Transcript X - Ms. Williams – Math

Properties or Parallel Lines
GEOMETRY LESSON 3-1
(For help, go to page 24 or Skills Handbook, page 720.)
Solve each equation.
1. x + 2x + 3x = 180
2. (w + 23) + (4w + 7) = 180
3. 90 = 2y – 30
4. 180 – 5y = 135
Write an equation and solve the problem.
5. The sum of m 1 and twice its complement is 146. Find m 1.
6. The measures of two supplementary angles are in the ratio 2 : 3.
Find their measures.
3-1
Properties or Parallel Lines
GEOMETRY LESSON 3-1
Solutions
1. Combine like terms: 6x = 180; divide both sides by 6: x = 30
2. Combine like terms: 5w + 30 = 180; Subtract 30 from both sides: 5w = 150;
divide both sides by 5: w = 30
3. Add 30 to both sides: 120 – 2y; divide both sides by 2: 60 = y
4. Subtract 180 from both sides: – 5y = –45; divide both sides by –5: y = 9
5. The complement of 1 is 90 – m 1; m 1 + 2(90 – m 1) = 146; distribute:
m 1 + 180 – 2m = 146; combine like terms: 180 – m 1 = 146; subtract
180 from both sides: –m 1 = –34; multiply both sides by –1: m 1 = 34
6. Let 2x represent the measure of one of the angles and 3x represent the measure
of the other angle. Then 2x + 3x = 180; combine like terms: 5x = 180; divide
both sides by 5: x = 36. Then 2x = 72 and 3x = 108.0.
3-1
Properties or Parallel Lines
GEOMETRY LESSON 3-1
Use the diagram above. Identify which angle forms a pair of same-side
interior angles with 1. Identify which angle forms a pair of corresponding
angles with 1.
Same-side interior angles are on the same side of transversal t
between lines p and q.
4, 8, and 5 are on the same side of the transversal as
but only 1 and 8 are interior.
So
1 and
8 are same-side interior angles.
3-1
1,
Properties or Parallel Lines
GEOMETRY LESSON 3-1
(continued)
Corresponding angles also lie on the same side of the transversal.
One angle must be an interior angle, and the other must be an exterior angle.
The angle corresponding to 1 must lie in the same position relative to
line q as 1 lies relative to line p. Because 1 is an interior angle, 1 and
are corresponding angles.
3-1
5
Properties or Parallel Lines
GEOMETRY LESSON 3-1
Compare 2 and the vertical angle of 1. Classify the angles
as alternate interior angles, same–side interior angles, or
corresponding angles.
The vertical angle of
1 is between the parallel runway segments.
2 is between the runway segments and on the opposite side of
the transversal runway.
Because alternate interior angles are not adjacent and lie between
the lines on opposite sides of the transversal, 2 and the vertical angle
of 1 are alternate interior angles.
3-1
Properties or Parallel Lines
GEOMETRY LESSON 3-1
Which theorem or postulate gives the reason that m
3 and
m
3+m
3+m
2 are adjacent angles that form a straight angle.
2 = 180 because of the Angle Addition Postulate.
3-1
2 = 180?
Properties or Parallel Lines
GEOMETRY LESSON 3-1
In the diagram above,
|| m. Find m 1 and then m
2.
1 and the 42° angle are corresponding angles. Because
m 1 = 42 by the Corresponding Angles Postulate.
|| m,
Because 1 and 2 are adjacent angles that form a straight angle,
m 1 + m 2 = 180 by the Angle Addition Postulate.
If you substitute 42 for m 1, the equation becomes 42 + m
Subtract 42 from each side to find m 2 = 138.
3-1
2 = 180.
Properties or Parallel Lines
GEOMETRY LESSON 3-1
In the diagram above,
|| m. Find the values of a, b, and c.
a = 65
Alternate Interior Angles Theorem
c = 40
Alternate Interior Angles Theorem
a + b + c = 180
Angle Addition Postulate
65 + b + 40 = 180
b = 75
Substitution Property of Equality
Subtraction Property of Equality
3-1
Properties or Parallel Lines
GEOMETRY LESSON 3-1
Pages 118-121 Exercises
1. PQ and SR with
transversal SQ;
alt. int. s
2. PS and QR with
transversal SQ;
alt. int. s
3. PS and QR with
transversal PQ;
same-side int. s
4. PS and QR with
transversal SR;
corr. s
5.
6.
7.
1 and 2: corr. s
3 and 4: alt. int.
5 and 6: corr. s
8. alt. int.
s
1 and 2: same-side
int. s
3 and 4: corr. s
5 and 6: corr. s
1 and 2: corr. s
3 and 4: same-side
int. s
5 and 6: alt. int.
3-1
s
s
9. a. 2
b. 1
c. corr.
10. a. Def. of
b. Def. of right
c. Corr. s of i lines are
d. Subst.
e. Def. of right
f. Def. of
.
Properties or Parallel Lines
GEOMETRY LESSON 3-1
11. m 1 = 75 because
14. 70; 70, 110
corr. s of || lines are ;
m 2 = 105 because
15. 25; 65
same-side int. s of ||
lines are suppl.
16. 20; 100, 80
12. m 1 = 120 because
17. m 1 = m 3 = m 6 =
corr. s of || lines are ;
m 8 = m 9 = m 11
m 2 = 60 because
= m 13 = m 15 = 52;
same-side int. s of ||
m 2=m 4=m 5=
lines are suppl.
m 7 = m 10 = m 12
= m 14 = 128
13. m 1 = 100 because
same-side int. s of ||
lines are suppl.; m 2 =
70 because alt. int. s of
|| lines have = measure.
3-1
18. You must find the
measure of one . All s
that are vert., corr., or
alt. int. to that will have
that measure. All other s
will be the suppl. of that
measure.
19. two
20. four
21. two
22. four
23. 32
Properties or Parallel Lines
GEOMETRY LESSON 3-1
24. x = 76, y = 37,
v = 42, w = 25
25. x = 135, y = 45
26. The s labeled are
corr. s and should be
. If you solve
2x – 60 = 60 – 2x, you
get x = 30. This would
be impossible since
2x – 60 and 60 – 2x
would equal 0.
27. Trans means across or
over. A transversal cuts
across other lines.
28. Answers may vary.
Sample: E illustrates corr. s ( 1 and 3,
2 and 4) and same-side int. s ( 1
and 2, 3 and 4);
I illustrates alt. int. s ( 1 and 4,
2 and 3) and same-side int. s
( 1 and 3, 2 and 4).
29. a. alt. int.
s
b. He knew that alt. int.
30. a. 57
b. same-side int.
3-1
s
s
of || lines are
.
Properties or Parallel Lines
GEOMETRY LESSON 3-1
31. a. If two lines are || and cut by a transversal,
then same-side ext. s are suppl.
b. Given: a || b
Prove: 4 and 5 are suppl.
1. a || b (Given)
2. m 5 + m 6 = 180 ( Add. Post.)
3. 4
6 (Corr. s are )
4. m 5 + m 4 = 180 (Subst.)
5.
4 and
5 are suppl. (Def. of suppl.)
32. 1. a || b (Given)
2. 1
2 (Vert.
3. 2
3 (Corr.
4.
1
33. Never; the two planes do
not intersect.
34. Sometimes; if they are ||.
35. Sometimes; they may be
skew.
36. Sometimes; they may be ||.
37. D
38. G
are
s are
s
.)
.)
39. D
40. I
3 (Trans. Prop.)
3-1
Properties or Parallel Lines
GEOMETRY LESSON 3-1
41. [2] a. First show that 1
7. Then show
that 7
5. Finally, show that
1
5 (OR other valid solution plan).
b.
1
7 because vert. s are .
7
5 because corr. s of || lines are
. Finally, by the Trans. Prop. of ,
1
5.
42. 121
43. 59
44. 29.5
45. (0.5, 7)
46. (–0.5, 3.5)
[1] incorrect sequence of steps OR incorrect
logical argument
47. (3, 3)
48. add 4; 20, 24
49. multiply by –2; 16, –32
50. subtract 7; –5, –12
3-1
Properties or Parallel Lines
GEOMETRY LESSON 3-1
In the diagram below, m || n. Use the diagram for Exercises 1–5.
1. Complete:
2. Complete:
6
and
and
4 are alternate interior angles.
8 are corresponding angles.
4
3. Suppose that m
3 = 37. Find m
4. Suppose that m
1 = x + 12 and m
6. 143
5 = 3x – 36. Find x. 24
5. If a transversal intersects two parallel lines, then same-side
exterior angles are supplementary. Write a Plan for Proof.
Given: m || n Prove: 2 and 7 are supplementary.
Show that m 2 = m 6. Then show that m 6 + m 7 = 180,
and substitute m 2 for m 6.
3-1
Proving Lines Parallel
GEOMETRY LESSON 3-2
(For help, go to page 24 and Lesson 2-1.)
Solve each equation.
1. 2x + 5 = 27
2. 8a – 12 = 20
3. x – 30 + 4x + 80 = 180
4. 9x – 7 = 3x + 29
Write the converse of each conditional statement.
Determine the truth value of the converse.
5. If a triangle is a right triangle, then it has a 90° angle.
6. If two angles are vertical angles, then they are congruent.
7. If two angles are same-side interior angles, then they are supplementary.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Solutions
1. Subtract 5 from both sides: 2x = 22; divide both sides by 2: x = 11
2. Add 12 to both sides: 8a = 32; divide both sides by 8: a = 4
3. Combine like terms: 5x + 50 = 180; subtract 50 from both sides: 5x = 130; divide
both sides by 5: x = 26
4. Add –3x + 7 to both sides: 6x = 36; divide both sides by 6: x = 6
5. Reverse the hypothesis and conclusion: If a triangle has a 90° angle,
then it is a right triangle. By definition of a right triangle, it is true.
6. Reverse the hypothesis and conclusion: If two angles are congruent, they are
vertical angles. A counterexample is the congruent base angles of an isosceles
triangle, which are not vertical angles. The converse is false.
7. Reverse the hypothesis and conclusion: If two angles are supplementary, then
they are the same-side interior angles of parallel lines. A counterexample is
two angles that form a straight angle. The converse is false.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
If two lines and a transversal form alternate interior
angles that are congruent, then the two lines are parallel.
Given:
1
Prove:
|| m
2
Write the flow proof below of the Alternate Interior Angles
Theorem as a paragraph proof.
By the Vertical Angles Theorem, 3
1. 1
2, so 3
2 by the
Transitive Property of Congruence. Because 3 and 2 are corresponding
angles, || m by the Converse of the Corresponding Angles Postulate.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Use the diagram above. Which lines, if any, must be parallel if
are supplementary?
It is given that
3 and
The diagram shows that
3 and
2 are supplementary.
4 and
2 are supplementary.
Because supplements of the same angle are congruent
(Congruent Supplements Theorem), 3
4.
Because 3 and 4 are congruent corresponding angles, EC || DK
by the Converse of the Corresponding Angles Postulate.
3-2
2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Use the diagram above. Which angle would you use with 1 to prove
the theorem In a plane, if two lines are perpendicular to the same line,
then they are parallel to each other (Theorem 3-6) using the Converse
of the Alternate Interior Angles Theorem instead of the Converse of the
Corresponding Angles Postulate?
By the Vertical Angles Theorem,
2 is congruent to its vertical angle.
Because 1
2, 1 is congruent to the vertical angle of
the Transitive Property of Congruence.
2 by
Because alternate interior angles are congruent, you can use the vertical
angle of 2 and the Converse of the Alternate Interior Angles Theorem to
prove that the lines are parallel.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Find the value of x for which
|| m.
The labeled angles are alternate interior angles.
If || m, the alternate interior angles are congruent,
and their measures are equal.
Write and solve the equation 5x – 66 = 14 + 3x.
5x – 66 = 14 + 3x
5x = 80 + 3x
Add 66 to each side.
2x = 80
Subtract 3x from each side.
x = 40
Divide each side by 2.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Suppose that the top and bottom pieces of a picture frame are
cut to make 60° angles with the exterior sides of the frame. At what
angle should the two sides be cut to ensure that opposite sides of the
frame will be parallel?
In order for the opposite sides of the frame to be parallel,
same-side interior angles must be supplementary.
Two 90° angles are supplementary, so find an adjacent angle that,
together with 60°, will form a 90° angle: 90° – 60° = 30°.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Pages 125-129 Exercises
1. BE || CG; Conv. of
Corr. s Post.
2. CA || HR; Conv. of
Corr. s Post.
3. JO || LM; if two lines
and a transversal
form same-side int. s
that are suppl., then
the lines are ||.
4. a || b; if two lines and a 8. a || b; Conv. of Corr.
s Post.
transversal form sameside int. s that are
9. none
suppl., then the lines
are ||.
10. a || b; Conv. of Alt. Int.
s Thm.
5. a || b; if two lines and a
transversal form same11. || m; Conv. of Corr.
side int. s that are
s Post.
suppl., then the lines
are ||.
12. none
6. none
13. a || b; Conv. of Corr.
s Post.
7. none
14. none
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
15.
|| m; Conv. of Alt.
Int. s Thm.
24. When the frame is put
together, each of the
frame is a right . Two
right s are suppl. By the
Conv. of the Same-Side
Int. s Thm., opp. sides
of the frame are ||.
18. 30
19. 50
16. a.
b.
c.
d.
Def. of
Given
All right s are .
Conv. of Corr. s
20. 59
21. 31
Post.
22. 5
17. a.
b.
c.
1
1
2
23. 20
d.
3
e. Conv. of Corr.
25. The corr. s are , so
the lines are || by the
Conv. of Corr. s Post.
26. a. Corr. s
b–c. 1, 3 (any order)
d. Conv. of Corr. s
s
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
27. 10; m 1 = m 2 = 70
33. PL || NA by Conv. of Same-Side Int.
28. 5; m 1 = m 2 = 50
34. none
29. 2.5; m 1 = m 2 = 30 35. PN || LA by Conv. of Same-Side Int.
s
Thm.
s
Thm.
30. 1.25; m 1 = m 2
= 10
36. Answers may vary. Sample: In the diagram,
AB BH and AB BD, but BH || BD. They intersect.
31. The corr.
are .
37. Reflexive: a || a; false; any line intersects itself.
Symmetric: If a || b, then b || a; true; b and a are
coplanar and do not intersect.
Transitive: In general, if a || b, and b || c, then a || c;
true; however, when a || b, and b || a, it does not
follow that a || a.
s
he draws
32. PL || NA and PN || LA
by Conv. of SameSide Int. s Thm.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
38. Reflexive: a a; false; lines are two lines that
intersect to form right s .
Symmetric: If a b, then b a; true; b and a
intersect to form right s .
Transitive: If a b, and b c, then a c; false; in a
plane, two lines to the same line are ||.
42. Answers may vary.
Sample: 3
11;
|| m by Conv. of the
Alt. Int. s Thm. and
j || k by Conv. of Corr.
s Post.
39. The corr. s are , and the oars are || by the Conv.
of Corr. s Post.
43. Answers may vary.
Sample: 3 and 12
are suppl.; j || k by the
Conv. of Corr. s Post.
40. Answers may vary. Sample:
Conv. of the Alt. Int. sThm.
3
9; j || k by
41. Answers may vary. Sample: 3
9; j || k by
Conv. of the Alt. Int. sThm. and || m by Conv. of
Same-Side Int. s Thm.
3-2
44. Vert. s Thm. and
Conv. of Corr. s Post.
Proving Lines Parallel
GEOMETRY LESSON 3-2
45. It is given that || m,
so 4
8 by Corr.
s Post. It is also
given that 12
8,
so 4
12 by
Trans. Prop. of .
So, j || k by the Conv.
of Corr. s Post.
47.
46.
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
48.
49.
50. a. Answers may vary.
Sample:
b. Given: a || b with
transversal e,
c bisects AOB,
d bisects AXZ.
c. Prove: c || d
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
50. (continued)
d. To prove that c || d,
show that 1
3.
1
3 if AOB
OXZ. AOB
OXZ by the
Corr. s Post.
e. 1. a || b (Given)
2.
AOB
AXZ (Corr.
s
Post.)
s)
3. m AOB = m AXZ (Def. of
4. m AOB = m 1 + m 2; m AXZ =
m 3 + m 4 ( Add. Post.)
5. c bisects AOB; d bisects AXZ. (Given)
6. m 1 = m 2; m 3 = m 4 (Def. of bisector)
7.
8.
9.
10.
11.
m 1 + m 2 = m 3 + m 4 (Trans. Prop. of
m 1 + m 1 = m 3 + m 3 (Subst.)
2m 1 = 2m 3 (Add. Prop.)
m 1 = m 3 (Div. Prop.)
c || d (Conv. of Corr. sPost.)
3-2
)
Proving Lines Parallel
GEOMETRY LESSON 3-2
51. C
52. F
53. B
54. [2] a. 136 + (x + 21)
= 180 so x = 23
(OR equivalent
equation
resulting in
x = 23).
54. (continued)
b. x + 21 = 2x so x =
21. Lines c and d
are not || because x
cannot = both 21
and 23 (OR
equivalent
explanation).
[1] incorrect
equations OR
incorrect solutions
55. (continued)
b. m 1 + m 3 = 180.
If 2x – 38 + 6x + 18
= 180, then x = 25.
The measures
are 2x – 38 = 12
and 25, but 12 =/
25. So a can’t be
|| to b.
[3] appropriate methods,
but with one
computational error
55. [4] a.
[2] incorrect diagram
solved correctly OR
correct diagram
solved incorrectly
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
55. (continued)
[1] correct answer
(lines a and b are
not ||), without
work shown
58. If you are west of the
Mississippi River, then
you are in Nebraska.
Original statement is
true, converse is false.
56. m 1 = 70 since it is a 59. If a circle has a radius
suppl. of the 110° .
of 4 cm, then it has a
m 2 = 110 since
diameter of 8 cm. Both
same-side int. s are
are true.
suppl.
60. If same-side int. s are
57. m 1 = 66 because
suppl., then a line
alt. int. s are . m 2
intersects a pair of ||
= 180 – 94 = 86
lines. Both are true.
because same-side
int. s are suppl.
3-2
61. If you form the past
tense of a verb, then
you add ed to the
verb. Original
statement is false,
converse is false.
62. If there are clouds in
the sky, then it is
raining. Original
statement is true,
converse is false.
63. 201.1 in.2
64. 28.3 cm2
Proving Lines Parallel
GEOMETRY LESSON 3-2
65. 63.6 ft2
66. 78.5 in.2
67. 6.2 m2
68. 4.5 m2
69. 17.7 ft2
70. 0.3 m2
3-2
Proving Lines Parallel
GEOMETRY LESSON 3-2
Use the diagram and the given information to
determine which lines, if any, are parallel. Justify your
answer with a theorem or postulate.
1.
6
3
m || n by Converse of Corresponding Angles Post.
2. 1 and 4 are supplementary.
a || b by Converse of Same-Side Interior Angles Theorem.
3. 2
4 No lines must be parallel.
Suppose that m 1 = 3x + 10, m 2 = 3x + 14,
and m 6 = x + 58 in the diagram above.
4. Find the value of x for which a || b. 24
5. Find the value of x for which m || n. 26
3-2
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
(For help, go to Lesson 1-4.)
Classify each angle as acute, right, or obtuse.
1.
2.
3.
Solve each equation.
4. 30 + 90 + x = 180
5. 55 + x + 105 = 180
6. x + 58 = 90
7. 32 + x = 90
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Solutions
1. The measure of the angle is 90°, so it is a right angle.
2. The measure of the angle is between 0° and 90°, so it is an acute angle.
3. The measure of the angle is between 0° and 90°, so it is an acute angle.
4. Combine like terms: 120 + x = 180; subtract 120 from both sides: x = 60
5. Combine like terms: x + 160 = 180; subtract 160 from both sides: x = 20
6. Subtract 58 from both sides: x = 32
7. Subtract 32 from both sides: x = 58
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Find m
Z.
48 + 67 + m
Z = 180
Triangle Angle-Sum Theorem
115 + m
Z = 180
Simplify.
Z = 65
Subtract 115 from each side.
m
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
In triangle ABC,
ACB is a right angle, and CD
Find the values of a, b, and c.
Find c first, using the fact that
m
ACB = 90
c + 70 = 90
c = 20
ACB is a right angle.
Definition of right angle
Angle Addition Postulate
Subtract 70 from each side.
3-3
AB.
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
(continued)
To find a, use
ADC.
a+m
ADC + c = 180
m ADC = 90
a + 90 + 20 = 180
a + 110 = 180
a = 70
To find b, use CDB.
70 + m CDB + b = 180
m
CDB = 90
70 + 90 + b = 180
160 + b = 180
b = 20
Triangle Angle-Sum Theorem
Definition of perpendicular lines
Substitute 90 for m ADC and 20 for c.
Simplify.
Subtract 110 from each side.
Triangle Angle-Sum Theorem
Definition of perpendicular lines
Substitute 90 for m
CDB.
Simplify.
Subtract 160 from each side.
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Classify the triangle by its sides and its angles.
The three sides of the triangle have three different lengths,
so the triangle is scalene.
One angle has a measure greater than 90, so the triangle is obtuse.
The triangle is an obtuse scalene triangle.
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Find m
m
1.
1 + 90 = 125
Exterior Angle Theorem
m
Subtract 90 from each side.
1 = 35
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Explain what happens to the angle formed by the back of the
chair and the armrest as you lower the back of the lounge chair.
The exterior angle and the angle formed by the back of the chair and
the armrest are adjacent angles, which together form a straight angle.
As one measure increases, the other measure decreases.
The angle formed by the back of the chair and the armrest increases as
you lower the back of the lounge chair.
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
Pages 135-139 Exercises
1. 30
9. 70
2. 83.1
10. 30
3. 90
11. 60
4. 71
12. acute, isosceles
5. 90
13. acute, equiangular,
equilateral
16.
6. x = 70; y = 110; z = 30
14. right, scalene
7. t = 60; w = 60
15. obtuse, isosceles
8. x = 80; y = 80
3-3
17. Not possible; a right
will always have one
longest side opp. the
right .
18.
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
19.
23.
27. 115.5
28. m 3 = 92; m 4 = 88
24. a.
5,
6,
8
29. x = 147, y = 33
20.
b.
1 and
1 and
1 and
3 for
2 for
2 for
5
6
8
30. a = 162, b = 18
31. x = 52.5;
52.5, 52.5, 75; acute
21.
c. They are
vert.
s
.
32. x = 7; 55, 35, 90; right
25. a. 2
22.
33. x = 37; 37, 65, 78;
acute
b. 6
26. 123
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
34. x = 38, y = 36, z = 90;
ABD: 36, 90, 54;
right; BCD: 90, 52,
38; right; ABC:
74, 52, 54; acute
35. a = 67, b = 58,
c = 125, d = 23,
e = 90; FGH:
58, 67, 55; acute;
FEH: 125, 32, 23;
obtuse; EFG: 67, 23,
90; right
37. 60; 180  3 = 60
39. eight
38. Yes, an equilateral is
isosc. Because if three
sides of a are ,
then at least two sides
are . No, the third
side of an isosc.
does not need to be
to the other two.
40. 32.5
36. x = 32, y = 62, z = 32,
w = 118; ILK:
118, 32, 30; obtuse
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
41. Check students’ work.
Answers may vary.
Sample: The two ext.
s Formed at vertex A
are vert. s and thus
have the same
measure.
42. 30 and 60
46. 103
50. 132; since the missing
is 68, the largest
ext. Is 180 – 48 =
132.
47. 32
48. a. 90
b. 180
c. 90
d. compl.
e. compl.
43. a. 40, 60, 80
b. acute
44. 160
45. 100
51. Check students’ work.
49. a.
b.
s
Add.
-Sum
52. a. 81
b. 45, 63, 72
c. acute
53. 120 or 60
c. Trans.
54. 135 or 45
d. Subtr.
55. 90
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
56. Greater than, because 63. Answers may vary. Sample: The measure of the
there are two s with
ext. is = to the sum of the measures of the two
measure 90 where the
remote int. s . Since these s are , the s formed
meridians the
by the bisector of the ext. s are to each of them.
equator.
Therefore, the bisector is || to the included side of
the remote int. s by the Conv. of the Alt Int. s Thm.
57. 1
3
64. B
58. 1
7
65. G
59. 1
66. B
1
60.
19
67. H
61. 0
62. 115
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
68. [2] a.
b. The correct equation is 2x = (2x – 40)
+ (x – 15) with the sol. x = 55. The three int.
s measure 70, 40, and 70.
69. (continued)
b. 1 to 68 ; since Y is
obtuse, its whole
number range is
from 91 to 158,
allowing the
measure of 1 for
m M when m Y =
158. When m Y =
91, then m M = 68.
[1] incorrect sketch, equation, OR solution
69. [2] a. 159; the sum of the three s of the is 180,
so m Y + m M + m F = 180. Since m F =
21, m Y + m M + 21 = 180. Subtr. 21 from
both sides results in m Y + m M = 159.
3-3
[1] incorrect answer to
part (a) or (b) OR
incorrect
computation in
either part
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
70. 53
71. 46
72. 7
73.
74.
3-3
Parallel Lines and the Triangle Angle-Sum Theorem
GEOMETRY LESSON 3-3
1. A triangle with a 90° angle has sides that are 3 cm, 4 cm,
and 5 cm long. Classify the triangle by its sides and angles.
scalene right triangle
Use the diagram for Exercises 2–6.
2. Find m 3 if m
2 = 70 and m
4 = 42. 68
3. Find m 5 if m
2 = 76 and m
3 = 90. 166
4. Find x if m
1 = 4x, m
5. Find x if m
2 = 10x, m
6. Find m
3 if m
3 = 2x + 28, and m
3 = 5x + 40, and m
1 = 125 and m
4 = 32. 30
4 = 3x – 4. 8
5 = 160. 105
3-3
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
(For help, go to Lesson 1-4 and 3-3.)
Find the measure of each angle of quadrilateral ABCD.
1.
2.
3.
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Solutions
1. m
DAB = 32 + 45 = 77; m
2. m
DAC = m
ACD = m
B = 65; m
D and m
BCD = 70 + 61 = 131; m
CAB = m
B=m
D = 87
BCA; by the Triangle
Angle-Sum Theorem, the sum of the measures of the angles is 180,
so each angle measures 180, or 60. So, m
3
m
B = 60, m
DAB = 60 + 60 = 120,
BCD = 60 + 60 = 120, and m
D = 60.
3. By the Triangle Angle-Sum Theorem m A + 55 + 55 = 180, so m
m ABC = 55 + 30 = 85; by the Triangle Angle-Sum Theorem,
m C + 30 + 25 = 180, so m C = 125; m ADC = 55 + 25 = 80
3-4
A = 70.
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Name the polygon. Then identify its vertices, sides,
and angles.
The polygon can be named clockwise or counterclockwise,
starting at any vertex.
Possible names are ABCDE and EDCBA.
Its vertices are A, B, C, D, and E.
Its sides are AB or BA, BC or CB, CD or DC, DE or ED, and EA or AE.
Its angles are named by the vertices, A (or EAB or BAE),
B (or ABC or CBA), C (or BCD or DCB),
D (or CDE or EDC), and E (or DEA or AED).
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Classify the polygon below by its sides. Identify it as convex
or concave.
Starting with any side, count the number of sides clockwise
around the figure. Because the polygon has 12 sides,
it is a dodecagon.
Think of the polygon as a star. If you draw a diagonal
connecting two points of the star that are next to each other,
that diagonal lies outside the polygon, so the dodecagon
is concave.
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Find the sum of the measures of the angles of a decagon.
A decagon has 10 sides, so n = 10.
Sum = (n – 2)(180)
Polygon Angle-Sum Theorem
= (10 – 2)(180)
Substitute 10 for n.
= 8 • 180
Simplify.
= 1440
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Find m
X in quadrilateral XYZW.
The figure has 4 sides, so n = 4.
m
X + m Y + m Z + m W = (4 – 2)(180)
m X + m Y + 90 + 100 = 360
m
X+m
m
m
Y + 190 = 360
X+m
X+m
2m
m
Polygon Angle-Sum Theorem
Substitute.
Simplify.
Y = 170
X = 170
X = 170
Subtract 190 from each side.
Substitute m X for m Y.
Simplify.
X = 85
Divide each side by 2.
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
A regular hexagon is inscribed in a rectangle. Explain how you
know that all the angles labeled 1 have equal measures.
Sample: The hexagon is regular, so all its angles are congruent.
An exterior angle is the supplement of a polygon’s angle because
they are adjacent angles that form a straight angle.
Because supplements of congruent angles are congruent, all the angles
marked 1 have equal measures.
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
Pages 147-150 Exercises
1. yes
2. No; it has no sides.
3. No; it is not a plane
figure.
4. No; two sides intersect
between endpoints.
6. KCLP; sides: KC, CL,
LP, PK; s : K, C,
L, P
12. 1800
7. HEPTAGN; sides: HE,
EP, PT, TA, AG, GN,
NH; s : H, E, P,
T, A, G, N
14. 3240
13. 1440
15. 180,000
16. 102
8. pentagon; convex
5. MWBFX; sides: MW,
9. decagon; concave
WB, BF, FX, XM ;
s:
M, W, B, F, 10. pentagon; concave
X
11. 1080
3-4
17. 103
18. 145
19. 37
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
20. 60, 60, 120, 120
31.
28.
21. 113, 119
22. 108; 72
23. 150; 30
32. 3
29.
33. 8
24. 160; 20
34. 13
25. 176.4; 3.6
26. 45, 45, 90
35. 18
30.
36. a. 3; 6
b. 4; 8
27.
c. 5; 10
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
37. octagon; m 1 = 135;
m 2 = 45
38. If you solve
(n – 2) 180 = 130,
n
you get n = 7.2. This
number is not an
integer.
39. 20-80-80; 50-50-80
40. 108; 5
43. 150; 12
44. 180 – x; 360
x
45. 4
5
46. a. n • 180
b. (n – 2)180
c. 180n – 180(n – 2) =
360
d. Polygon Ext.
-Sum Thm.
41. 144; 10
48. w = 72, x = 59, y = 49,
z = 121;
49. x = 36, 2x = 72,
3x = 108, 4x = 144;
quad.
50–53. Answers may
vary. Samples are
given
50.
51.
47. y = 103; z = 70; quad.
42. 162; 20
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
52.
55. Answers may vary.
Sample: The figure is a
convex equilateral
quad. The sum of its s
is 2 • 180 or 360.
b.
53.
56. octagon
54. Yes; the sum of the
57. a. (20, 162), (40, 171),
measures of s at the
(60, 174), (80,175.5),
int. point is 360. The
(100, 176.4),
sum of the measures
(120, 177),
of all the s is 180n.
(140, 177.4),
180n – 360 =
(160, 177.75),
(n – 2)180
(180, 178),
(200, 178.2)
3-4
c. It is very close to 180
d. No, two sides cannot
be collinear.
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
58. a. [180(n – 2)] ÷ n = 180n – 360 = 180 – 360 .
n
n
62.
b. As n gets larger, the size of the angles get closer
to 180. The more sides it has, the closer the
polygon is to a circle.
63. Not possible; opp. and
adj. sides would
59. 36
overlap.
60–63. Answers may vary. Samples are given.
64. 4140
60.
65. 20
66. 225
61. Not possible; opp. sides would overlap.
67. 360
68. 157.5
3-4
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
69. 27
78. Div. Prop.
70. 10
79. Trans. Prop.
71. 120, 25
80. RT, RK
72. 50, 40
81.
73. 104, 76, 35, 69
82. Answers may vary.
Sample: BR and TK.
BRT,
BRK
74. Distr. Prop.
75. Subst. Prop.
83. Answers may vary.
Sample: BRM
76. Reflexive Prop. of
84.
TRM
77. Symm. Prop. of
85.
TRK
3-4
86. R
The Polygon Angle-Sum Theorems
GEOMETRY LESSON 3-4
For Exercises 1 and 2, if the figure is a polygon, name it by its
vertices and identify its sides. If the figure is not a polygon,
explain why not.
quadrilateral ABCD;
1.
not a polygon because
two sides intersect at
a point other than
endpoints
2.
AB, BC, CD, DA
3. Find the sum of the measures of the angles in an octagon.
1080
4. A pentagon has two right angles, a 100° angle and a
120° angle. What is the measure of its fifth angle?
140
ABCDEFGHIJ is a regular decagon.
5. Find m
6.
ABC. 144
XBC is an exterior angle at vertex B. Find m
3-4
XBC.
36
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
(For help, go to Page 151.)
Find the slope of the line that contains each pair of points.
1. A(–2, 2), B(4, –2)
2. P(3, 0), X(0, –5)
3. R(–3, –4), S(5, –4)
4. K(–3, 3), T(–3, 1)
5. C(0, 1), D(3, 3)
6. E(–1, 4), F(3, –2)
7. G(–8, –9), H(–3, –5)
8. L(7, –10), M(1, –4)
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Solutions
y –y
–2 – 2
1. m = x2 – x1 = 4 – (–2)
2
1
y2 – y1
2. m = x2 – x1
=
y –y
–5 – 0
0–3
–4 – (–4)
3. m = x2 – x1 = 5 – (–3)
2
1
y –y
1–3
4. m = x2 – x1 =
2
1
– 3 – (–3)
y2 – y1
–4
2
=–
6
3
5
–5
= –3 = 3
=
0
=0
8
–2
=
, which is undefined; There is no slope.
0
=
3–1
5. m = x – x =
=2
2
1
3–0
3
y2 – y1
–2 – 4
6. m = x – x = 3 – (–1)
2
1
y2 – y1
7. m = x – x
2
1
y –y
–5 – (–9)
= – 3 – (–8)
8. m = x2 – x1 =
2
1
–4 – (–10)
1–7
=
–6
4
=
3
=– 2
4
5
=
6
=–1
–6
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Use the slope and y-intercept to graph the line y = –2x + 9.
When an equation is written in the form y = mx + b, m is the slope
and b is the y-intercept.
In the equation y = –2x + 9, the slope is –2 and the y-intercept is 9.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Use the x-intercept and y-intercept to graph 5x – 6y = 30.
To find the x-intercept, substitute
0 for y and solve for x.
To find the y-intercept, substitute
0 for x and solve for y.
5x – 6y = 30
5x – 6(0) = 30
5x – 0 = 30
5x = 30
x =6
5x – 6y = 30
5(0) – 6y = 30
0 – 6y = 30
–6y = 30
y = –5
The x-intercept is 6.
A point on the line is (6, 0).
The y-intercept is –5.
A point on the line is (0, –5).
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
(continued)
Plot (6, 0) and (0, –5). Draw the line containing the two points.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Transform the equation –6x + 3y = 12 to slope-intercept form,
then graph the resulting equation.
Step 1: Transform the equation
to slope-intercept form.
–6x + 3y = 12
3y = 6x + 12
3y
6x
12
=
+
3
3
3
Step 2: Use the y-intercept and
the slope to plot two points
and draw the line containing them.
Add 6x to each side.
Divide each side by 3.
y = 2x + 4
The y-intercept is 4 and the slope is 2.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Write an equation in point-slope form of the line with slope –8
that contains P(3, –6).
y – y1 = m(x – x1)
y – (–6) = –8(x – 3)
y + 6 = –8(x – 3)
Use point-slope form.
Substitute –8 for m and (3, –6) for (x1, y1).
Simplify.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Write an equation in point-slope form of the line that contains
the points G(4, –9) and H(–1, 1).
Step 1: Find the slope.
y –y
m = x2 – x1
2
1
Use the formula for slope.
1 – (–9)
m = –1 – 4
Substitute (4, –9) for (x1, y1) and (–1, 1) for (x2, y2).
10
m = –5
m = –2
Simplify.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
(continued)
Step 2: Select one of the points.
Write the equation in point-slope form.
y – y1 = m(x – x1)
y – (–9) = –2(x – 4)
y + 9 = –2(x – 4)
Point-slope form
Substitute –2 for m and (4, –9) for (x1, y1).
Simplify.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Write equations for the horizontal line and the vertical line that
contain A(–7, –5).
Every point on the horizontal line through A(–7, –5)
has the same y-coordinate, –5, as point A.
The equation of the line is y = –5.
It crosses the y-axis at (0, –5).
Every point on the vertical line through A(–7, –5)
has the same x-coordinate, –7, as point A.
The equation of the line is x = –7.
It crosses the x-axis at (–7, 0).
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
1.
4.
7.
5.
8.
2.
6.
3.
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
9.
11. y = 2x + 1
13. y = –2x + 4
14. y = –2x + 4
10.
12. y = x + 1
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
15. y = – 1 x + 1
3
19. y – 5 = –1(x + 3)
25. y – 6 = 1(x – 2)
20. y + 6 = –4(x + 2)
26. y – 4 = 1(x + 4)
1
21. y – 1 = 2 (x – 6)
16.
22. y – 4 = 1(x – 0) or
y–4=x
23–28. Equations may
vary from the pt.
chosen. Samples are
given.
17. y – 3 = 2(x – 2)
23. y – 5 =
18. y + 1 = 3(x – 4)
3
(x – 0)
5
24. y – 2 = – 1 (x – 6)
2
3-5
1
(x + 1)
2
2
28. y – 10 = 3 (x – 8)
27. y – 0 =
29. a. y = 7
b. x = 4
30. a. y = –2
b. x = 3
31. a. y = –1
b. x = 0
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
32. a. y = 4
b. x = 6
36.
39. No; a line with no
slope is a vertical line.
0 slope is a horizontal
line.
37.
40. a. m = 0; it is a
horizontal line.
33.
b. y = 0
34.
41. a. Undefined; it is a
vertical line.
38. a. 0.05
35.
b. the cost per min
c. 4.95
d. the initial charge for
a call
3-5
b. x = 0
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
42–44. Answers may
vary. Samples are
given.
42. The eq. is in standard
form; change to
slope-intercept form,
because it is easy to
graph the eq. from
that form.
43. The eq. is in slope-int.
form; use slope-int.
form, because the eq.
is already in that form.
44. The eq. is in pointslope form; use pointslope form, because
the eq. is already in
that form.
46.
45.
The slopes are all
different and the
y-intercepts are the
same.
47. Check students’ work.
The slopes are the
same and the
y-intercepts are
different.
3-5
48.
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
49.
50.
3
1
3
1
52. 10 = 0.3, 12 = 0.083; > ;
10 12
it is possible only if the ramp zigzags.
53. The y-intercepts are the same, and the lines have
the same steepness. One line rises from left to right
while the other falls from left to right.
54. Answers may vary. Sample: x = 5, y – 6 = 2(x – 5),
y=x+1
55. (2, 0), (0, 4); m =
51.
0 – 4 –4
=
= –2
2–0
2
y – 0 = –2(x – 2), 2x + y = 4 or y = –2x + 4
56. a. y – 0 = 5 (x – 0) or y = 5 x
2
2
b. y – 5 = – 5 (x – 2) or y = – 5 x + 10
2
3-5
2
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
56. (continued)
c. The abs. value of
the slopes is the
same, but one
slope is pos. and
the other is neg.
One y-int. is at
(0, 0) and the other
is at (0, 10).
60. Yes; the slope of
JK = the slope of KL.
68. C
61. y – 2 = 3(x + 2);
3x – y = –8
1
62. y – 5 = 2 (x – 5);
x – 2y = –5
2
57. Yes; the slope of
AB = the slope of BC.
58. No; the slope of
DE =/ the slope of EF.
59. Yes; the slope of
GH = the slope of HI.
67. C
63. y – 6 = 3 (x – 2);
2x – 3y = –14
64. D
65. G
66. B
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
3
69. [2] a. Line a: y = x + 12 OR equivalent equation;
2
Line b: 3y + 4x = 19 OR equivalent equation
b.
72. 1620
73. 2160
74. yes
75. No; || lines never
intersect, but they are
not skew.
point of intersection: (22, 9)
[1] at least one correct eq. or graph
76. No; all obtuse
two acute s .
s
have
77. a = 5; m MPR = 30
70. 1260
78. a = 4; m MPR = 21
71. 540
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
79. a = 2; m QPR = 8
80. a = 6; m MPQ = 21
3-5
Lines in the Coordinate Plane
GEOMETRY LESSON 3-5
Three points are on a coordinate plane: A(1, 5), B(–2, –4), and C(6, –4).
1. Find the x-intercept and the y-intercept of the line 5x + 4y = –80.
x-intercept: –16, y-intercept: –20
2. Write an equation in point-slope form of the line with slope –1
that contains point C.
y + 4 = –1(x – 6)
3. Write an equation in point-slope form of the line that contains
points A and B.
y – 5 = 3(x – 1) or y + 4 = 3(x + 2)
4. Write an equation of the line that contains B and C.
y = –4
5. Graph and label the equations of the lines in
Exercises 2–4 above.
3-5
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
(For help, go to page 151 and Lesson 3-5.)
Find the slope of the line through each pair of points.
1. F(2, 5), B(–2, 3)
2. H(0, –5), D(2, 0)
3. E(1, 1), F(2, –4)
Find the slope of each line.
4. y = 2x – 5
5. x + y = 20
6. 2x – 3y = 6
7. x = y
8. y = 7
9. y = 2 x + 7
3
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Solutions
1. m =
y2 – y1
x2 – x1
3–5
= –2–2
= –2 = 1
–4
2. m =
y2 – y1
x2 – x1
= 0 – (–5) = 5
3. m =
y2 – y1
x2 – x1
= –4–1
2–0
2 –1
2
2
–5
= 1
= –5
4. The equation is in slope-intercept form and the slope is the coefficient
of x, which is 2.
5. Rewrite the equation in slope-intercept form: y = – x + 20; the slope
is the coefficient of x, which is – 1.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Solutions (continued)
6. Rewrite the equation in slope-intercept form: y = 2 x – 2; the slope
3
2
is the coefficient of x, which is .
3
7. The equation in slope-intercept form is y = x + 0; the slope is the
coefficient of x, which is 1.
8. The equation in slope-intercept form is y = 0x + 7; the slope is the
coefficient of x, which is 0.
9. The equation is in slope-intercept form and the slope is the coefficient
of x, which is 2 .
3
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Line 1 contains P(0, 3) and Q(–2, 5). Line
and S(3, –10). Are lines 1 and 2 parallel? Explain.
2
contains R(0, –7)
Find and compare the slopes of the lines.
Slope of line
Slope of line
y2 – y1
5–3
y2 – y1
–10 –(–7)
3–0
1
= x –x =
–2 – 0
2
1
2
= x –x =
2
1
=
2
–2
= –1
= –3 = –1
3
Each line has slope –1.
The y-intercepts are 3 and –7.
The lines have the same slope and different y-intercepts, so they are parallel.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Are the lines y = –5x + 4 and x = –5y + 4 parallel? Explain.
The equation y = –5x + 4 is in slope-intercept form.
Write the equation x = –5y + 4 in slope-intercept form.
x = –5y + 4
x – 4 = –5y
Subtract 4 from each side.
–1x+ 4 =y
5
Divide each side by –5.
5
y=–1x+ 4
5
5
The line x = –5y + 4 has slope – 1 .
5
The line y = –5x + 4 has slope –5.
The lines are not parallel because their slopes are not equal.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Write an equation in point-slope form for the line
parallel to 6x – 3y = 9 that contains (–5, –8).
Step 1: To find the slope of the line, rewrite the equation
in slope-intercept form.
6x – 3y = 9
–3y = –6x + 9
y = 2x – 3
Subtract 6x from each side.
Divide each side by –3.
The line 6x – 3y = 9 has slope 2.
Step 2: Use point-slope form to write an equation for the new line.
y – y1 = m(x – x1)
y – (–8) = 2(x – (–5))
y + 8 = 2(x + 5)
Substitute 2 for m and (–5, –8) for (x1, y1).
Simplify.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Line 1 contains M(0, 8) and N(4, –6). Line 2 contains P(–2, 9)
and Q(5, 7). Are lines 1 and 2 perpendicular? Explain.
Step 1: Find the slope of each line.
m1 = slope of line
m2 = slope of line
y –y
1
= x2 – x1 =
2
1
–6 – 8
4–0
y –y
7–9
2
= x2 – x1 =
5 – (–2)
2
1
=
–14
7
=–
4
2
=
–2
= –2
7
7
Step 2: Find the product of the slopes.
m1 • m2 = –
7
2
•– =1
2
7
Lines 1 and 2 are not perpendicular because the product
of their slopes is not –1.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Write an equation for a line perpendicular to
5x + 2y = 1 that contains (10, 0).
Step 1: To find the slope of the given line, rewrite the equation
in slope-intercept form. 5x + 2y = 1
2y = –5x + 1
Subtract 5x from each side.
y=–5 x+ 1
Divide each side by 2.
2
2
The line 5x + 2y = 1 has slope – 5 .
2
Step 2: Find the slope of a line perpendicular to 5x + 2y = 1.
Let m be the slope of the perpendicular line.
– 5 m = –1
The product of the slopes of perpendicular lines is –1.
2
m = –1 • ( – 2 )
Multiply each side by – 2 .
m= 2
5
5
5
Simplify.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
(continued)
Step 3: Use point-slope form, y – y1 = m(x – x1), to write an equation
for the new line.
y – 0 = 2 (x – 10)
5
y = 2 (x – 10)
5
Substitute 2 for m and (10, 0) for (x1, y1).
5
Simplify.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
The equation for a line containing a lead strip is y = 1 x – 9.
2
Write an equation for a line perpendicular to it that contains (1, 7).
Step 1: Identify the slope of the given line.
1
y = 2x – 9
slope
Step 2: Find the slope of the line perpendicular to the given line.
Let m be the slope of the perpendicular line.
1
m = –1
2
m = –2
The product of the slopes of perpendicular lines is –1.
Multiply each side by 2.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
(continued)
Step 3: Use point-slope form to write an equation for the new line.
y – y1 = m(x – x1)
y – 7 = –2(x – 1)
Substitute –2 for m and (1, 7) for (x1, y1).
Step 4: Write the equation in slope-intercept form.
y – 7 = –2(x – 1)
y – 7 = –2x + 2
y = –2x + 9
Use the Distributive Property.
Add 7 to each side.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
Pages 161-164 Exercises
1.
2.
Yes; both slopes = – 1.
2
1
No; the slope of 1 = 3 ,
and the slope of 2 = 1 .
2
3. No; the slope of
1
= 3,
2
and the slope of
2
= 2.
4. Yes; both slopes = 4.
5. Yes; both slopes = 0.
3
6. Yes; the lines both
have a slope of 2 but
different y-intercepts.
10. No; one slope = – 4
and the other slope
= –3.
7. Yes; the lines both
have a slope of 3 but
11. Yes; the lines both
have a slope of – 2
5
but different
y-intercepts.
4
different y-intercepts.
8. Yes; the lines both
have a slope of –1 but
different y-intercepts.
9. No; one slope = 7 and
the other slope = –7.
3-6
12. y – 3 = –2(x – 0) or
y – 3 = –2x
13. y – 0 = 1 (x – 6) or
3
1
y = (x – 6)
3
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
1
2
14. y – 4 = (x + 2)
15. y + 2 = – 3 (x – 6)
2
16. Yes; the slope of 1 =
– 1 , and the slope of
2
1
• 2 = –1.
2 = 2; –
19. Yes; the slope of 1 = 25. No; 1 • 2 =/ –1.
2
–1, and the slope of 2
= 1; –1 • 1 = –1.
26. Yes; 1 • (–1) = –1.
20–23. Answers may
vary. Samples are
given.
2
3
17. Yes; the slope of 1 =
– 3 , and the slope of
2
2 ; – 3 • 2 = –1.
2 =
3
2
3
18. No; the slope of 1 =
–1, and the slope of
2
4
5
= ; –1 •
4
=/ –1.
5
20. y – 6 = – 2 (x – 6)
21. y = –2(x – 4)
22. y – 4 =
1
(x – 4)
2
4
x
5
3
24. y = – x
2
23. y =
3-6
27. Yes; one is vertical
and the other is
horizontal.
28. No; – 1 • (–3) =/ –1.
2
2 3
29. Yes; – • = –1.
3 2
30. No; 2 • – 7 =/ –1.
7
4
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
31. slope of AB = slope of
2
CD = ; AB || CD
3
slope of BC = slope of
AD = –3; BC || AD
32. slope of AB = slope of
CD = – 3 ; AB || CD
4
slope of BC = slope of
AD = 1; BC || AD
33. slope of AB = 1 ; slope
2
1
of CD = ; AB || CD
4
slope of BC = –1;
slope of AD 5 – 1 ;
2
BC || AD
34. slope of AB = slope of
CD = 0; AB || CD
slope of BC = 3 and
slope of AD = 3 ;
2
BC || AD
35. Answers may vary.
Sample: y = 4 x + 5,
5
5
y=– x+5
4
36. No; two II lines with the
same y-intercept are
the same line.
3-6
37. RS and VU are
horizontal with slope
= 0; RS II VU;
slope of RW = slope of
UT = 1; RW || UT;
slope of WV = slope of
ST = –1; WV || ST
38. No; because no pairs
of slopes have a
product of –1.
39. The lines will have the
same slope.
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
40. When lines are , the
product of their
slopes is –1. So, two
lines to the same
line must have the
same slope.
3
41. a. y + 20 = 4 (x – 35)
b. because you are
given a point and
can quickly find the
slope
42. ||
44. neither
45.
46.
(7 – 9)2 + (11 – 1)2 =
(13 – 3)2 + (7 – 5)2 =
47. AC: d =
BD: d =
AC
104
104
BD
1
1
48. slope of AC = –5; slope of BD = ; since –5 • = –1,
5
5
AC BD; midpoint AC = (8, 6); midpoint BD = (8, 6);
since the midpoints are the same, the diagonals
bisect each other.
43.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
49. a–b. Answers
may vary.
Sample:
51. B
52. I
53. C
54. [2] a. slope of line c:
1 – (–2)
3
1
=
=
–
–4 – 2
–6
2
; slope of line
b. 0
c. The other possible
locations for S are
(–2, 3) and (8, 7).
50. y – 5 = 1 (x – 4)
3
[1] at least one correct slope
1
1
55. y – 3 = – 2 (x – 0) or y – 3 = – 2 x
56. y – 2 = 5 (x + 4)
3
3-6
to c: 2
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
57. y + 2 =
3
(x – 3)
4
58. Refl. Prop. of
59. Mult. Prop. of =
60. Dist. Prop.
61. Symm. Prop. Of
62. If you are in geometry
class, then you are at
school.
63. If you travel to
Switzerland, then you
have a passport.
3-6
Slopes of Parallel and Perpendicular Lines
GEOMETRY LESSON 3-6
1. Are lines
1
and
2
parallel? Explain.
Yes; the lines have the same slope
and different y-intercepts.
2. Are the lines x + 4y = 8 and
2x + 6y = 16 parallel? Explain.
No; their slopes are not equal.
3. Write an equation in point-slope form for the line parallel
to –18x + 2y = 7 that contains (3, 1). y – 1 = 9(x – 3)
2
4. Are the lines y = 3 x + 5 and 3x + 2y = 10 perpendicular? Explain.
Yes; the product of their slopes is –1.
5. Write an equation in point-slope form for the line
perpendicular to y = – 1 x – 2 that contains (–5, –8).
6
y + 8 = 6(x + 5)
3-6
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
(For help, go to Lesson 1-5.)
Use a straightedge to draw each figure. Then use a straightedge
and compass to construct a figure congruent to it.
1. a segment
2. an obtuse angle
3. an acute angle
Use a straightedge to draw each figure. Then use a straightedge
and compass to bisect it.
4. a segment
5. an acute angle
3-7
6. an obtuse angle
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Solutions
1.
2.
3.
4.
5.
6.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Examine the diagram below. Explain how to
construct 1 congruent to H.
Use the method learned for constructing congruent angles.
Step 1: With the compass point
on point H, draw an arc that
intersects the sides of H.
Step 2: With the same compass
setting, put the compass point
on point N. Draw an arc.
Step 3: Put the compass point below point N where the arc intersects HN.
Open the compass to the length where the arc intersects line .
Keeping the same compass setting, put the compass point above point N
where the arc intersects HN. Draw an arc to locate a point.
Step 4: Use a straightedge to draw line m through the point you located and point N.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Construct a quadrilateral with both pairs of sides parallel.
Step 1: Draw point A and two rays with
endpoints at A. Label point B on
one ray and point C on the other ray.
Step 2: Construct a ray parallel to AC
through point B.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
(continued)
Step 3: Construct a ray parallel to AC
through point C.
Step 4: Label point D where the ray parallel
to AC intersects the ray parallel to AB.
Quadrilateral ABDC has both pairs of
opposite sides parallel.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
In constructing a perpendicular to line at point P, why must
you open the compass wider to make the second arc?
With the compass tip on A and B, the same compass setting would
make arcs that intersect at point P on line . Without another point,
you could not draw a unique line.
With the compass tip on A and B, a smaller compass setting would make arcs
that do not intersect at all. Once again, without another point, you could not
draw a unique line.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Examine the construction. At what special point does RG
meet line ?
Point R is the same distance from point E as it is from point F
because the arc was made with one compass opening.
Point G is the same distance from point E as it is from point F
because both arcs were made with the same compass opening.
This means that RG intersects line at the midpoint of EF, and RG
is the perpendicular bisector of EF.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Pages 168-170 Exercises
5–7. Constructions may
vary. Samples using
the following
segments are shown.
3.
1.
|| AB
|| AB
5.
4.
2.
6.
|| AB
|| AB
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
7.
11.
9.
AB
8.
RS
10.
12.
AB
RS
RS
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
13.
15. Construct a alt. int.
then draw the || line.
; 17.
16. a–b
RS
18.
14.
1
2
17–25. Constructions may
vary. Samples are
given.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
19.
23.
21. a.
b. The sides are ||
and .
20.
c. Check students’
work.
22.
3-7
24. a.
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
24. (continued)
b. Answers may vary.
Sample:
26. a–b. Check students’
work.
25. a–c.
c. p || m; in a plane,
two lines to a
third line are ||.
c. One; if the lengths for
the 3 sides are given,
only one is possible;
many different quad.
are possible because
the s formed by the
sides can vary.
d. The sides of the
smaller are half
the length of the
sides of the larger
that they are || to.
27–28. Answers may
vary. Samples are
given.
27.
e. Check students’
work.
The quad. is a rectangle.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
28.
30.
32.
33.
The quad. is a square.
29.
31.
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
34.
36. Not possible; the
39. (continued)
smaller sides would
b. All points on the arc
meet at the midpoint of
with center G are the
the longer side, forming
same dist. from G, so
a segment.
GR = GT. The is
isosc. because an
37. A
isosc. must have at
least two sides of the
38. I
same length.
35. Not possible; if
39. [2] a.
a = 2b = 2c, then
2a = 2b + 2c or
a = b + c. The smaller
sides would meet at
the midpoint of the
longer side, forming a
segment.
[1] incorrect sketch OR
incorrect
classification
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
40. [2] a. I, IV, III, I
b. (III): location of
compass at
points C and G;
(I): same
as c and the
intersection
points of with
arcs drawn in
(III)
41. No; the slopes are
different.
42. No; the slopes are
different.
43. Yes; the slopes are
both – 1 .
3
44. 10
45. 8.9
[1] incorrect sequence
OR incorrect
location of
compass point
46. EB
47. DF
3-7
Constructing Parallel and Perpendicular Lines
GEOMETRY LESSON 3-7
Draw a figure similar to the one given. Then complete the construction.
1. Construct a line through D that is parallel to XY.
Answers may vary. Sample given:
2. Construct a quadrilateral with one pair of parallel sides of lengths p and q.
Answers may vary. Sample given:
3. Construct the line perpendicular
to line m at point Z.
4. Construct the perpendicular
to line n through point O.
3-7
Parallel and Perpendicular Lines
GEOMETRY CHAPTER 3
Page 176
1. acute isosceles
5. m 1 = 85 because
corr. s are ; m 2 =
95 because same-side
int. s are suppl.
2. obtuse scalene
3. m 1 = 65 because
corr. s are ; m 2 =
65 because vert. s
are .
11. 5
12. 25
13. 6
6. m 1 = 70 because
corr. s are ; m 2 =
110 because sameside int. s are suppl.
7. yes
4. m 1 = 85 because
alt. int. s are ; m 2 8. yes
= 110 because sameside int. s are suppl.
9. no
10. no
3-A
14. 75
15.
Parallel and Perpendicular Lines
GEOMETRY CHAPTER 3
16.
20. y + 1 = –5(x – 3) or
y = –5x + 14
24.
25. ||
17. Answers may vary.
Sample: Z
18. Check students’ work.
21. Draw segments
connecting
nonconsecutive
vertices. If no points of
the segments are
outside the polygon,
then it is convex.
Otherwise, the polygon
is concave.
19. a. Given
b. Corr. s are .
c. Given
22. 109
d. Trans. Prop.
e. If corr. s are ,
23. x = 85; y = 100; z = 100
then the lines are ||.
3-A
26. neither
27.
28. 30