
y=—
2
So, the perpendicular lines intersect at (10, − 23).
13
So, the perpendicular lines intersect at −1, —
.
2
⋅
5
y = −— 10 + 2
2
y = −25 + 2
Find the distance from (15, −21) to (10, −23).
(
Find the distance from
——
distance =
= √(−5)2 + (−2)2
—
—
= √25 + 4 = √ 29 ≈ 5.39 units
)
) (
)
13
5 to −1, —
.
2
√( ( ) ) (
———
———
distance = √(10 − 15)2 + (−23 − (−21))2
(
1
−—4 ,
2
1
−1 − −—
4
13
+ —−5
2
√( ) ( )
√( ) ( ) √
)
2
———
=
−4
4
1
4
—+—
2
13 10
+ —−—
2
2
——
2
—
−3 2
3 2
9
9
+ — = —+—
4
2
16 4
—
—
36
9
45
= — + — = — ≈ 1.68 units
16 16
16
=
√
—
√
25. Because the slopes are opposites but not reciprocals, their
product does not equal −1. Lines 1 and 2 are neither parallel
nor perpendicular.
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Geometry
Worked-Out Solutions
91
Chapter 3
26. Parallel lines have the same slope, not the same y-intercept.
y = 2x + 1,
(3, 4)
8
8−0
slope = — = — = 1
1 − (−7) 8
4 = 2(3) + b
−2 = b
The slope of the perpendicular line is −1.
The line y = 2x − 2 is parallel to the line y = 2x + 1.
( −42+ 4 3 −2 1 )
27. midpoint = —, — = (0, 1)
1
−4
−1 − 3
slope = — = — = −—
2
4 − (−4)
8
The slope of the perpendicular line is 2.
y = 2x + b
y = −1x + b
⋅
4 = −1 (−3) + b
4=3+b
1=b
Because m = − 1 and b = 1, the equation of the
— is y = −x + 1.
perpendicular bisector of PQ
31. In order to divide the segment in the ratio 1 to 4, partition the
segment into 1 + 4, or 5 congruent parts. P is —15 of the way
⋅
1=2 0+b
1=b
( −72+ 1 0 +2 8 ) ( −62 82 )
30. midpoint = —, — = —, — = (−3, 4)
from point A to point B.
Because m = 2 and b = 1, the equation of the perpendicular
— is y = 2x + 1.
bisector of PQ
−5 + 3 −5 + 3
28. midpoint = —, — = (−1, −1)
2
2
3 − (−5) 8
slope = — = — = 1
3 − (−5) 8
(
)
The slope of the perpendicular line is −1.
y = −1x + b
2 − (−2) 4
slope = — = —
5 − (−4) 9
To find the coordinates of P, add —15 (or 0.2) of the run to the
x-coordinate and add —15 (or 0.2) of the rise to the y-coordinate.
⋅
⋅
run = 9 0.2 = 1.8,
rise = 4 0.2 = 0.8,
x = −4 + 1.8 = −2.2
y = −2 + 0.8 = −1.2
The point on the graph that represents the school is
(−2.2, −1.2)
—
⋅
1
2
6−4
2
2 − 6 −4
4−1 3
—
slope of RS = — = — = 3
6−5 1
1
1 − 3 −2
—
slope of ST = — = — = −—
2
5−1
4
6−3 3
—
slope of QT = — = — = 3
2−1 1
32. slope of QR = — = — = −—
−1 = −1 (−1) + b
−1 = 1 + b
−2 = b
Because m = −1 and b = −2, the equation of the
— is y = −x − 2.
perpendicular bisector of PQ
( 0 +2 6 2 −2 2 )
29. midpoint = —, — = (3, 0)
2
−2 − 2 −4
slope = — = — = −—
3
6−0
6
The slope of the perpendicular line is —32 .
3
y = —x + b
2
3
0=— 3+b
2
9
0=—+b
2
9
−— = b
2
⋅
9
Because m = —32 and b = −—2 , the equation of the
— is y = —3x − —9.
perpendicular bisector of PQ
2
2
— RS
— because they
Quadrilateral QRST is a parallelogram. QT
—
—
have the same slope (m = 3), and ST QR because they have
1
the same slope m = −—2 .
(
)
—
8−6 2
5−0 5
−1 − 8 −9
—
slope of MN = — = — = 9
4−5
−1
7
6+1
—
slope of LN = — = —
0 − 4 −4
33. slope of LM = — = —
Triangle LMN is not a right triangle because the slopes of
2 7
the sides are —, −—, and 9. No combination of the products
5 4
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.
92
Geometry
Worked-Out Solutions
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Chapter 3
2
38. The angles of the quadrilateral are all right angles because
35. Train tracks: y = −—3 , P(2, 2)
The slope of the perpendicular line is
3
—2 .
y = —32x + b
⋅
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
2 = —32 2 + b
the ratio YP to PX is 5 to 3. This is the same point P in
both cases.
2=3+b
−1 = b
Because m = —32 and b = −1, the equation of the line that
represents the new road is y = —32x − 1.
36. The distance between the gazebo and the nature trail
is 42.4 feet.
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
− 4, gazebo: (−6, 4)
4x − 4x − 9 = 1
The slope of the perpendicular line is −3.
−9 = 1
Nature trail: y =
1
—3 x
−9 ≠ 1
y = −3x + b
⋅
4 = −3 (−6) + b
Because there is no solution, the lines do not intersect and
are, therefore, parallel.
4 = 18 + b
b. Solve the second equation for y.
−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
Substitute the result for y in the first equation.
Equation 1
3(2x − 18) + 4x = 16
1
⋅
—3 x − 4 = −3x − 14
⋅
⋅
⋅
3 —13x − 3 4 = 3 (−3x) − 3 14
x − 12 = −9x − 42
−y = −2x + 18
y = 2x − 18
Equation 1
y = −3x − 14
2x − y = 18
6x − 54 + 4x = 16
10x − 54 = 16
10x = 70
10x − 12 = −42
x=7
10x = −30
⋅
y = 2x − 18
x = −3
y = —13 (−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.
⋅
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
infinitely many solutions, and the lines are the same.
⋅
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.
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Geometry
Worked-Out Solutions
93
Chapter 3
42. Given ax + by = 0,
Find the distance between (x0, y0) and
(x0, y0)
(
ax + by = 0
by = −ax
a
y = −—x
b
a
The slope is −—.
b
b
The perpendicular line has slope —.
a
Let c be the y-intercept.
b
y = —x + c
a
b
y0 = —x0 + c
a
b
y0 − —x0 = c
a
ay0 − bx0
—=c
a
ay0 − bx0
b
y = —x + —
a
a
Find the point of intersection of the line ax + by = 0 and the
perpendicular line.
ay0 − bx0
b
y = —x + —
a
a
ay
−
bx0
a
0
b
−—x = —x + —
a
b
a
ax bx ay0 − bx0
−— = — + —
a
b
a
ax bx ay0 − bx0
−— − — = —
a
b
a
−
bx0
ay
ax bx
0
− —+— =—
a
b
a
ay0 − bx0
a2x + b2x
− — =—
a
ab
2
2
−
bx0
ay
a +b
0
− — x=—
a
ab
ay0 − bx0
ab
x = — −—
a
a2 + b2
ay0 − bx0
b
x = — −—
1
a2 + b2
(
)
)
(
((
))
(
(
)(
)(
b(ay0 − bx0)
x = −——
a 2 + b2
(
)
)
)
(
(
a (ay0 − bx0)
y=— —
1 a2 + b2
(
a(ay0 − bx0)
y = ——
a2 + b2
94
Geometry
Worked-Out Solutions
b(ay0 − bx0)
x0 − −——
a2 + b2
2
a(ay0 − bx0)
+ y0 − ——
a2 + b2
)
2
——————
=
√(
√(
√(
x0(a2 + b2) + b(ay0 − bx0)
2
———
+
a2
b2
)
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)
) (
(y0b2 + abx0)(y0b2 + abx0)
——— + ———
(a2 + b2)2
(a2 + b2)2
)
——————
=
√(
√(
a2(x 2a2 + 2x aby + b2y 2)
(a + b )
) (
b2(y 2b2 + 2abx y + a2x 2)
(a + b )
0
0
0
0
0
0 0
0
+ ———
———
2
2 2
2
2 2
)
——————
=
a2(a2x02 + 2abx0y0 + b2y02)
) (
b2(a2x02 + 2abx0y0 + b2y02)
——— + ———
(a2 + b2)2
√
√
√
(a2 + b2)2
———
(a2 + b2)(a2x02 + 2abx0y0 + b2y02)
= ———
(a2 + b2)2
———
(a2x02 + 2abx0y0 + b2y02)
= ———
(a2 + b2)
——
((ax0 + by0)2) ∣ ax0 + by0 ∣
= —— = —
—
(a2 + b2)
√a2 + b2
So, the distance between the line ax + by = 0 and the
ay0 − bx0 ∣ ax0 + by0 ∣
b
perpendicular line y = —x + — is —
— .
a
a
√a2 + b2
Sample answer: Use the line and point from Monitoring
Progress Exercise 7 on page 159.
∣ (2)(−1) + (1)(6) ∣
∣ −2 + 6 ∣
∣4∣
4
√22 + 12
√5
√5
√5
−2 − k
−2 − k
−7 − (−1) −7 + 1
−2 − k −(2 + k) (2 + k)
=—=—=—
−6
−6
6
)
)
(
distance =
43. slope = — = —
)
b(ay0 − bx0) a(ay0 − bx0)
Point of intersection: −——
, ——
a2 + b2
a2 + b2
√( (
—————
=—
=—
——
—
— = —
— ≈ 1.79 units
—
a
y = −—x
b
a b(ay0 − bx0)
y = −— −——
b
a2 + b2
)
)) (
) (
) (
)
b(ay0 − bx0) a(ay0 − bx0)
−——
, —— .
a2 + b2
a2 + b2
The slope of the parallel line is 1.
2−k
1=—
6
6=2+k
4=k
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)
Chapter 3
−2
7−k
0−2
7−k
49. If lines x and y are vertical lines and they are cut by any
44. slope = — = —
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).
The slope of the given line is 1. The slope of the
perpendicular line is −1.
−2
−1 = —
7−k
−1(7 − k) = −2
50. If lines x and y are horizontal, then by definition 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.
−7 + k = −2
k=5
51. By definition, the x-axis is perpendicular to the y-axis. Let
45. Using points A(3, 2) and B(6, 8), find the coordinates of
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.
point P that lies beyond point B along ⃗
AB so that the ratio of
AB 3
AB to BP is 3 to 2. In order to keep the ratio, — = —,
BP 2
2
solve this ratio for BP to get BP = —AB. Next, find the rise
3
and run from point A to point B. Leave the slope in terms of
8 − 2 6 rise
rise and run and do not simplify. mAB = — = — = —.
6 − 3 3 run
2
Add — of the run to the x-coordinate of B, which is
3
2
2
— 3 + 6 = 8. Add — of the rise to the y-coordinate of B,
3
3
2
which is — 6 + 8 = 12. So, the coordinates of P are (8, 12).
3
⋅
Maintaining Mathematical Proficiency
52.
7
6
5
4
3
2
⋅
1
46. The slope of the perpendicular line is −—2 .
y=
1
−—2x
53.
y
A(3, 6)
1
−3 −2−1
−1
Use the y-intercept of y = 2x + 5, (0, 5).
1 2 3 x
−2
−3
−4 B(0, −4)
−5
1
+b
y
3
2
1 2 3 x
1
y = −—2x + b
1
54.
⋅
5 = −—2 0 + b
5=b
4
3
2
55.
y
1
1
1
y = −—2x + 5
−1
1
Find the intersection of the perpendicular lines y = −—2 x + 5
and y = 2x.
y
2
−3 −2−1
C(5, 0)
1 2 x
D(−1, −2)
1 2 3 4 5 x
−3
−2
1
2x = −—2x + 5
56.
4x = −x + 10
y=x+9
5x = 10
x=2
x
⋅
y=2 2=4
——
distance = √(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.
⋅
⋅
⋅
−2
−1
0
−2 + 9 = 7
−1 + 9 = 8
0+9=9
1
2
y = x + 9 1 + 9 = 10
Find the distance between (0, 5) and (2, 4).
⋅
x
57.
x
−2
y = x − —34
x
y=x−
2 + 9 = 11
−1
11
7
−2 − —34 = −—
−1 − —34 = −—4
4
1
3
—4
0
3
—4
3
0 − —34 = −—4
2
1− =
1
—4
3
5
2 − —4 = —4
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.
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Geometry
Worked-Out Solutions
95
Chapter 3
3.4 – 3.5 What Did You Learn? (p. 163)
1. You can find the distance between two lines only if the two
lines are parallel. If you choose a point on one line and find
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 setting to mark a point on the
other perpendicular line that is the same distance from
the other endpoint. Connect these two points to construct
the fourth segment of the rectangle. This segment should
be congruent and parallel to the original segment and
perpendicular to the other two constructed segments.
7. 58° + 2x° = 180°
x = 61
2y° = 58°
Corresponding Angles Theorem
(Thm. 3.2)
y = 29
The values are x = 61 and y = 29.
8. (6x + 32)° = 116°
Chapter 3 Review (pp. 164–166)
1. All angles may or may not be right angles, and lines that
appear perpendicular to ⃖⃗
QR are ⃖⃗
QL, ⃖⃗
RM, ⃖⃗
QP, and ⃖⃗
RN.
⃖⃗ are ⃖⃗
2. The lines that appear parallel to QR
JK , ⃖⃗
ML, and ⃖⃗
NP.
3. The lines that appear skew to ⃖⃗
QR are ⃖⃗
KP, ⃖⃗
KL, ⃖⃗
JN , and ⃖⃗
JM.
4. The plane that appears parallel to LMQ is plane JKPN, which
can be defined by any three of these four vertices.
5. 35° + x° = 180°
Definition of supplementary angles
x = 145
y = 35
x = 14
(5y − 21)° + (6x + 32)° = 180°
⋅
The values are x = 145 and y = 35.
Consecutive Interior Angles Theorem
(Thm. 3.4)
y = 132
(5x − 17)° = 48°
Definition of
supplementary angles
5y − 21 + 6 14 + 32 = 180
5y − 21 + 84 + 32 = 180
5y + 95 = 180
5y = 85
y = 17
The values are x = 14 and y = 17.
9. By the Consecutive Interior Angles Converse (Thm. 3.8),
m n when the marked angles are supplementary.
x° + 73° = 180°
x = 107
The lines are parallel when x = 107.
10. By the Alternate Exterior Angles Converse (Thm. 3.6), m n
when the marked angles are congruent.
147° = (x + 14)°
133 = x
The lines are parallel when x = 133.
Corresponding Angles Theorem
(Thm. 3.2)
6. 48° + y° = 180°
Alternate Exterior Angles Theorem
(Thm. 3.3)
6x = 84
3. Because the distance from your house to the school is
one-fourth of the distance from the school to the movie
theater, you have to use five congruent segments. Four of
the segments are between your school and the movie theater
and one is between your house and your school.
Definition of supplementary angles
2x = 122
11. Use the Vertical Angles Congruence Theorem (Thm. 2.6) and
the Consecutive Interior Angles Converse (Thm. 3.8).
3x° + (2x + 20)° = 180°
5x + 20 = 180
5x = 160
x = 32
Alternate Interior Angles Theorem
(Thm. 3.2)
5x = 65
x = 13
The values are x = 13 and y = 132.
The lines are parallel when x = 32.
12. Use the Corresponding Angles Converse (Thm. 3.5).
(7x − 11)° = (4x + 58)°
3x = 69
x = 23
The lines are parallel when x = 23.
13. x y; Because x ⊥ z and y ⊥ z, lines x and y are parallel by the
Lines Perpendicular to a Transversal Theorem (Thm. 3.12).
96
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Chapter 3
14. none; The only thing that can be concluded in this diagram
is that x ⊥ z and w ⊥ y. In order to say that lines are parallel,
you need to know something about both of the intersections
between the two lines and a transversal.
15. ℓ m n, a b; Because a ⊥ n and b ⊥ n, lines a and b are
parallel by the Lines Perpendicular to a Transversal Theorem
(Thm. 3.12). Because m ⊥ a and n ⊥ a, lines m and n are
parallel by the Lines Perpendicular to a Transversal Theorem
(Thm. 3.12). Because ℓ⊥ b and n ⊥ b, linesℓand n are
parallel by the Lines Perpendicular to a Transversal Theorem
(Thm. 3.12). Because ℓ n and m n, linesℓand m are
parallel by the Transitive Property of Parallel Lines
(Thm. 3.9).
16. a b; Because a ⊥ n and b ⊥ n, lines a and b are parallel by the
Lines Perpendicular to a Transversal Theorem (Thm. 3.12).
17. The slope of the parallel line is −1.
−1 = b
Because m = −1 and b = −1, an equation of the line is
y = −x − 1.
y=
5=
−1 =
+b
⋅6 + b
−1 = 3 + b
−4 = b
Because m = —12 and b = −4, the equation of the line is
y = —12 x − 4.
22. The slope of the perpendicular line is 2.
y = 2x + b
⋅
3=2 0+b
3=0+b
3=b
Because m = 2 and b = 3, an equation of the line is
y = 2x + 3.
23. The slope of the perpendicular line is −—4 .
−4 = −3 + b
1
—2 x
1
—2
y=
1
—2 x
1
—2
1
y = −x + b
18. The slope of the parallel line is
1
21. The slope of the perpendicular line is —2 .
1
—2 .
+b
⋅ (−6) + b
1
y = −—4 x
1
2 = −—4
+b
⋅8 + b
2 = −2 + b
4=b
1
Because m = −—4 and b = 4, an equation of the line is
1
y = −—4 x + 4.
24. The slope of the perpendicular line is −7.
5 = −3 + b
y = −7x + b
8=b
Because m = —12 and b = 8, an equation of the line is
y = —12x + 8.
19. The slope of the parallel line is 3.
y = 3x + b
⋅
5 = −7 (−1) + b
5=7+b
−2 = b
Because m = −7 and b = −2, an equation of the line
is y = −7x − 2.
⋅
0=3 2+b
0=6+b
−6 = b
Because m = 3 and b = −6, an equation of the line is
y = 3x − 6.
1
20. The slope of the parallel line is —3 .
y = —13x + b
⋅
−1 = —13 3 + b
−1 = 1 + b
−2 = b
Because m = —13 and b = −2, an equation of the line is
y = —13x − 2.
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Geometry
Worked-Out Solutions
97
Chapter 3
(
)
4
3
distance = √ ( − − (−2) ) + ( − 3 )
5
5
4
3 15
= √( − + 2 ) + ( − )
5
5
5
4 10
12
= √( − + ) + ( )
5
5
5
6
12
= √( ) + ( )
5
5
36 144
180
=
=√ +
√ 25
25
25
4 3
Find the distance from −—, — to (−2, 3).
5 5
25. The slope of the perpendicular line is 1.
y=x+b
———
⋅
−1 = 1 2 + b
2
—
−1 = 2 + b
2
—
——
2
—
−3 = b
The line perpendicular to y = −x + 4 is y = x − 3. Find the
point of intersection.
y = −x + 4
Equation 1
y=x−3
Equation 2
( )
—
2
2
—
2
—
—
—
—
—
(Thm. 2.6); y = 61 by the Alternate Exterior Angles
Theorem (Thm. 3.3).
( )
7
1
distance = √ ( − 2 ) + ( − (−1) )
2
2
7 4
1 2
= √( − ) + ( + )
2 2
2 2
3
3
9 9
18
≈ 2.12 units
= √( ) + ( ) = √ + =
2
2
4 4 √4
———
2
2
—
——
2
—
—
—
2
1. x = 61 by the Vertical Angles Congruence Theorem
7 1
Find the distance from —, — to (2, −1).
2 2
2
—
Chapter 3 Test (p. 167)
7 1
So, the perpendicular lines intersect at —, — .
2 2
—
—
≈ 2.68 units
−7 7
x=—=—
−2 2
7 6 1
7
y=—−3=—−—=—
2
2 2 2
—
2
——
—
−2x = −7
—
—
——
−x + 4 = x − 3
—
—
2
—
2
—
—
—
—
—
26. The slope of the perpendicular line is −2.
y = −2x + b
2. 8x° = 96°
Corresponding Angles Theorem (Thm. 3.1)
x = 12
96° + (11y + 7)° = 180° Linear Pair Postulate (Post. 2.8)
11y + 103 = 180
11y = 77
y=7
3. (8x + 2)° = 42°
Alternate Interior Angles Theorem
(Thm. 3.2)
8x = 40
⋅
3 = −2 (−2) + b
x=5
3=4+b
−1 = b
1
The line perpendicular to y = —x + 1 is y = −2x − 1. Find
2
the point of intersection.
y = −2x − 1 Equation 1
1
y = —x + 1 Equation 2
2
1
−2x − 1 = —x + 1
2
−4x − 2 = x + 2
42° + [6(2y − 3)]° = 180°
Consecutive Interior Angles
Theorem (Post. 3.4)
42 + 12y − 18 = 180
12y + 24 = 180
12y = 156
y = 13
−5x − 2 = 2
−5x = 4
−5x
−5
4
−5
4
4
x = — = −—
5
−5
—=—
y = −2
⋅ ( −—5 ) − 1 = —85 − —55 = —35
4
(
)
4 3
So, the perpendicular lines intersect at −—, — .
5 5
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Chapter 3
4. The slope of y = −x is −1, so the line perpendicular to
y = −x will have a slope of 1.
———
distance = √(−3 − 0)2 + (7 − (−2))2
——
y=x+b
= √(−3)2 + (9)2
4=3+b
= √9 + 81 = √ 90 ≈ 9.49 units
—
1=b
The line perpendicular to y = −x is y = x + 1.
y = −x
Equation 1
y=x+1
Equation 2
x=6
8. Use the Vertical Angles Congruence Theorem (Thm. 2.6) and
−2x = 1
1
−2
—x = —
−2
−2
1
x = −—
2
1
1
y = − −— = —
2
2
the Consecutive Interior Angles Converse (Thm. 3.8).
(11x + 33)° + (6x − 6)° = 180°
17x + 27 = 180
17x = 153
( )
x=9
(
)
)
1 1
So, the perpendicular lines intersect at −—, — .
2 2
1 1
Find the distance from (3, 4) to −—, — .
2 2
(
√( ( ) ) ( )
———
2
1
3 − −—
2
√( ) (
√( ) ( )
1
+ 4−—
2
2
——
=
1
2
6
2
—+—
2
8 1
+ —−—
2 2
—
=
=
2
7
2
—
7
+ —
2
49
4
49
4
)
2
—+—=
1
9. a. The slope of the parallel line is 2.
y = 2x + b
2 = 2(−5) + b
2 = −10 + b
12 = b
Because m = 2 and b = 12, an equation of the parallel
line is y = 2x + 12.
1
b. The slope of the perpendicular line is −—2 .
2
—
√
Alternate Exterior Angles Converse
(Thm. 3.7)
4x = 24
−x = x + 1
distance =
6. x = 97 by the Corresponding Angles Converse (Thm. 3.5).
7. 8x° = (4x + 24)°
Find the point of intersection.
—
1
√984 ≈ 4.95 units
—
—
1
5. The slope of y = —3 x − 2 is —3 , so the line perpendicular to
y = —13x − 2 will have a slope of −3.
y = −3x + b
y = −—2x + b
1
2 = −—2(−5) + b
2 = —52 + b
4 = 5 + 2b
−1 = 2b
1
⋅
7 = −3 (−3) + b
7=9+b
−—2 = b
1
Because m = −—2 and b = − —12, an equation of the
1
perpendicular line is y = −—2 x − —12.
−2 = b
The line perpendicular to y = —13 x − 2 is y = −3x − 2.
Find the point of intersection.
y = −3x − 2 Equation 1
y = —13x − 2
−3x − 2 =
Equation 2
1
—3 x
−2
−9x − 6 = x − 6
−10x − 6 = −6
−10x = 0
x=0
y=
( ⋅ 0 ) − 2 = −2
1
—3
So, the perpendicular lines intersect at (0, −2).
Find the distance from (−3, 7) to (0, −2).
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99
Chapter 3
1
10. a. The slope of the parallel line is −—3 .
y=
−9 =
−9 =
1
−—3x + b
1
−—3(−1) +
1
—3 + b
d. Find the distance from the meeting point, (150, 200), to
the subway, (450, 300).
———
distance = √ (450 − 150)2 + (300 − 200)2
b
——
= √ 3002 + 1002
—
= √ 100,000
−27 = 1 + 3b
≈ 316
−28 = 3b
28
−—
3 =b
1
28
Because m = −—3 and b = −—
3 , an equation of the parallel
1
28
line is y = −—3x − —
.
3
b. The slope of the perpendicular line is 3.
y = 3x + b
−9 = 3(−1) + b
−9 = −3 + b
−6 = b
Because m = 3 and b = −6, an equation of the
perpendicular line is y = 3x − 6.
11. The student assumes k ℓ and is trying to use the
Perpendicular Transversal Theorem (Thm. 3.11).
12. a. Line q passes through the points (100, 350) and (200, 50).
−300
50 − 350
The slope of line q is — = — = −3.
200 − 100
100
y = mx + b
y = −3x + b
50 = −3(200) + b
50 = −600 + b
650 = b
The equation of line q is y = −3x + 650.
b. Line p passes through the points (0, 150) and (450, 300).
300 − 150 150 1
The slope of line p is — = — = —.
450 − 0
450 3
y = mx + b
1
y = —x + b
3
1
150 = —(0) + b
3
150 = b
1
The equation of line p is y = —x +150.
3
c. Find the point of intersection of line q and line p.
y = −3x + 650
Equation 1
1
Equation 2
y = —x + 150
3
1
−3x + 650 = —x + 150
3
−9x + 1950 = x + 450
−10x + 1950 = 450
−10x = −1500
x = 150
The distance from the meeting point and the subway is
about 316 yards.
13. a. Sample answer: A pair of skew lines is ⃖⃗
AB and ⃖⃗
LM
because the lines are non-intersecting, non-coplanar, and
non-parallel.
b. Sample answer: A pair of perpendicular lines is ⃖⃗
EF and
⃖⃗
IJ because the lines intersect at a right angle.
⃖⃗ and ⃖⃗
c. A pair of paralel lines is CD
EF because the lines are
perpendicular to the same transversal.
d. A pair of congruent corresponding angles is ∠ 1 and ∠ 3
because the angles are corresponding and ⃖⃗
CD ⃖⃗
EF.
e. A pair of congruent alternate interior angles is ∠ 2 and ∠ 3
because the angles are alternate interior and ⃖⃗
CD ⃖⃗
EF.
Chapter 3 Standards Assessment (pp. 168–169)
1. Every point on the red arc in Step 1 is the same distance
from point A. Because the same compass setting is used,
every point on the red arc in Step 2 is the same distance from
point B as all of the points in the blue arc are from point A.
— ⊥ AB
— because the shortest distance from a point to
Also, CD
a line is the perpendicular segment that connects the point
— are
to that line. So, points C and D and every point on CD
equidistant from points A and B, which means that M is the
— by definition.
midpoint of AB
2. x + 2y = 10
2y = −x + 10
1
y = −—2 x + 5
1
a. The slope of the parallel line is −—2 .
y=
−5 =
1
−—2 x + b
1
−—2 (4) +
b
−5 = −2 + b
−3 = b
1
Because m = −—2 and b = −3, an equation of the parallel
line is y =
1
−—2x
− 3.
b. The slope of the perpendicular line is 2.
y = 2x + b
−1 = 2(2) + b
−1 = 4 + b
−5 = b
Because m = 2 and b = −5, an equation of the
perpendicular line is y = 2x − 5.
y = −3(150) + 650 = 200
The coordinates (150, 200) represent the meeting point.
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Chapter 3
3. a. The angles are supplementary angles because
44° + 136° = 180°.
b. The angles are adjacent angles because they have a
common side and a common vertex.
c. The angles are vertical angles because they are
non-adjacent and share a common vertex.
d. The angles are complementary angles because
150 + 150 400 + 100
midpoint = —, —
2
2
300 500
= —, — = (150, 250)
2
2
The midpoint of the line segment joining the two houses
is (150, 250).
(
(
)
)
b. School: (400, 300)
42° + 48° = 90°.
Find the distance from the midpoint to the school.
4. a. The length of the field is 360 feet.
———
distance = √(400 − 150)2 + (300 − 250)2
b. The perimeter of the field is
⋅
9. a. Friend’s house: (150, 400), your house: (150, 100)
⋅
—
2 160 + 2 360 = 1040 feet.
c. The area of the field is 160
⋅ 360 = 57,600 square feet.
The cost at $2.69 per square foot is
57,600 2.69 = $154,944. Because this is greater than
$150,000, the school does not have enough money.
⋅
= √2502 + 502
—
= √65,000
≈ 255 yd
You and your friend walk about 255 yards together.
5. Given ∠ 1 ≅ ∠ 3
Prove ∠ 2 ≅ ∠ 4
1
3
2
4
STATEMENTS
REASONS
1. ∠ 1 ≅ ∠ 3
1. Given
2. ∠ 1 ≅ ∠ 2
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. ∠ 2 ≅ ∠ 3
3. Transitive Property of Congruence
4. ∠ 3 ≅ ∠ 4
4. Vertical Angles Congruence
Theorem (Thm. 2.6)
5. ∠ 2 ≅ ∠ 4
5. Transitive Property of Congruence
6. yes; Because 141° + 39° = 180°, the marked angles are
supplementary. They are consecutive interior angels, so m n
by the Consecutive Angels Converse (Thm. 3.5).
7. D; Skew lines are lines that are non-coplanar,
non-intersecting, and non-parallel.
8. a. ∠ 4 ≅ ∠ 5 by the Alternate Interior Angles Theorem
(Thm. 3.2).
b. ∠ 2 ≅ ∠ 6 by the Corresponding Angles Theorem
(Thm. 3.1).
c. ∠ 1 ≅ ∠ 8 by the Alternate Exterior Angles Theorem
(Thm. 3.3).
d. m∠ 6 + m∠ 4 = 180° by the Consecutive Interior Angles
Theorem (Thm. 3.4).
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101