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Congruence and Transformations
.....................................................................
8
INVESTIGATION 8A THE CONGRUENCE
RELATIONSHIP
8.01
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Getting Started
1. Answers will vary. Possible answers: Two figures are the
same if they have the same shape and size. Two figures
are the same if you can cut one out and fit it exactly on
top of the other.
2. Answers may vary. Sample: All four figures are
squares.
3. Answers may vary. Sample: Squares ABCD and EFGH
seem to be identical in every way, except that they have
different names. Square NOPQ is positioned differently,
but if you rotated it, it would match up with squares
ABCD and EFGH perfectly. Square VWXY has shorter
sides than the other squares (hence, a smaller area).
4. Answers may vary. Sample: The = sign might stand for
“same size,” since this symbol is used with numbers and
size is a measurable quantity. The ∼ sign might stand for
“same shape,” as this symbol is used to express
similarity, when objects have the same shape.
On Your Own
5. Answers will vary. Possible Response:
One way to determine if two line segments are
congruent is by measuring their lengths with a ruler to
see if they are equal. Another is by placing one segment
on top of the other to see if they perfectly match up.
You can test to see if two angles are congruent by
measuring the angles with a protractor to see if they are
equal. You also could line up one angle on top of another
(matching vertex to vertex and one of the rays) to see if
they perfectly match up. With this test, it does not matter
if the sides of the angle are drawn to be the same length;
all that matters is that the sides line up. This is because
the sides of the angle are rays that continue forever, no
matter how long they are drawn.
For two triangles, you could measure all the angles of
each triangle with a protractor and all the sides with a
ruler to see if all the corresponding measurements are the
same. Or, you could place one triangle on top of the other
to see if they match up perfectly.
For any solid figure (box, cone, or cylinder), you could
first compare their volumes: if the volumes are different,
pages 305–331
the shapes cannot be congruent. However, if the volumes
are the same this DOES NOT tell you that the shapes are
congruent. You would need another test after this to see
if they indeed have the same shape.
For two rectangular solids, you could measure all the
corresponding edges to see if they are equal. Since they
are boxes, the angles must all be right angles. This
means you do not need to check that the angles are the
same size.
Right cones (where the apex of the cone is directly
above the center of the base circle) and right cylinders
(where the two circle bases are exactly lined up) are
defined by their radii and their heights. You could
measure the radii of the bases and the heights of each to
check that they are the same. This would tell you if they
are congruent.
6. Yes, the cut-and-move test will work for line
segments.
7. In extending the superimposability test for congruence to
3-dimensional objects, you can imagine that two
3-dimensional objects are congruent if they have the
same size and shape, and take up the same space. One
conceivable test for 3-dimensional congruence is to wrap
one shape and see if the the same wrapper fits around the
second shape. In doing this, it is important to recognize
that a wrapper could fit two non-congruent shapes;
however, if you are careful in how you construct the
wrapper and you don’t distort, refold, or cut the wrapper
when placing it on the second shape, this could be a fair
test.
For more uniform solids, like cubes and spheres, you
could just check that they have the same volume. (This is
because all cubes are similar, as are all spheres, so you
know they have the same shape. You need to check that
they have the same size.) This can be accomplished by
filling the shapes with water, or by submerging them in
water to see how much they displace. See the sidenote to
problem 5 for an example of how this method would fail
for non-uniform solids.
Maintain Your Skills
8.
9.
10.
11.
12.
Mathematics I Solutions Manual
Triangle TRL is congruent to triangle MTV.
Segment TO is congruent to segment BE.
Square BARK is congruent to square MEOW.
Triangle LMO is congruent to triangle L M O .
Angle ABC is congruent to angle ABD.
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8.02
Length, Measure, and Congruence
Check Your Understanding
1. (a) Two numbers are mentioned: The length of segment
JK and the length of segment RS.
(b) Two segments are mentioned: JK and RS.
(c) Two segments and two numbers are mentioned. The
segments are JK and RS; the numbers are their
lengths.
(d) Four points and two numbers are mentioned. The
points are J, K, R, and S; the numbers are the
distances between the points.
2. (a) The sentence “ NPQ = 56.6◦ ” is confusing. The
symbol NPQ refers to the the angle itself, not its
measure. What is meant here is that the measure of
the object NPQ is 56.6◦ .
(b) The correct way to write the sentence is
m NPQ = 56.6◦ .
3. Yes, the angles are congruent. This is similar to asking,
“Are two line segments congruent, if they are the same
length?” In both cases, the shape of the object is
determined by its measure. The sides of an angle are rays,
which never end. So, even if the rays are drawn so that
they appear to be different lengths, they really are
congruent. The only thing that matters is the “space”
between the rays, which is the measure of the
angle.
4. Yes, the angles have the same measure. Congruent
angles, by definition, must have the same shape and the
same size. Size, for angles, is the measure of the angle.
5. True; this is called The Transitive Property of
Congruence. What it means is that all three triangles have
the same size and shape. You can think of it as stacking
up the three triangles, one on top of another, on top of
another. All three would match up perfectly.
6. Students responses may vary. Some of the statements
called nonsensical in the solutions below could possibly
be considered false under some interpretations.
(a) True; F and E are midpoints of AD and BD. This
makes FD = 12 AD and ED = 12 BD. Since
AD = BD, it follows that 12 AD = 12 BD.
(b) Nonsensical; two segments cannot be equal, they can
only be congruent.
(c) Nonsensical; A segment cannot equal a measurement.
You could say FD = 1.5 cm.
(d) Nonsensical; An angle cannot equal a measurement.
You could say that ACD is a right angle, or that
m ACD = 90◦ .
(e) Nonsensical; Two triangles cannot be equal, they can
only be congruent.
(f) True; DC is an altitude, which means m ACD = 90◦
and m BCD = 90◦ . And an angle that measures 90◦
is a right angle.
(g) True; By the same reasoning as in statement (a), we
know that FA and BE have the same length, and
hence are congruent.
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(h) False; These segments do not have the same length.
We know that FA = 12 DA, and DA = BD. This
makes FA half the length of BD.
(i) Nonsensical; Two angles cannot be equal, they can
only have the same measurement.
(j) True; The altitude of an isosceles triangle bisects the
vertex angle. To convince yourself it is true, fold
along altitude DC. The angles match up perfectly.
(k) Nonsensical; Two measurements cannot be
congruent.
(l) True; Fold along altitude DC. The angles match up
perfectly.
(m) Nonsensical; A triangle cannot be congruent to an
angle.
(n) Nonsensical; Once again, a triangle cannot be
congruent to an angle.
On Your Own
7. (a) Two figures are congruent when they have the same
shape and same size, regardless of location or
orientation.
(b) The ∼
= symbol means “is congruent to.”
(c) The ⊥ symbol means “is perpendicular to.”
8. Measures of angles are equal, not congruent. The correct
answer is B.
9. Not all equilateral triangles are congruent. They have the
same shape (they are similar), but they may have different
side lengths. If the sides of two equilateral triangles have
the same lengths, then the triangles are congruent.
10. There is only one domino which can be made with two
squares, since the squares must fully coincide along an
edge. All others are congruent to this one.
11. With three squares, you can make two different
trominoes: an L-shape or a strip of three squares in a row.
One way to find these is to add on one square to the
domino in Exercise 10. By placing the one square
alongside each edge of the domino, the only two shapes
ever formed are the L-shape and the strip shown below.
12. There are five different shapes that can be made. You can
think of this as trying to find how many different ways
there are to add one square to each of the tromino shapes
above.
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(there are 10 pairs to check). If these distances are the
same, they must be congruent.
• Two snowflakes The snowflakes appear to be
rotated and translated images. Trace one snowflake
onto another piece of paper. You will be able to rotate
the tracing so that it will line up perfectly with the other
snowflake.
13. Below are two examples of an eight-square polyomino
made using a T tetromino and another 4-square
polyomino.
This 8-square polyomino can only be made in these two
ways. Here’s one that can be made in only one way.
Maintain Your Skills
14. The shapes in each pair appear to be congruent. The
cut-and-move strategy works for all pairs (it is harder
with the overlapped ones, but can be done if you first
trace one of the shapes onto another paper.) Here are
some other strategies:
• Two Circles Measure the diameters (or radii, or
circumferences) of the circles. If they are the same, the
circles must be congruent. (All circles have the same
shape; the only consideration is the size.)
• Two Artists The artists appear to be reflections of each
other, across a vertical line half-way between them.
Fold the paper to see if one artist will perfectly match
up with the other.
• Two Pentagons The pentagons appear to be regular; if
this is the case, measure one side of each pentagon to
see if they are the same. Otherwise, measure all pairs
of corresponding sides, as well as all pairs of
corresponding angles to see that they are the same.
• Two bent arrows The arrows appear to be rotated and
translated images. Trace one arrow onto another piece
of paper and rotate it until it lines up perfectly with the
other arrow.
• Two stars The stars appear to be translated images.
Measure the distances between corresponding vertices
8.03
Corresponding Parts
Check Your Understanding
1. The correct statement is part (d), DFA ∼
= ECG. In this
statement, the order of the vertices indicates which
vertices are corresponding. Notice how D and E are both
listed first. This means that they are corresponding
vertices of the triangles. Likewise, F and C are
corresponding vertices, and A and G are corresponding
vertices. The question becomes, “Why is this the correct
correspondence?”
Since we know the triangles are congruent, we need to
match them up according to the information shown in the
diagram. In the two triangles, D and E are marked
congruent, so D and E are corresponding vertices. Since
EC and DF are marked congruent, C and F must also be
corresponding vertices. This means A and G correspond
as well. So the vertices DFA in order correspond to the
vertices ECG.
For this reason, none of the other congruence
statements are correct.
2. There are six correct statements, each obtained by
permuting the vertices. They are:
• DFA ∼
= ECG
• FAD ∼
= CGE
• AFD ∼
= GCE
• DAF ∼
= EGC
• FDA ∼
= CEG
• ADF ∼
= GEC
Notice that there are other answers. The statement
DFA ∼
= ECG also yields the following six
statements regarding the sides and angles of the two
triangles:
∼ CE
• FD =
• FA ∼
= CG
• FD ∼
= EC
• D∼
= E
• F∼
= C
• A∼
= G.
3. Possible answer: If two figures are congruent, then one is
really just a copy of the other, although it may be flipped,
rotated, or translated. That means that every angle,
segment, or point on the first figure has a “match” on the
second figure. These matching pairs are called
corresponding parts. If two figures are congruent, you
can match up these corresponding pairs and they will be
congruent to each other.
Mathematics I Solutions Manual
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4. Below is an example of a pair of congruent triangles and
the corresponding congruence statements.
B
D
A
F
E
C
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(d) In the triangle BCE, BCD ∼
= ECD. Notice
that congruent triangles are always listed with
corresponding vertices in order.
10. (a) m E = 50◦ ; m A = 80◦ ; m D = 90◦ .
Since E ∼
= H and m H = 50◦ , it follows that
◦
m E = 50 . Since A ∼
= F and m F = 80◦ , it
follows that m A = 80◦ . Finding m D is a little
more involved. First, notice that B ∼
= JCH,
making m JCH = 40◦ . Next, notice that JCH
and DCE are vertical angles, making them
congruent as well. So, m DCE = 40◦ . Because
all the angles in a triangle must add up to 180◦ ,
m D = 180◦ − 50◦ − 40◦ = 90◦
(b)
ABC ∼
= DEF.
The following pairs of corresponding parts are
congruent: A ∼
= D, B ∼
= E, C ∼
= F , AB ∼
= DE,
∼
∼
AC = DF, and BC = EF.
E
D
B
F
On Your Own
5. C ∼
= D, A ∼
= O, T ∼
= G, CA ∼
= DO, AT ∼
= OG,
∼
and CT = DG
6. The correct answer is A. Since DFG ∼
= CHK, the
vertex G corresponds to vertex K. Therefore m FGD ∼
=
m CKH. In each of the other three options, the
supposed congruences do not features corresponding
parts.
7. The following pairs of triangles might be congruent:
C
A
G
• AFD and AEB
• FBD and EDB
• EGD and FGB.
J
H
The corrected picture
8. (a) If two polygons are congruent, they must have the
same area. This is what it means to say you can place
one exactly on top of the other.
(b) Just because two polygons have the same area, they
are not necessarily congruent. A 4 × 4 square has
the same area as a 8 × 2 rectangle, but they are not
congruent.
9. (a) The three congruent triangles are ABC, DBC, and
DEC.
(b)
A
Maintain Your Skills
11. In the picture below, pentagons ABCDE and FGHIJ are
congruent.
B
G
C
H
C
F
A
I
D
B
E
J
D
E
(c) In the quadrilateral ABDC, DBC ∼
= ABC.
Mathematics I Solutions Manual
There are 10 pairs of corresponding parts.
The congruent sides are: AB ∼
= FG, BC ∼
= GH ,
CD ∼
= HI , DE ∼
= IJ , and EA ∼
= JF.
The congruent angles are: A ∼
= F, B ∼
= G,
C ∼
= H, D ∼
= I, and E ∼
= J.
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8.04 Triangle Congruence
Check Your Understanding
1. The triangles may not be congruent. You can construct
many different triangles with these angle measures. The
triangles can be different sizes. One example of two
non-congruent triangles is the following:
A
A
8 cm
5 cm
8 cm
5 cm
B
B
C
C
If you knew one side length, and where it is in relation to
the angles, then there would be only one triangle you
could construct, because it would satisfy the ASA
Triangle Congruence Postulate.
2. There are exactly two different triangles that can be made
with this information. Triangles ABC 1 , and ABC 2 shown
below, fit the information:
C2
7 cm
C1
7 cm
B
60°
8 cm
A
To choose between these two possibilites, we would
need any other piece of information about the triangle.
If we knew m B or m C, then the information would
satisfy ASA. If we knew AC, it would satisfy SSS
and SAS.
3. The two triangles are congruent by SSS . They have two
pairs of congruent sides (by the tick marks): AB ∼
= AD,
and BC ∼
CD.
The
triangles
also
share
AC.
=
4. These triangles can be proven congruent, but you need to
know that AB ∼
= AD. This additional fact makes
ABC ∼
ADC
by SSS or by SAS.
=
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5. Given any square, you can cut along the diagonal,
producing two right triangles. One of the triangles can be
reflected about the diagonal and will fit exactly on top of
the other. Here are two ways to prove they are congruent,
using the triangle congruence postulates:
Using SSS: Suppose ABCD is a square, with diagonal
DB. By the definition of a square, you know that there are
four congruent sides. Specifically AB ∼
= CB and
AD ∼
= CD. Notice that ABD and CBD share side
BD. Since they have three pairs of congruent sides,
ABD ∼
= CBD.
Using SAS: As before, by definition of a square,
AB ∼
= CB and AD ∼
= CD. Also, by definition of a square
A and C are right angles. This makes A ∼
= C. Since
the triangles have two pairs of congruent sides, and a pair
of congruent, included angles, ABD ∼
= CBD.
6. It is unclear whether the 14-inch and the 8-inch sides are
the two sides of the triangle adjacent to the 30◦ angle at
the tip of the pennant, or if one of those two sides is
opposite the 30◦ angle.
7. Sides AO and BO are congruent, as they are both radii of
the circle. Triangles COA and COB also share side OC
and angle OCB. The two triangles are definitely not
congruent, however, as one is contained inside the other
(at the very least, BOC is contained inside AOC,
making it smaller). Thus, knowing SSA information is
not enough to uniquely determine a triangle.
On Your Own
8. There is not enough information, as only two sides are
given.
9. There is enough information. By the SSS postulate,
ABC ∼
= DEF.
10. There is enough information. By the SAS postulate
ABC ∼
= DBC. (Notice that BC is a common side.)
11. There is not enough information. Notice that the two
triangles have two pairs of congruent sides, and a pair of
congruent angles. In ABC, the marked angle is an
included angle (between the two sides), while in BCD
it is a non-included angle (not between the two sides).
This means they are not corresponding angles - there is
no way you could rotate, flip, or slide these two triangles
to make the congruent angles and the congruent sides
match up. It is possible the triangles are congruent, but
you would need to know that either ABC ∼
= DBC or
∼
DC = AC.
12. There is enough information. By the SAS postulate,
ABC ∼
= YXC.
13. The only congruence statements you can make about the
parts of ACD and BCD are ACD ∼
= BCD and
∼
CD = CD. You would need at least one more pair of
congruent sides, or one more pair of congruent angles to
prove ACD ∼
= BCD. The correct answer is A.
14. Triangles ABD and CBD are congruent. The
fact that BD is the perpendicular bisector of AC yields
Mathematics I Solutions Manual
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two pieces of information:
K
AD = DC and m BDA = m BDC = 90◦ .
E
This, combined with the fact that both triangles share side
BD lets you apply the SAS postulate.
15. (a) You cannot prove any new triangles congruent.
AB = AC is not actually new information. From
Exercise 14, we know ABD ∼
= ACD, and
from this we can conclude that AB = AC. No new
information means no new triangles are
congruent.
(b) You can prove that ADE ∼
= ADF by SAS.
This is the same argument as the one used in
Exercise 14. Because AD is the perpendicular
bisector of EF, we know that ED = FD. We already
know AD is a common side and ADF ∼
= ADE.
You can also prove that AEB ∼
= AFC and
AEC ∼
= AFB. The two known congruences,
ADB ∼
= ADC and ADE ∼
= ADF, yield
DB = DC and DE = DF by CPCTC. These
statements imply that BF = EC and BE = FC.
CPCTC also gives that AB = AC and AE = AF.
Now apply SSS to obtain the two congruences.
(c) You can prove ADE ∼
= ADF by ASA.
( EAD ∼
= FAD, AD is a common side, and
ADB ∼
= ADC.)
As in the solution to part (b), you can also prove
AEB ∼
= AFC and AEC ∼
= AFB.
16. (a) Yes, ABC ∼
= DEF.
m C = 180 − 72 − 42 = 66◦ = m F . So, the
triangles are congruent by the ASA Postulate.
(b) Suppose you have two triangles with two pairs of
corresponding angles congruent and a pair of
non-included sides congruent. Because all the angles
in the triangles must add up to 180◦ , it follows that
the third pair of angles is congruent. Now, the given
congruent sides are included between a pair of
congruent angles. So, by the ASA Postulate, the
triangles are congruent. This shows that AAS proves
triangles congruent.
I
T
KIT ∼
= KET, by the same SSS argument used for the
square in Exercise 5. However, KEI is not congruent to
TEI, as KE = ET, by definition of a kite.
8A
MATHEMATICAL REFLECTIONS
1. (a) ASA, SSS, and SAS
(b) The ASA postulate says that if two triangles have two
congruent pairs of angles, and if the side between
those two pairs of angles are congruent, then the two
triangles are congruent.
The SSS postulate says that if two triangles have
all three corresponding sides congruent to each other,
then the triangles are congruent.
The SAS postulate says that if two triangles have
two congruent pairs of sides, and if the angle between
those two pairs of sides are congruent, then the two
triangles are congruent.
(c) Below are possible illustrations for each postulate.
SSS
ASA
Maintain Your Skills
17. A diagonal of a rectangle divides it into two congruent
triangles. Both of the arguments used for the square in
Exercise 5 apply to the rectangle as well.
A diagonal of a parallelogram divides it into two
congruent triangles as well. The SSS argument used for
the square in Exercise 5 applies here.
A diagonal of a trapezoid does not divide it into
congruent triangles. The bases of the trapezoid are
not equal. This makes a pair of corresponding sides
of the triangles not congruent. Since all six pairs of
corresponding parts must be congruent for the triangles
to be congruent, these triangles cannot be.
One of the diagonals of a kite will divide it into
congruent triangles. Consider kite KITE.
SAS
2. (a) Does not make sense; MA and FL represent lengths,
it should say MA = FL or MA ∼
= FL.
(b) Makes sense
(c) Makes sense
(d) Makes sense
(e) Does not make sense; It should say m AMT =
m LFO or AMT ∼
= LFO.
3. Yes, the additional information is sufficient to show that
the two triangles are congruent. You know that CB ∼
= FD,
and since CA ∼
= CB and FD ∼
= FE you can conclude that
CA ∼
= FE. Together with the fact that ACB ∼
= DFE,
∼
this tells you that ABC = DEF by SAS.
Mathematics I Solutions Manual
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4. Here is the figure described in the exercise:
B
O
C
D
Some pairs of congruent triangles include:
• AOB ∼
= DOC by SAS ( AOB ∼
= DOC because
they are vertical angles.)
• AOC ∼
= DOB by SAS ( AOC ∼
= DOB because
they are vertical angles.)
• Using CPCTC on the previous triangle congruences,
you have AB ∼
= DC and BD ∼
= CA, so ABD ∼
=
DCA by SSS
• Similarly, BAC ∼
= CDB by SSS
6.
7.
8.
BD ∼
= AC is a consequence of the fact that
AOC ∼
= DOB and CPCTC.
CD ∼
= AB is a consequence of the fact that AOB ∼
=
DOC and CPCTC.
You’re given that OA ∼
= OB and AOD ∼
= BOD.
You also know that OD ∼
OD.
You
can
conclude that
=
∼
AOD ∼
BOD
by
SAS.
Then
AD
BD
by CPCTC.
=
=
Two figures are congruent if they have the same size and
shape. Their positions and orientations can be different.
For plane figures, you can try to superimpose one figure
on another to test to see if they are congruent. There are
also many ways of comparing two figures quantitatively
to see if they are congruent (like measuring segments and
angles).
There are many good reasons to keep track of the
corresponding parts in congruent figures. For one, when
you know two figures are congruent, you can infer
congruences among the parts (segments, angles) if you
know which parts are corresponding. For another reason,
in Exercise 11 of Lesson 8.04, we found a pair of
congruent angles was not a pair of corresponding angles;
this emphasizes the importance of matching up
corresponding parts when looking for triangle
congruences.
You could measure all the side lengths and all the
angles, to see if all six pairs of corresponding parts are
congruent. You could use one of the congruence
postulates: ASA, SAS, or SSS. You could also use AAS.
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INVESTIGATION 8B TRANSFORMATIONS
8.05
A
5.
2013/2/27
Getting Started
For You to Explore
1. (a) AX ∼
= BX because when the paper is folded on ,
they coincide. That means that the segments have
equal length.
(b) m AXC = 90◦ because when the paper is folded on
you can see that AXC ∼
= BXC because they
coincide. These two angles have equal measure and
they’re supplementary, so both have measure 90◦ .
(c) ABC is isosceles, because when the paper is folded
on you can see that AC ∼
= BC.
(d) is the perpendicular bisector of AB.
2. All of the segments from a point in the pre-image to the
corresponding point in the image appear to be parallel to
each other. They also appear to be perpendicular to the
fold line (and bisected by it). In fact, in the first In-Class
Experiment in the lesson, you saw that the line of
reflection is the perpendicular bisector of any segment
joining a point in the pre-image to the corresponding
point in the image after reflection.
3.
A
B J
I
A
C K
G
M
(c)
(b)
D
H
BN
L
D
CO
P
(a)
E
F
(−3, −4), (−1, −4), (−1, −1), (−3, −1)
(3, 4), (1, 4), (1, 1), (3, 1)
(1, 4), (−1, 4), (−1, 1), (1, −1)
(i) clockwise
(ii) counterclockwise
(iii) counterclockwise
(iv) counterclockwise
(b) All the paths for the images go in the same direction
(counterclockwise), but the path for the pre-image
goes in the opposite direction (clockwise).
(a)
(b)
(c)
4. (a)
5. (a)
(b)
(c)
(d)
(e)
(f)
12
6
4
2
6
3
On Your Own
6. The image must be congruent to the pre-image, because
when the paper is folded the two figures coincide exactly.
Mathematics I Solutions Manual
• Chapter 8, page 311
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7. (a) mPQ = 1, mQR = − 12 , mRS = 1, mPQ = − 12 . This
means that PQ is parallel to RS, and QR is parallel to
SP, so PQRS is a parallelogram because its opposite
sides are parallel.
(b) Here is a graph of the reflection of PQRS over the
line with equation y = −2. Note that since S is on
the line of reflection, it is its own image.
(c) Yes. mP Q = −1, mQ R = 12 , mR S = −1, mP Q = 12
This means that P Q is parallel to R S and Q R is
parallel to SP , so P Q R S is a parallelogram
because its opposite sides are parallel.
(d) No, they’re not. In each case the slope in the image is
equal to the opposite of the slope of the
corresponding segment in the pre-image.
8. (a) Graphs will vary.
(b) You may have chosen to label your quadrilateral in a
clockwise orientation, but you may have chosen to
label it with counter-clockwise orientation. Either is
correct.
(c) One way to check your reflection is to see if the line
with equation x = −2 is, in fact, the perpendicular
bisector of any segment between a point in the
pre-image and a corresponding point in the image.
(d) The image will have opposite orientation to the
pre-image.
(e) They’re always different.
9.
'
'
A = (2, −5), B = (4, −2)
10.
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Maintain Your Skills
11. (a) As long as the line of reflection doesn’t intersect the
triangle, the image will have no points in common
with the original triangle.
(b) The line of reflection can have only one point in
common with the original triangle if you want the
image to have one point in common with the original
triangle. The only way to make this happen is to have
the line of reflection contain one vertex of the
triangle, but no other point of the triangle. This
vertex will be the point that the original triangle and
the image have in common.
(c) The line of reflection must have two points in
common with the triangle. This will happen as long
as the line contains any point of the interior of the
triangle. The two points where the line of reflection
intersects the triangle will also be points in
the image.
(d) The line of reflection must contain one of the sides of
the triangle. Then the image will have that side in
common with the original triangle.
12. If you fold DEF so that point E matches up with point
F , the reflected image will coincide completely with
the original triangle. (This fold line is also called a line
of symmetry for the triangle.) If the triangle is
equilateral, then all three lines that are perpendicular
bisectors work.
8.06
Reflections
Check Your Understanding
1. (a) Always true. This is a consequence of the fact that the
image of a figure after reflection over a line is
congruent to its pre-image.
(b) Always true. This is a consequence of the fact that the
image of a figure after reflection over a line is
congruent to its pre-image.
(c) Always true. This is a consequence of the fact that the
image of a figure after reflection over a line is
congruent to its pre-image.
(d) Sometimes true. If the segment in question is parallel
to the line of reflection its image will also be parallel
to the line of reflection, and so it will have the same
slope as the pre-image segment. Also, if a segment is
perpendicular to the line of reflection, its image will
still be perpendicular to the line of reflection.
Instances where the slope of the image is not equal to
the slope of the pre-image occur in many of the other
exercises in this section.
(e) Always true. This is a consequence of the fact that the
image of a figure after reflection over a line is
congruent to its pre-image.
(f) Always true. This is a consequence of the fact that
the image of a figure after reflection over a line is
congruent to its pre-image.
Mathematics I Solutions Manual
• Chapter 8, page 312
“000200010271723958_CH08_p305-331”
2.
(a) A = (6, 5), B = (4, 2)
(b) A = (10, 5), B = (12, 2)
(c) The segment has been translated (slid) 8 spaces to the
right.
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triangle that Y cannot be a right angle. Therefore,
←
→
←
→
PY is not perpendicular to . It follows that PX is
the unique line through P and perpendicular to .
(b) Label the intersection of m and C. Draw a circle
with center C and radius CP. The circle intersects m
at two points: P and the desired point P . Since is
the perpendicular bisector of PP , part (a) implies
that P must be on m. Further, since we must have
CP = CP , it must be on the circle. Therefore, P can
only be the point of intersection different from P.
6. Given segment AB and line : Segment A B is
constructed so that is the perpendicular bisector of
segments AA and BB . Segments AA and BB intersect in points M and N respectively. We can mark the figure
to show this information:
3. Connect P and P , and let Q be the intersection point
with . Since is the perpendicular bisector of PP , it
follows that PQ = P Q and m PQO = 90◦ = m P QO.
Inasmuch as OQ = OQ, OQP ∼
= OQP by SAS.
∼
Finally, POR = P OR, by CPCTC.
4. If the segment is contained by the line of reflection, then
all of its points are fixed, and it is its own image after
reflection over that line. Since the segment was parallel to
the line before the reflection, it is still parallel after the
reflection.
In the second case, you’re given AB parallel to a line with no points in common with . The image of AB after
reflection over is A B , and segments AA and BB
intersect at points M and N respectively.
The image after reflection is always congruent to its
pre-image, so you know that AB ∼
= A B . You know that is the perpendicular bisector of AA , so m AMN = 90◦ .
Because AB is parallel to , you also know that
m MAB = 90◦ , since same-side interior angles must be
supplementary.
m A MN = m MA B = 90◦ because these two
angles are the images after reflection over of angles
AMN and MAB, and reflection preserves angle
measure. Since A MN and MA B must then be
supplementary, you can conclude that A B is parallel to .
2013/2/27
We can see that BNM ∼
= B NM by SAS. So,
∼
BM = B M and BMN ∼
= B MN by CPCTC. We
know that m AMB = 90◦ − m BMN and m A MB =
90◦ − m B MN , so we can conclude that AMB ∼
=
A MB . Then we know that AMB ∼
= A MB by SAS.
That means that AB ∼
= A B by CPCTC.
In the case where AB is perpendicular to , using the
fact that is a bisector, we get AN = A N and
BN = B N . By subtraction, AB = A B . In the case where
AB intersects the line at a point C, from the previous
cases we conclude that AC = A C and CB = CB . By
addition, AB = A B .
7.
B (3, 3)
A (1, 2)
B'' (3, 0)
A' (1, –1) = A''
B' (3, –2)
On Your Own
5. (a) Draw a line perpendicular to through P. Label the
intersection of with this line X. Now draw another
line through P that intersects at the point Y . Then
PXY is a right triangle with right angle X. It
follows from the sum of the interior angles of a
Mathematics I Solutions Manual
(a) A = (1, −1), B = (3, −2)
(b) A = (1, −1), B = (3, 0)
(c) The segment has been translated (slid) down 3 spaces.
• Chapter 8, page 313
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8.
'
'
''
''
This segment has been rotated (turned) 180◦ around the
point (4, −1).
9.
x+
x–
3y =
3y =
6
6
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12. The y-axis is a line of symmetry for the graph because
reflection over the y-axis maps every point (x, y) to the
corresponding point (−x, y) and (x)2 = (−x)2 because
of the properties of multiplying signed numbers.
13. The y-axis is a line of symmetry for the graph because
reflection over the y-axis maps every point (x, y) to the
corresponding point (−x, y) and (x)2 + 2 = (−x)2 + 2.
14. This graph has a different line of symmetry—the line
with equation x = −1. You can see this by folding the
graph (or visualizing such a fold). So now you need to
think about what the image of a point (x, y) will be after
reflection over the line x = −1. Imagine a point (x, y) in
the first quadrant. If you reflect it over the line with
equation x = −1, it will map to the point (−x − 2, y).
One way to see this is to realize that this reflection has no
effect on the y-coordinate of a point, but that the
x-coordinates of the original point, xo , and of its image
after the reflection, xi must average to −1 because you
want the midpoint of the segment with endpoints xo and
i
=
xi to lie on the line with equation x = −1. So xo +x
2
−1, which can be simplified to get xo + xi = −2 and then
xi = −2 − xo . To assure yourself that this mapping holds
even when the x-coordinate of a point is not positive, try
it for a few more points.
So now you need to show that
x2 + 2x + 1 = (−2 − x)2 + 2(−2 − x) + 1
Multiply out the right side:
10. Say the reflection of P(a, b) is Q(c, d). Since the slope of
y = x is 1 (goes up by an angle of 45◦ ), a perpendicular
line has slope −1 (goes down by an angle of 45◦ ). This
implies (1) d−b
= −1. Since y = x bisects PQ, the
c−a
midpoint of PQ is on y = x, and so has equal
= b+d
. From (1) and (2)
coordinates. This implies (2) a+c
2
2
we have a − c = d − b and a + c = b + d, respectively.
Add these and divide by two to get a = d. It follows that
c = b. Hence, the reflection is obtained by switching
coordinates. The correct answer is C.
11. The y-axis is a line of symmetry for the graph of the
equation y = |x|. Imagine folding the graph along the
y-axis. It will coincide with itself. Points along the arm of
the graph on the right map to points on the left arm,
points on the left arm map to points on the right. To prove
this conjecture, however, you need to show two things.
• |x| = | − x|. (This is true because of the properties of
the absolute value function.)
• Reflection of a point (x, y) over the y-axis maps it to
(−x, y) and vice versa. (This is true because the y-axis
is the perpendicular bisector of every segment with
endpoints (x, y) and (−x, y). Such a segment is
horizontal, so the y-axis is certainly perpendicular to
the segment. Also, the point (0, y) is the midpoint of
the segment and lies on the y-axis.)
Thus, reflection of the graph of the equation y = |x| over
the y-axis maps the graph onto itself and the y-axis is a
line of symmetry for the graph.
x2 + 2x + 1 = 4 + 4x + x2 − 4 − 2x + 1
Combine like terms and you have it:
x2 + 2x + 1 = x2 + 2x + 1
Now, since you know that when the graph is reflected
about the line with equation x = −1, that under this
reflection (x, y) → (−2 − x, y). Since both of these
points are on the graph of the original parabola, you can
conclude that the reflection maps the parabola onto itself.
This means that the line with equation x = −1 is a line of
symmetry for the parabola.
15. (a) Yes. The image after two reflections is no longer
backward. A rotation of the original pre-image F
around the intersection of lines r and s would map F
onto s(r(F)).
(b) No. After three reflections, the image of the original
F is backward. This means that no slide or turn could
map the original pre-image onto it without including
some kind of flip or reflection.
(c) Yes. r(F) is also a backward F , so you should be able
to visualize a slide or turn (depending on whether the
line t you chose is parallel to or intersects line s).
(d) Yes. After four reflections w(t(s(r(F)))) is not
backward, so the original pre-image F could be slid
Mathematics I Solutions Manual
• Chapter 8, page 314
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or turned to coincide with it. No flip (reflection) is
necessary.
(e) After an odd number of reflections, the final image is
backward. After an even number of reflection, the
final image is not backward. If the final image is not
backward, there is a slide or a rotation that can map
the original pre-image onto the final image.
16. There is only one line of symmetry for this combined
figure—the line with equation y = −5. It maps both the
circle onto itself and the chord onto itself.
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On Your Own
4.
8
6
B = (2, 5) B'
B''
4
2
A = (1, 2)
–5
5
17. (a) A, H, I, M, O, T, U, V, W, X, and Y appear to have a
vertical line of symmetry.
(b) B, C, D, E, H, I, K, O, and X appear to have a
horizontal line of symmetry.
(c) H, I, O, and X appear to have both.
(d) There are no letters that appear to have lines of
symmetry that are neither horizontal nor vertical.
(e) H, I, N, O, S, X, and Z appear to have 180◦ rotational
symmetry. (They look the same when the paper is
turned upside-down.)
10
15
x=6
x=3
–2
Maintain Your Skills
A''
A'
–4
The single transformation that maps AB onto A B is a
translation 6 spaces to the right.
5. You can show that quadrilateral AA B B is a
parallelogram in many ways. Here’s one way, using the
slope of the opposite sides: mAA = mBB = 0 so AA is
parallel to BB . mAB = mA B = 3 so AB is parallel to
A B . Since opposite sides are parallel, the quadrilateral
is a parallelogram.
6. (a) The equation becomes
y = −(x + 3)2 + 2(x + 3)
y = −(x2 + 6x + 9) + 2x + 6
8.07 Translations
y = −x2 − 6x + 2x − 9 + 6
Check Your Understanding
y = −x2 − 4x − 3
1. (a) You could slide a copy AKLJ so that it coincided
with A K L J without turning or flipping it.
(b) Move it over 8 spaces to the right and up 6 spaces.
(c) Add 8 to each x-coordinate and 6 to each
y-coordinate of the points in the original
quadrilateral, and you’ll get the corresponding points
in the image quadrilateral, or in symbols: (x, y) →
(x + 8, y + 6).
2. The figure you get will be a triangle that is congruent to
and in the same orientation as your original triangle. It
will be 10 spaces to the right and 6 spaces up from your
original triangle.
3. After the translation, (6, 7), the new x and y are too large,
so substitute x − 6 for x and y − 7 for y in the equation
for the line s.
The graph is three spaces to the left of the original.
x
⫺4
2
4
⫺4
(b) The graph will be translated two spaces to the right.
2
y
x
⫺2 O
2
4
⫺2
2(x − 6) + (y − 7) = 3
2x − 12 + y − 7 = 3
2x + y = 22
Mathematics I Solutions Manual
⫺2
⫺2
⫺4
You can check your solution by finding two points on the
original line s, such as (0, 3) and (1, 1). Translate these
points by (6, 7) to get (6, 10) and (7, 8), and check to see
that these two points are on your translated line.
2(6) + (10) = 22 and 2(7) + (8) = 22.
y
2
⫺4
Its equation is
y = −(x − 2)2 + 2(x − 2)
y = −x2 + 4x − 4 + 2x − 4
y = −x2 + 6x − 8
• Chapter 8, page 315
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(c) The equation becomes y − 2 = −x2 + 2x or y =
−x2 + 2x + 2. The new graph is two spaces up from
the original.
y
2
x
⫺4
⫺2 O
2
4
⫺2
⫺4
(d) The graph will be translated three spaces down.
2
y
x
⫺4
⫺2 O
2
4
⫺2
⫺4
x2 + (y + 1)2 = 4
This equation can also be written as
x2 + y2 + 2y = 3
As a check of your work, you can translate the points
you found for the original circle through the same
translation as you used to obtain the image circle, and
they should satisfy the equation for the image circle.
For example, (2, 0) → (2, −1). That translated point
is on the image circle, because (2)2 + (−1 + 1)2 = 4
is true.
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(b) x2 + (y − 1)2 = 4 or x2 + y2 − 2y = 3
(c) (x − 6)2 + y2 = 4 or x2 + y2 − 12x = −32
(d) (x − 3)2 + (y − 4)2 = 4 or x2 + y2 − 6x − 8y = −21
8. To find points on the circle, you can choose a value for x
or y and solve the equation to find the other
corresponding coordinate. If you use x = 0, you get
y2 + 2y = 3 or y2 + 2y − 3 = 0. This equation can be
solved using either factoring or the quadratic formula.
Factoring, you would get (y + 3)(y − 1) = 0, so y = 1 or
−3. The √quadratic formula would
give
√
−2+ 4+12
−2− 4+12
y=
= 1 or y =
= −3 Either
2
2
method allows you to find that (0, 1) and (0, −3) are on
the circle. Because these two points are on a vertical line
together, you know that the center of the circle must have
the y-coordinate of the midpoint of the segment between
them. So the y-coordinate of the center is –1.
√Similarly
you√can substitute y = 0 to find the points ( 3, 0) and
(− 3, 0). Because these two points are on a horizontal
line together, you know that the center of the circle must
have the x-coordinate of the midpoint of the segment
between them. So the x-coordinate of the center is 0 and
the center is (−1, 0).
(a) When the circle is translated by (0, −2), the equation
of the image is x2 + (y + 2)2 + 2x = 3 which can be
written as (x + 1)2 + (y + 2)2 = 4 or x2 + y2 +
2x + 4y = −1.
(b) When the circle is translated by (0, 2), the equation
of the image is x2 + (y − 2)2 + 2x = 3 which can be
written as (x + 1)2 + (y − 2)2 = 4 or x2 + y2 +
2x − 4y = −1.
(c) When the circle is translated by (3, 0), the equation of
the image is (x − 3)2 + y2 + 2(x − 3) = 3. This gives
you x2 − 6x + 9 + y2 + 2x − 6 = 3. which can be
written as (x − 2)2 + (y)2 = 4 or x2 + y2 − 4x = 0.
(d) When the circle is translated by (1, 4), the equation of
the image is (x − 1)2 + (y − 4)2 + 2(x − 1) = 3. This
gives you x2 − 2x + 1 + y2 − 8y + 16 + 2x − 2 = 3
which can be written as (x)2 + (y − 4)2 = 4 or
x2 + y2 − 8y = −12.
Its equation is y + 3 = −x2 + 2x or y = −x2 +
2x − 3.
7. You can find points on the circle by choosing a value for
x so that x2 ≤ 4, substituting that value for x in
the equation of the circle and solving for y. Some of the
simplest points to locate may be the places where the
circle intersects the coordinate axes. These points are
(2, 0), (0, 2), (−2, 0) and (0, −2). To show that the
center is (0, 0), you just need to confirm that all of these
points are equidistant from that center. Since each is 2
units away from (0,0), that is the center of the circle.
(a) When the circle is translated by (0, −1), you can
follow the process from episode 30. Think of the
original circle as the image of the circle you want
after the opposite translation—by (0, 1). The
coordinates of a point (x, y) after this translation
would map to (x, y + 1). You know an equation that
is true for any of the points on this image circle, so if
you substitute the coordinates of this point, you’ll get
a true equation.
2013/2/27
9. First you get the point (5, 2c − 3). Then, that is reflected
to (5, 2d − (2c − 3)). The correct answer choice is D.
Maintain Your Skills
10. (a) An equation for a circle centered at (4, 3) with radius
5 is (x − 4)2 + (y − 3)2 = 25 or x2 + y2 − 8x −
6y = 0.
(b) An equation for a circle centered at (−2, 6) with
radius 5 is (x + 2)2 + (y − 6)2 = 25 or x2 + y2 +
4x − 12y = −15.
(c) An equation for a circle centered at (5, −1) with
radius 5 is (x − 5)2 + (y + 1)2 = 25 or x2 + y2 −
10x + 2y = −1.
11. (a) The translation (−4, −3) would translate the circle in
part (a) so that it was centered at the origin. That
translation maps points (x, y) to (x − 4, y − 3).
(b) The translation (2, −6) would translate the circle in
part (b) so that it was centered at the origin. That
Mathematics I Solutions Manual
• Chapter 8, page 316
“000200010271723958_CH08_p305-331”
translation maps points (x, y) to (x + 2, y − 6).
(c) The translation (−5, 1) would translate the circle in
part (c) so that it was centered at the origin. That
translation maps points (x, y) to (x − 5, y + 1).
12. (a) Substituting (x + 4, y + 3) into the equation
(x − 4)2 + (y − 3)2 = 25 gives (x + 4 − 4)2 +
(y + 3 − 3)2 = 25, which is just x2 + y2 = 25. Or, if
you substitute into the other form, you’ll get
(x + 4)2 + (y + 3)2 − 8(x + 4) − 6(y + 3) = 0
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(c) For a regular polygon with an odd number
of sides, each line of symmetry passes through one
vertex and the midpoint of the opposite side. For a
regular polygon with an even number of sides, half of
the lines of symmetry go through an opposite pair of
vertices, and the other half of the lines of symmetry
go through the midpoints of an opposite pair
of sides.
3.
O
x + 8x + 16 + y + 6y + 9 − 8x − 32 − 6y − 18 = 0
2
2
x2 + y2 = 25
B
(b) (x − 2 + 2)2 + (y + 6 − 6)2 = 25 gives x2 + y2 = 25
immediately. If you use the other form of the
equation of the circle, you’ll get
B'
(x − 2) + (y + 6) + 4(x − 2) − 12(y + 6) = −15
2
2
x2 − 4x + 4 + y2 + 12y + 36 + 4x −
A
8 − 12y − 72 = −15
x + y2 = 25
2
(c) (x + 5 − 5)2 + (y − 1 + 1)2 = 25 gives x2 + y2 = 25
immediately. If you use the other form, you’ll get
(x + 5)2 + (y − 1)2 − 10(x + 5) + 2(y − 1) = −1
A'
4.
x2 + 10x + 25 + y2 − 2y + 1 − 10x −
50 + 2y − 2 = −1
B
B''
x + y = 25
2
2
B'
A''
8.08
Rotations
A
r
A'
1. (a) A (4, 1), B (2, 3), C (4, 5), D (5, 2.5)
(b) A (4, −1), B (2, −3), C (4, −5), D (5, −2.5)
(c) Theorem 8.1 states that the composition of two
reflections about intersecting lines produces a
rotation. Its center is the intersection of the lines. The
angle of rotation is equal to twice the measure of the
angle formed by two lines.
(d) a rotation of 180◦ around the origin
(e) If (x, y) are the coordinates of a point of ABCD, then
the corresponding point of A B C D has
coordinates (−x, −y).
2. (a)
Shape
Number of Number of Total Ways
Rotations Reflections
to Map
Figure
Onto Itself
Eq. Triangle
3
3
6
Square
4
4
8
Regular
5
5
10
Pentagon
Regular
6
6
12
Hexagon
(b) If n is the number of sides, there are n lines of
symmetry.
s
the point of intersection of r and s; about 115◦
counterclockwise; yes, by measuring one of the acute
angles formed by r and s and doubling the result.
5. (a) Answers may vary. Sample:
(b) A pentagon with exactly two lines of symmetry does
not exist. If a pentagon had exactly two lines of
symmetry, it would have a rotational symmetry other
than 360 degrees, since a rotation is a composition of
two reflections. The only pentagon with rotational
symmetry other than 360 degrees is a regular
pentagon, which has more than two lines of
symmetry. Therefore, no pentagon has exactly two
lines of symmetry.
6. (a) Check students’ work.
(b) Check students’ work.
Mathematics I Solutions Manual
• Chapter 8, page 317
“000200010271723958_CH08_p305-331”
(c) Answers may vary depending on lines of symmetry
selected. If students selected both diagonal lines of
symmetry, or the horizontal and vertical line of
symmetry, the answer is 180-degree rotation. If
students selected one diagonal line of symmetry and
either the horizontal or vertical line of symmetry, the
answer is 90-degree rotation, but that rotation may be
clockwise or counterclockwise depending on the
selection.
7.
O
B
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page 318
←
→
←
→
11. Answers may vary. Sample: Draw OA and OA .
Construct the line k that bisects AOA . If you reflect
←
→
ABC over OA and then reflect the image over k, you
get A B C .
12. D
13. (a) Any could have been first.
(b) Answers may vary. Sample: The figure will have a
finite number of F’s symmetrically spaced with
respect to the origin and be symmetric with respect to
both lines.
(c) There are several repeated rotations.
(d) Answers may vary. Sample: The overall figure has
both reflection and rotational symmetry.
C'
C
8.09
A
2013/2/27
Congruence and Isometry
B'
A'
8. Check students’ work. Answers may vary. Sample:
P
9. CA = CB, so A and B are on the circle with center C and
radius CA; the measure of the angle of rotation is the
same as m ACB.
10.
1. An isometry can be thought of as a function that maps
points in the plane to points in the plane. Formally, an
isometry is a function from R2 to R2 .
2. Answers may vary. Sample: You would need a
description or ordered list of the transformations that
make up the isometry.
3. (a) any point on the circle with center P and radius AB
(b) Answers may vary. Sample: a description of the
transformations that make up the isometry,
including the order they were performed in
4. Answers may very. Check students’ work.
5. (a) Isometry; the composition of a reflection and a
translation is an isometry.
(b) Not an isometry; the transformation scales the
figure, which does not preserve length.
(c) Not an isometry; the transformation stretches the
figure, which does not preserve length.
(d) Isometry; the transformation is a translation.
(e) Isometry; the transformation is a translation.
(f) Not an isometry; the transformation scales the
figure, which does not preserve length.
6. Triangles may vary. Check students’ work. Sample
explanation: Yes; the transformation is an isometry,
which maps figures onto congruent figures.
7. No; a transformation that takes one circle to the other
does not preserve length.
8. (a) Answers may vary. Sample: Rotate ABC until BC
is parallel to B C . Then translate ABC until point
C is concurrent with point C . Then reflect so that
point B is concurrent with point B .
(b) Check students’ work.
(c) Point A is a fixed distance from point B and a fixed
distance from point C. Point A1 is the same distance
from point B and the same distance from point C as
point A. Therefore, point A1 coincides with point A.
(d) Check students’ work.
9. Check students’ work. Recall that two shapes are
congruent if there is an isometry that maps one shape
onto the other.
10. (a) 4
(b) Check students’ work.
Mathematics I Solutions Manual
• Chapter 8, page 318
“000200010271723958_CH08_p305-331”
11. (a) 1
(b) Rotate ABC until AB is parallel to A B . Then
translate ABC until point C is concurrent with
point C .
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3. Refer to the following figure.
P''
10
8
O''
6
y=5
8B
MATHEMATICAL REFLECTIONS
4
1. Here is a figure.
2 O'
P
y=1
6
B
–5
B’ = (5, 5)
P'
5
10
–2
4
The coordinates of the end points of O P are
yO = yO + 2 times the distance of O from line y = 1.
2
–5
5
=0+2·1=2
10
xO = xO = 0
–2
A
O
A ' = (12, –2)
yP = YP − 2 times the distance of P from line y = 1.
–4
=2−2·1=0
–6
xP = xP = 3.
You can calculate the coordinates of the image A B as
follows:
xA = xA + 2 (distance of A from the line of reflection)
The coordinates of O P are
yO = yO + 2 times the distance of O from line y = 5.
=2+2·3=8
xO = xO = 0
= −6 + 2 · 9 = 12
y
A
yP = yP + 2 times the distance of P from line y = 5.
= yA
= 0 + 2 · 5 = 10
xP = xP = 3.
xB = xB + 2 (distance of B from the line of reflection)
The composition of these two reflections is a translation.
A rule in the form (x, y) → (, ) that summarizes this
transformation is
= 1 + 2 · 2 = 5.
(x, y) → (x + 0, y + 8).
2. Here is a graph of the parabola with equation y =
2x2 + 3 and the translated parabola.
10
8
6
4
2
–10
–5
5
10
–2
The equation of the translated parabola is y = 2x2 −
4x + 7. You can obtain this equation by substituting
x = x − 1, y = y − 2 in the original equation and then
writing x , y simply without their superscripts, since the
meaning of x was “the coordinate of the new x.”
4. Yes, you can obtain a rotation by composing two
reflections over lines that are not parallel.
5. The reflected object is congruent to its pre-image. This
means that many of its properties, including angle
measures, segment measures, collinearity, parallel lines,
perpendicular lines, and so on will be the same. The
image will, however have a different orientation. In
other words, if you walk around the pre-image’s vertices
in alphabetical order and your path appears to be
clockwise to a person observing your walk from above,
then if you walk around the image’s vertices in
alphabetical order, your path will appear to be
counterclockwise.
6. When you compose two reflections, you will get either a
translated image or a rotated image. If the lines you
reflected about were parallel, the composition is a
translation. If they were intersecting, the composition is
a rotation about the intersection point through an angle
that measures twice as much as the angle formed by the
two lines. In either case, the orientation of the final
image will be the same as that of the pre-image.
Mathematics I Solutions Manual
• Chapter 8, page 319
“000200010271723958_CH08_p305-331”
When you compose a third reflection, the result will
be equivalent to either a reflection or a combination of a
reflection and a translation. If you haven’t come across
that case yet, don’t worry about it, but the composition
of three reflections will not always be equivalent to a
single reflection. However, the image after a
composition of three reflections will always have
opposite orientation to the original pre-image.
The fourth reflection will create a composition that
will be equivalent to either a translation or a rotation.
The orientation of the final image will be the same as
that of the pre-image.
7. A = (2, 1), B = (−2, 3), C = (2, 3)
INVESTIGATION 8C
8.10
GEOMETRY IN THE
COORDINATE PLANE
Getting Started
For You to Explore
1. (a) The coordinates of the fourth vertex are (2, −6).
(b) Points inside the rectangle have x-coordinates such
that 2 < x < 8, and y-coordinates such that
−6 < y < 5.
(c) Points outside the rectangle have x-coordinates such
that x < 2 or x > 8, or y-coordinates such that
y < −6 or y > 5.
(d) Points that lie on the rectangle will fall into one of
four categories.
• Points along the top side will have coordinates
(x, y) such that 2 ≤ x ≤ 8 and y = 5.
• For points along the right side, x = 8 and
−6 ≤ y ≤ 5.
• For points along the bottom side, 2 ≤ x ≤ 8 and
y = −6.
• For points along the left side, x = 2 and
−6 ≤ y ≤ 5.
(e) If a point’s x-coordinate is between 2 and 8 and its
y-coordinate is between −6 and 5, then that point is
inside this rectangle.
2. (a) 391
(b) (2, 200.5)
3. (a) one—the line with equation x = −1
(b) I = (−1, −5)
(c)
R
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4. (a) (−2,
√ 1).
√
√
(b) 22 + 62 = 40 = 2 10.
On Your Own
5. The number midway between −1 and 3 is 1, so the
midpoint is (1, 1).
6. (a) CD = 17
(b) The coordinates of the midpoint of CD are (3.5, −7).
(c) DE = 10
(d) The coordinates
of the√
midpoint of DE are (12, −2).
√
(e) CE = 172 + 102 = 389
(f) The
coordinates
of CE are
of 7the midpoint
−5+12
−7+3
=
,
,
−2
.
2
2
2
7. (a) The coordinates of A are (5, 5) and B are (5, 0).
(b) AB = 5
(c) The coordinates of the midpoint of AB are (5, 2.5).
(d) The area of AOB = 12 (5)(5) = 25
= 12.5
2
√
√
2
2
(e) AO = 5 + 5 = 5 2
(f) The coordinates of the midpoint of AO are (2.5, 2.5).
8. Solutions will vary. Here is an example of a puzzle that
draws CMEP:
• Start at (−4, 2), connect to (−6, 2), (−6, −2), and
then to (−4, −2).
• Start at (−3, −2), connect to (−3, 2), (−1, 1 12 ), (0, 2),
and then to (0, −2).
• Start at (3, 2), connect to (1, 2), (1, −2), and then to
(3, −2).
• Start at (3, 0) and connect to (1, 0).
• Start at (4, 0), connect to (6, 0), (6, 2), (4, 2), and
finally to (4, −2).
9. Take It Further
(a) The only vertical line through the origin is the y-axis.
It touches (it is a boundary of) all four quadrants, but
doesn’t really pass through any.
(b) Any other vertical line must pass through exactly two
quadrants.
(c) A slant line through the origin must pass through
exactly two quadrants.
(d) Any other slant line must pass through exactly three
quadrants.
10. Take It Further Any line must pass through at least two
quadrants because at least one of its two coordinates is
changing. That changing coordinate must eventually
assume both positive and negative values. No line can
pass through all four quadrants. A slanting line that passes
through only two quadrants must pass through the origin.
Maintain Your Skills
11. (a)
I
TI = 4, RI = 14, TR =
R
√
√
42 + 142 = 212
A
B
(x, y)
(x + 3, y) (−x, y) (x, −y)
(2x, 2y)
(2, 1)
(5, 1)
(−4, 0) (−1, 0)
(−2, 1) (2, −1)
(4,0)
(−4, 0)
(4, 2)
(−8, 0)
(−5, 4) (−2, 4)
(5, 4)
Mathematics I Solutions Manual
C
D
E
F
x
2,
y
2
1
2
G
(−y, x)
(−1, 2)
1,
(−2, 0) (0, −4)
(−5, −4) (−10, 8) − 52 , 2 (−4, −5)
• Chapter 8, page 320
“000200010271723958_CH08_p305-331”
(b) The triangle described by the three points in column
B is identical (congruent) to the triangle described
by the three points in A, but translated three units to
the right.
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y
10
E
y
6
A
A
B
10
–10
–6
6
x
x
–6
–10
(c) “Triangle C” is a reflection of “triangle A” about the
y-axis.
y
y
6
6
A
A
C
F
6 x
–6
–6
6
x
–6
–6
(d) Triangle D is a reflection of triangle A about the
x-axis. Triangles E and F are scalings of triangle A.
They are similar to triangle A (the same shape as A),
but twice and half its size respectively. The vertices
of E are twice as far from the origin as the vertices of
A, and the vertices of F are half as far from the origin
as the vertices of A.
Triangle G is congruent to triangle A, but rotated 90◦
counterclockwise about the origin.
y
y
6
6
A
A
–6
6
x
6 x
–6
D
G
–6
Mathematics I Solutions Manual
–6
• Chapter 8, page 321
“000200010271723958_CH08_p305-331”
8.11
Check Your Understanding
1. The midpoint of any diameter is the center of the circle.
(a) (−33, 561)
(b) (3x, 0.5y)
2. (a) The lengths of the sides of ABC are computed
below.
AB = (8 − 1)2 + (4 − 2)2
√
√
= 49 + 4 = 53
(8 − (−2))2 + (4 − 6)2
√
√
√
= 100 + 4 = 104 = 2 26
(1 − (−2))2 + (2 − 6)2
√
√
= 9 + 16 = 25 = 5
AC =
Opposite sides have the same slope, so they’re parallel,
so this is a parallelogram.
Here’s the length method:
• The lengthof the segment between points (3, 5) and
2
2
(4,
−1) is (−1 −√5) + (5 − 4) =
2
2
(−6) + (1) = 37.
• The length ofthe segment between points (4, −1) and
2
2
(−1,
−2) is (−2 −√(−1)) + (−1 − 5) =
2
2
(−1) + (−5) = 26.
• The length of the
segment between points (−1, −2)
and
(−2,
4)
is
(4 − (−2))2 + (−2 − (−1))2 =
√
(6)2 + (−1)2 = 37.
• The length
of the segment between points (−2, 4) and
(3,
5)
is
(5 − 4)2 + (3 − (−2))2 =
√
2
(1) + (5)2 = 26.
(b) To find the lengths of the medians, first find the
midpoints of each side. The midpoint of BC is M1 = (5, 3). The midpoint of CA is M2 = 4, − 21 .
The midpoint of AB is M3 = 3, 4 12 . Now, the length
of each median can be computed.
AM 1 = (1 − 3)2 + (2 − 5)2
√
√
= 4 + 9 = 13
1 2
8− −
+ (4 − 4)2
2
√
72.25 = 8.5
CM 3 =
−2 − 4
1
2
2
+ (6 − 3)2
√
√
42.25 + 9 = 51.25
√
AB = 34 = A B
√
BC = 29 = B C √
CA = 13 = C A
=
3.
ABC ∼
= A B C by SSS.
4. There are other tests for congruence, but SSS is the
easiest one to use when you’re just looking at the
coordinates of the vertices. For each triangle, find the
distance between each pair of the three vertices and then
compare corresponding sides. The two triangles are
congruent when the corresponding sides are congruent.
5. • The
coordinates
of the midpoint of ST are
−2+8
, 8+2
= (3, 5).
2
2
• The
of the midpoint of TA are
coordinates
8+0 2+(−4)
= (4, −1).
, 2
2
• The
coordinates
0+−2
of the midpoint of AR are
−4+0
= (−1, −2).
,
2
2
page 322
• The slope of the segment between points (3, 5) and
(4, −1) is −1−5
= −6
= −6.
4−3
1
• The slope of the segment between points (4, −1) and
(−1, −2) is −2−(−1)
= −1
= 15 .
−1−4
−5
• The slope of the segment between points (−1, −2) and
4−(−2)
6
(−2, 4) is −2−(−1)
= −1
= −6.
• The slope of the segment between points (−2, 4) and
5−4
(3,5) is 3−(−2)
= 51 .
BC =
=
1:46
• The
coordinates
−2+−2
of the midpoint of RS are
0+8
= (−2, 4).
,
2
2
There are two ways to show that the quadrilateral through
points (3, 5), (4, −1), (−1, −2), and (−2, 4) is a
parallelogram. We might find the slope of the four sides
and see that opposite sides have the same slope, or we
might find the lengths of the four sides and see that
opposite sides have the same length.
Here’s the slope method:
Midpoint and Distance Formulas
BM 2 =
2013/2/27
Opposite sides have the same length, so they’re
congruent, and this is a parallelogram.
6. Here the problem is solved for one set of four particular
points, A = (3, 1), B = (4, 7), C = (11, 3), and
D = (8, −5). The midpoints of the four sides of the
quadrilateral are: MAB = (3.5, 4), MBC = (7.5, 5),
MCD = (9.5, −1), and MAD = (5.5, −2).
To show that the shape you get when you connect these
midpoints is a parallelogram, show that you have two
pairs of congruent opposite sides.
√
d(MAB , MBC ) = d(MAD , MCD ) = 17
and
√
d(MAD , MAB ) = d(MBC , MCD ) = 2 10
As in Exercise 5, you could also find the slopes of
opposite sides
√ and show that they are equal.
7. • A = (2, 4√ 6)
• B = (5, 5 3)
• C = (6,√8) √
• D = (5 2, 5 2)
• E = (10, 0)
• F = (0, −10)
• G = (−8, −6)
Mathematics I Solutions Manual
• Chapter 8, page 322
“000200010271723958_CH08_p305-331”
On Your Own
8. Plotting the three known points shows that (−1, 5) and
(3, −3) are the two diagonally-opposite vertices.
(a) So the center of the square must be midway between
them, at (1, 1).
(b) That must also be the midpoint of the diagonal
between the second vertex, (5, 3), and the fourth
vertex, so the fourth vertex must be at (−3, −1).
9. These points are hard to plot, but a rough sketch will help
you solve the problem. Such a sketch will show that
(−214, −86) and (186, 114) are opposite vertices of the
square. (You can verify this by calculating distances
between these three pairs of points.)
(a) The center of the square must be the midpoint of the
segment
these
opposite vertices, or
−214+186between
−86+114
=
(−14,
14).
,
2
2
(b) This center must also be the midpoint of a segment
from the third given point (−114, 214) to the fourth
vertex of the square. Traveling from the given point
to the center takes you 100 units to the right and 200
units down. If you travel 100 units right and 200
units down from the center, you’ll get to the fourth
vertex of the square. That point will have coordinates
(86, −186).
You can check your work by finding the lengths of
the sides of your figure. If they’re all equal, your
figure is a square.
10. To move from D to F , you move 2.5 to the right and up
1, so to move from F to E you would do the same,
landing at E = (7, 18).
11. E is the
ofAB, so its coordinates are
midpoint
E = 110+116
, 15+23
= (113, 19). To get from C to E
2
2
you move 3 to the right and down 4, so to move from E
to D you would do the same, landing at D = (116, 15).
12. There are many possibilities; these are just examples:
(a) (21, 7) and (46, 7)
(b) (−16, −13) and (−16, 12)
(c) (0, 0) and (15, 20)
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page 323
(b) Answers may vary. Sample: (0, 0), (6, 0), (9, 0), (9,
10), (9, −99), (−10, −100)
(c) The point must have x-coordinate equal to 10. In
other words, it must lie on the line with equation
x = 10, which is the perpendicular bisector of PQ).
17. Let the vertices of the triangle be A = (x1 , y1 ),
B = (x2 , y2 ), and C = (x3 , y3 ). Find the midpoints of
two sides:
x1 + x2 y1 + y2
MAB =
,
.
2
2
x2 + x3 y2 + y3
MBC =
,
.
2
2
The length of AC, the third side of the triangle, is
AC = (x3 − x1 )2 + (y3 − y1 )2 .
The length of segment MAB MBC is
d(MAB , MBC )
x2 + x 3
x 1 + x2 2
y 1 + y2 2
y2 + y3
=
−
−
+
.
2
2
2
2
This expression simplifies to
1
(x3 − x1 )2 + (y3 − y1 )2
d(MAB , MBC ) =
= AC.
2
2
Maintain Your Skills
18.
19.
20.
21.
22.
−13
−13
−13
−13
−13
8.12
Parallel Lines and Collinear Points
Check Your Understanding
13. There are many possibilities; these are just examples:
•
•
•
•
2013/2/27
1. (a) parallel, because
=
2
2
3
3
1
(b) parallel, because 31 = 13
(c) They are the same line.
√
(d) not parallel, because 22 =
(6, 10) and (10, 10)
(8, 0) and (8, 20)
(7, 9) and (9, 11)
(7, 7) and (9, 13)
√
14. The coordinates of the midpoint are the averages of the
coordinates of the endpoints, and the average of two
values is half their sum. So, if (−2, 1.5) is the midpoint,
then twice that, (−4, 3), is the pair of coordinates
representing the “sum” of the coordinates of the
endpoints. If (−4, 3) is the sum, and (−7, −2) is one of
the endpoints, then (3, 5) must be the other.
15. Use
the distance formula to get
√
(3.5 + 8.5)2 + (−2 − 14)2 cm = 400 = 20 cm. Then
convert to meters: 20 · 100
m = 2000 m. The correct
1
answer choice is D.
16. (a) Answers may vary. Sample: (10, 0), (10, 1), (10, 10),
(10, 110), (10, −11), (10, −47)
4
4
(e) parallel, because 2 1 2 = √42
2. There are many possible solutions, but one way is to
notice that traveling from A to B involves moving to the
right 3 and down 4. If you move to the right 3 and down 4
from B you’ll stay on the line and locate the point
(11, −7). This method allows you to find as many points
on the line as you want to, but it won’t work as a general
test to see if a point is on this line.
You could also write an equation for the line through
points A and B. Then, any point whose coordinates
satisfy the equation is on the line. This would be a
stronger method, because you would have a way to test
any point to see if it’s on the line or not. We know that the
slope between any point (x, y) that’s on this line and A
Mathematics I Solutions Manual
• Chapter 8, page 323
“000200010271723958_CH08_p305-331”
must be the same as the slope between points A and B. In
= − 43 . This simplifies to
symbols, that would be y−1
x−5
4x + 3y = 23.
3. As in Exercise 2, there are many possible solutions.
Probably the most straightforward is to write an equation
for the line through A and B. Then you can generate as
many points on the line as you want by substituting
values for x and finding the corresponding y values. You
can also test points to see whether they are solutions to
the equation or not.
50
4. The slope between points R and S is 20−(−30)
= 120
=
80−(−40)
−30−(−14)
5
. The slope between points P and R is −40−(−4) =
12
−16
= 49 . Since the two slopes aren’t equal, P is not
−36
collinear with R and S.
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Maintain Your Skills
10. Your solution will depend on the points you choose to
test. Here’s an example testing the point (20, 0).
• The
distance from (−40, −30) to (80, 20) is
2
(20 − (−30))2 =
√(80 − (−40)) + √
14400 + 2500 = 16900 = 130.
• The
distance from (−40, −30) to (20,√0) is
2
2
√(20 − (−40))
√ + (0 − (−30)) = 3600 + 900 =
4500 = 30 5.
• The
20) is
distance from (20, 0) to (80,
√
2 + (20 − 0)2 =
(80
−
20)
3600
+ 400 =
√
√
4000 = 20 10.
√
√
• Since 130 = 30 5 + 20 10, the point (20, 0) is not
collinear with our target points.
Here’s an example testing the point (20, −5).
On Your Own
• The
distance from (−40, −30) to (80, 20) is
2
(20 − (−30))2 =
√(80 − (−40)) + √
14400 + 2500 = 16900 = 130.
• The
distance from (−40, −30) to (20, −5) is
2 + (−5 − (−30))2 =
√(20 − (−40)) √
3600 + 625 = 4225 = 65.
• The
20) is
distance from (20, −5) to (80,
√
2
2
√(80 − 20) + (20 − (−5)) = 3600 + 625 =
4225 = 65.
• Since 130 = 65 + 65, the point (20, −5) is collinear
with our target points.
5. Two vertices of the equilateral triangle are (1, 0) and
(9, 0), so all three sides must be 8 units long. Since the
third point of the triangle is equidistant from the two that
are known, it must lie on the perpendicular bisector of the
segment from (1, 0) to (9, 0), which is the line with
equation x = 5. Since the x-coordinate of the third point
must be 5, you can find the y-coordinate by using the
distance formula.
8 = (5 − 1)2 + (y − 0)2
Square both sides of the equation, and make a note that
you need to check for any extra, irrelevant solutions that
may result.
√
√
64 = 16 + y2 ⇒ y2 = 48 ⇒ y = 4 3 or − 4 3
6.
7.
8.
9.
11. Take It Further. This is the equation students should
start with:
(r1 − p1 )2 + (r2 − p2 )2 + (s1 − p1 )2 + (s2 − p2 )2
= (s1 − r1 )2 + (s2 − r2 )2 .
There√are two solutions.
√ The third point might be
(5, 4 3) or (5, −4 3). Both solutions make sense,
because the third point can be above or below the other
two.
Lines , p, q, and r all meet at the origin. Lines m and n
are parallel. Lines m and n intersect lines , p, q, and r
elsewhere. No two of the lines are the same.
11−10
1
The slope between (60, 10) and (10, 11) is 10−60
= − 50
.
10−9
1
The slope between (110, 9) and (60, 10) is 60−110 = − 50 .
Therefore (110, 9) is collinear with (60, 10) and (10, 11).
You could find an equation for the line and test the points.
Alternatively, you can simply observe that as x decreases,
y increases. So when x = 2, y must be greater than 4. The
correct answer choice is A.
Line a has equation x + y = 1, line b has
equation
x − y = 0. They intersect at the point 12 , 21 .
a
b
(0, 1)
Square both members and isolate the remaining square
root in the left side of the equation. This is the equation
after simplification in the right member.
(r1 −p1 )2 (s1 − p1 )2 + (r1 − p1 )2 (s2 − p2 )2
2
+(r2 − p2 )2 (s1 − p1 )2 + (r2 − p2 )2 (s2 − p2 )2
= −2p21 − 2p22 + 2r1 p1 + 2r2 p2 + 2s1 p1 + 2s2 p2
− 2s1 r1 − 2s2 r2
Now look at the first member as
(r1 s1 − s1 p1 − r1 p1 + p21 )2 +
(r1 s2 − s2 p1 − r1 p2 + p1 p2 )2 +
(r2 s1 − p2 s1 − p1 r2 + p2 p1 )2 +
(r2 s2 − s2 p2 − r2 p2 + p22 )2
and look at the second member of the equation as
[(s1 r1 − s1 p1 − r1 p1 + p21 ) + (r2 s2 − s2 p2 − r2 p2 + p22 )]2 .
So by simplifying equal terms the equation becomes
(m, m)
(r1 s2 − s2 p1 − r1 p2 + p1 p2 )2 +
(r2 s1 − p2 s1 − p1 r2 + p2 p1 )2 =
(1, 0)
2(s1 r1 − s1 p1 − r1 p1 + p21 )(p22 − r2 p2 − s2 p2 + r2 s2 )
Mathematics I Solutions Manual
• Chapter 8, page 324
“000200010271723958_CH08_p305-331”
but noticing that
1:46
page 325
circle). In general the equation of a line of slope m
through a point (a, b) can be written in the form
m = y−b
. (In other words, it’s all the points that have
x−a
slope m when paired with the point (a, b).) So you can
leave m as the unknown and calculate the distance of a
line m = y−0
or mx − y = 3m from the origin. To find
x−3
this distance, you’ll write the equation of the line
perpendicular to this line that passes through the origin.
Its equation will be of the form x + my = d, and since the
line passes through (0, 0), d = 0. So the perpendicular to
mx − y = 3m through the origin is x + my = 0. Solve
this system of two equations to find the intersection point:
(s1 r1 − s1 p1 − r1 p1 + p21 )(p22 − r2 p2 − s2 p2 + s2 r2 ) =
(r1 s2 − s2 p1 − r1 p2 + p1 p2 )(r2 s1 − p2 s1 − p1 r2 + p2 p1 )
the equation becomes
[(r1 s2 − s2 p1 − r1 p2 + p1 p2 ) −
(r2 s1 − p2 s1 − p1 r2 + p2 p1 )]2 = 0.
The solution is
(r1 − p1 )(s2 − p2 ) = (r2 − p2 )(s1 − p1 ).
This is the equation that the coordinates of P have to
satisfy for P to be collinear with R and S.
8.13
2013/2/27
mx − y = 3m
x + my = 0
Multiply the first equation by m:
Perpendicular Lines
m2 x − my = 3m2
x + my = 0
Check Your Understanding
1. The slope, m, of a line with equation ax + by = c is − ba .
All the lines perpendicular to this line will have slope ba ,
and so the equation of any line perpendicular to the line
with equation ax + by = c will be of the form bx −
ay = d, where d is any real number. This means that the
equations of all the perpendicular lines to x + y = 3 can
be written in the form x − y = q, where q ∈ R.
2. The equations of all the perpendicular lines to x + 2y =
4 can be written in the form 2x − y = q, where q ∈ R.
Since the line has to contain (1, 0), you can find q by
solving the equation: 2 − 0 = q. So the line you are
looking for has equation 2x − y = 2.
3. The equation of any line perpendicular to s will be of the
form 4x + 3y = d. Since you know that P is on the line
you want, to find d, substitute the coordinates of P into
the equation 4(3) + 3(2) = d = 18. So the equation of
the line perpendicular to s through P is 4x + 3y = 18.
Then you can find the intersection of the two
perpendicular lines by solving the system of equations
4x + 3y = 18
3x − 4y = 8
Multiply the top equation by 4 and the bottom equation
by 3 to get
2
Add the two to get (m2 + 1)x = 3m2 or x = m3m
2 +1 . Then
substitute that value into either equation to find y.
3m2
+ my = 0, so y = m−3m
2 +1 .
m2 +1
Then
find
the
distance
from
this point to the origin, which
2 2 2
3m
3m
+ 1+m
= √3m2 . You can now
− 1+m
is
2
2
m +1
determine m by making this distance equal to 1 (the
length of the radius).
3m
=1
√
m2 + 1
9m2 = m2 + 1
1
m2 =
8
Which gives two solutions m =
tangents are y −
=
−3
√
8
or − √18 . So the two
and y +
√1 x
8
=
√3 .
8
On Your Own
5. There are many possible solutions for these problems,
and those that follow are just examples that pass through
the origin.
(a)
(b)
(c)
(d)
16x + 12y = 72
9x − 12y = 24
Now, add the two equations to get 25x = 96, which
means that x = 288
= 3.84. To find y, use either line’s
75
equation. 4(3.84) + 3(y) = 18, gives you y = 22
= 0.88.
288 22 25
So the intersection is at the point H = 75 , 25 . The
2 2
288
distance between H and P is
− 3 + 22
−2 =
75
25
2
√
21 2
1
+ 28
= 25
441 + 784 = 35
= 75 .
25
25
25
4. The distance of a tangent line to the center of a circle is
the length of the radius of the circle. Therefore, you are
looking for equations of the two lines through a point that
have a given distance from the origin (the center of the
√1 x
8
√1
8
6. (a)
(b)
(c)
(d)
7. (a)
Mathematics I Solutions Manual
x−y =0
x+y =0
3x − 19y = 0
x + 4y = 0
3x + 2y = 0
x−y =3
x − 2y = 2
3x + 2y = 7
The equation of the line through (1, −2) and
perpendicular to 3x − y = −5 will be of the form
x + 3y = q. By substituting x = 1 and y = −2, you
get 1 + 3(−2) = q or q = −5, so the equation of the
line through (1, −2) and perpendicular to 3x − y =
−5 is x + 3y = −5.
To find the point of intersection, add that equation
to 3(3x − y = −5). You get 10x = −20 or x = −2
• Chapter 8, page 325
“000200010271723958_CH08_p305-331”
1:46
page 326
2 2
point. That is 13 − 73
+ −4 − 89
= 63
This
25
25
5
is the height of the
triangle.
63
The area is 12 (5) 63
= 2.
5
(b) Again you can choose any base. Here is an example
of the calculations for the base AC.
√
AC = (3 − 1)2 + (2 − 1)2 = 5.
for its x-coordinate. Substitute x = −2 into either of
the line equations to find the y-coordinate, which
turns out to be y = −1. Now, you just need to find
the distance between the points (1, −2) and
(−2,
−1). Using the distance formula,
√ you get
2
2
(1 − (−2)) + (−1 − (−1)) = 10, which is the
distance from (1, −2) to 3x − y = −5.
(b) The equation of the line through (−1, 0) and
perpendicular to y = 5x − 1 isx + 5y= −1. The
4 −6
intersection of the two lines is 26
, 26 . The distance
between the point and the line is, again, the distance
between the point
the intersection of the
(−1, 0) and
2
2
4
two lines, or
− (−1) + −6
−0 =
26 √
26
√
2
(30)2 +(−6)2
30 2
+ −6
=
= 3 1326
26
26
26
(c) The equation of the line through points (3, 1) and
(7, 4) is 3x − 4y = 5. The equation of the line
through (2, 2) and perpendicular to 3x − 4y = 5 is
4x + 3y = 14. To find the intersection of the two
lines, add 3 times the first to 4 times the second and
solve
71 22for
x. The intersection of the two lines is
. The distance between the point and the line
,
25
25
71
2 22
2 21 2 −28 2
is
−
2
+
−
2
=
+ 25 =
25
25
√ 25
√
(21)2 +(28)2
= 1225
= 75
25
25
(d) The equation of the line through (1,2) and
perpendicular to 20x − 21y = 58 is 21x + 20y = 61.
To find the intersection of the two lines add 20 times
the first to 21 times the second and solvefor x. The
intersection of the two lines is 2441
, 2 . The
841 841
distance
between the point and the line is
2 2
2
2441
− 1 + 841
−2 =
841
√
−1680 2
(1600)2 +(−1680)2
1600 2
+
=
= 80
841
841
841
29
2013/2/27
The line through points A and C has equation
3−1
= y−1
, which can be simplified to 2x − y = 1.
2−1
x−1
The perpendicular to 2x − y = 1 through (4, 2) has
the form x + 2y = d, and d = (4) + 2(2) = 8. The
complete equation for this perpendicular line is
x + 2y = 8.
Now you need to solve the system
2x − y = 1
x + 2y = 8
Multiply the top equation by 2:
4x − 2y = 2
x + 2y = 8
Add the two equations to get 5x = 10, so x = 2. This
means that the two lines intersect at point C = (2, 3),
so ABC is a right triangle
with its right angle at C.
2
(2 − 3)
(4 − 2)2 =
Thus,
the
height
is
BC
=
√ +√
√
1
5. The area of the triangle is 2 ( 5)( 5) = 52 .
9. The slope of the line with equation y = − 12 x + 3 is − 12 .
The negative reciprocal of the slope is therefore 2. The
point-slope form of an equation that passes through (4, 1)
with slope 2 is y−1
= 2. Rearrange this equation to get
x−4
2x − y = 7. The correct answer choice is D.
10. Here is a graph of quadrilateral ABCD.
8. (a) Choose any base and find its length. Then find the
distance from the third point to this base, and that
will be the height of your triangle. Then compute the
area as A = 12 bh. For example, if the base is AB, that
√
length is (4 − 1)2 + (5 − 1)2 = 9 + 16 = 5. The
equation of the line through the points (1, 1) and
(4, 5) is 5−1
= y−1
, which can be simplified to
x−1
4−1
4x − 3y = 1. A perpendicular to this line would have
the form 3x + 4y = d, and since you want the point
(13, −4) to be on the line, d = 3(13) + 4(−4) = 23.
So now you need to solve the system
4x − 3y = 1
3x + 4y = 23
Multiply the top equation by 4 and the bottom
equation by 3 to get
16x − 12y = 4
9x + 12y = 69
Add the two equations
get 25x = 73 or
to
4 73
25 −1
= 2.92. So
= y and y = 89
= 3.56.
x = 73
3
25
25
Now find the distance between (13, −4) and this
Mathematics I Solutions Manual
To see how many right angles it has, you should find the
slopes of its sides and see which adjacent sides have
slopes that are opposite reciprocals of each other.
3−4
1
√ =√
3− 2
2−3
−4 − 3
7
= √
=√
− 2−3
2+3
−3 + 4
1
=
√ =√
−3 + 2
2−3
4+3
7
=√
=√
2+3
2+3
mAB =
mBC
mCD
mDA
• Chapter 8, page 326
“000200010271723958_CH08_p305-331”
It’s hard to tell if these slopes are opposite reciprocals or
not, so you can check by multiplying two slopes and
seeing if their product is −1.
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1:46
page 327
12. (a) Here is a figure.
8
B
1
7
√
√
2−3
2+3
7
(mAB )(mBC )
= −1
√
√
2+3 2−3 2−9
(mAB )(mBC ) =
6
C
M
4
A
2
This means that quadrilateral ABCD is a rectangle and
has 4 right angles.
11. (a) Here is a figure.
–5
8
5
–2
C
6
(b) The coordinates of the midpoint, M, of AB are
A+B
2+2 3+7
M=
=
,
= (2, 5).
2
2
2
4
2
A
(c) Since yC = yM , an equation of the line through M
and C will be y = k where k is a constant. In this
case, k = yC = yM = 5.
(d) Since an equation of the line through AB is x = 2,
←
→
then AB is perpendicular to CM.
B
–5
5
–2
Maintain Your Skills
(b) Since AB is parallel to the x-axis, then an equation
for the line containing the altitude relative to it is
x = k, where k is a constant. Since the line x = k
goes through point C, then k = xC = 3.
(c) To find the coordinates of H, you have to solve a
system of two equations, where one equation is an
equation of the line through AB and the other is
an equation for the triangle’s height through C. An
equation for the line through AB is simply y = 1
(since yA = yB = 1), so the system is
13. (a) Perpendiculars to 2x + 3y = 4 are of the form
3x − 2y = d. If you want the perpendicular to pass
through (1, 2), then d = 3(1) − 2(2) = −1 and the
equation of the perpendicular is 3x − 2y = −1. To
find the intersection of the two lines solve this
system:
Multiply the top equation by 2 and the bottom
equation by 3 to get
y=1
x = 3.
4x + 6y = 8
9x − 6y = −3
No solving has to be done because of the nice form of
these two equations. H = (3, 1).
(d) Let’s find the coordinates of the midpoint of AB,
which we’ll call M for now.
A+B
=
M=
2
2x + 3y = 4
3x − 2y = −1
1+5 1+1
,
2
2
= (3, 1)
Since the coordinates of H and M are the same, then
M ≡ H. This means that H is the midpoint of AB.
Mathematics I Solutions Manual
5
Add the two equations to get 13x = 5 or x = 13
.
5
Then substitute to find y, getting 2 13 + 3y = 4,
which gives y = 14
. Then the distance from the point
13
(1, 2) to this point is the
from (1, 2) to the
distance
2
5 2
line. That distance is 1 − 13 + 2 − 14
=
13
√
64
144
4 13
+ 169 = 13 .
169
(b) Perpendiculars to 2x + 3y = 4 are still of the
form 3x − 2y = d. If you want the perpendicular
to pass through (xP , yP ), then d = 3(xP ) −
2(yP ) and the equation of the perpendicular is
• Chapter 8, page 327
“000200010271723958_CH08_p305-331”
3x − 2y = 3(xP ) − (yP ). To find the intersection of
the two lines solve this system:
2013/2/27
2x + 3y = 4
Multiply the top equation by 2 and the bottom
equation by 3 to get
ax + by = c
bx − ay = b(xP ) − a(yP )
4x + 6y = 8
9x − 6y = 9(xP ) − 6(yP )
Add the two equations to get 13x = 8 +
9(xP ) − 6(yP ) or x = 8+9(xP13)−6(yP ) . Then substitute
to find y. To do this with these “messier” numbers,
you might want to first solve one of the two equations
for y. For example, if 2x + 3y = 4, then y = −2
x + 43 .
3
4
−2 8 + 9(xP ) − 6(yP )
+
y=
3
13
3
−16 − 18(xP ) + 12(yP )
4
y=
+
39
3
Multiply the top equation by a and the bottom
equation by b to get:
a2 x + aby = ac
b2 x − aby = b2 (xP ) − ab(yP )
Add the two equations to get (a2 + b2 )x = ac +
2
P )−ab(yP )
b2 (xP ) − ab(yP ) or x = ac+b (x
. Then you
a2 +b2
need to substitute for x in one of the equations to
find y. To do this with these messier numbers, you
might want to first solve one of the two equations
for y. For example, if ax + by = c, then y = ba x + bc .
2
P )−ab(yP )
+ bc . This is
So in this case, y = ba ac+b (x
a2 +b2
not going to be simple!
Now, to find the distance from (xP , yP ) to this
point, you need to use the distance formula. Here it is!
Lovely!
So now the distance from the point (xP , yP ) to this
point is the distance from (xP , yP ) to the line. That
distance is
xP −
8+9(xP )−6(yP ) 2
13
2
−16−18(xP )+12(yP )
+ yP −
− 43
39
So, you can use this rather complicated expression to
find the distance from any point (xP , yP ) to the line
2x + 3y = 4.
(c) Perpendiculars to ax + by = c are of the form
bx − ay = d. If you want the perpendicular to pass
through (1, 2), then d = b(1) − a(2) = b − 2a and
the equation of the perpendicular is bx − ay =
b − 2a. To find the intersection of the two lines solve
this system:
xP −
8C
a2 x + aby = ac
Add the two equations to get (a2 + b2 )x = ac +
2 −2ab
b2 − 2ab or x = ac+b
. Then you need to
a2 +b2
substitute for x in one of the equations to
find y. To do this with these “messier” numbers, you
might want to first solve one of the two equations for
y. For example, if ax + by = c, then y = −a
x + bc .
b
2
−2ab
+
Then substitute to find y, getting ba ac+b
a2 +b2
2
which gives y = 2a a+cb−ab
. Then the distance from
2 +b2
the point (1, 2) to this point is the distance from (1, 2)
to
is
the line. That distance
1−
ac+b2 −2ab
a2 +b2
2
+ 2−
2a2 +cb−ab
a2 +b2
2
+ yP −
a
b
ac+b2 (xP )−ab(yP )
a2 +b2
2
−
c
b
MATHEMATICAL REFLECTIONS
The distance of A from B is
2 2
1 1 2
AB =
− (−3) +
−
3
4 2
121
1
1√
1√
+
=
=
121 · 16 + 9 =
1945.
9
16
12
12
Multiply the top equation by a and the bottom
equation by b to get
b2 x − aby = b2 − 2ab
ac+b2 (xP )−ab(yP )
a2 +b2
1. The distance of A from O is
2 1
1
1√
2
AO = (−3) +
= 9+ =
37.
2
4
2
ax + by = c
bx − ay = b − 2a
c
,
b
page 328
(d) Perpendiculars to ax + by = c are of the form
bx − ay = d. If you want the perpendicular to pass
through (xP , yP ), then d = b(xP ) − a(yP ). The
equation of the perpendicular is bx − ay =
b(xP ) − a(yP ).
To find the intersection of the two lines solve this
system:
3x − 2y = 3(xP ) − 2(yP )
1:46
2. The coordinates of the midpoint of segment JK are:
0−4 3−6
3
M=
,
= −2, −
.
2
2
2
3. To decide whether these points are collinear, you can
calculate the slope between two pairs of them and see if it
is the same. We will take the first and the second and call
their slope s1 and then the first and the third and call their
slope s2 .
2
. So, you can
use this rather complicated expression to find the
distance from any point (1, 2) to the line ax + by = c.
Mathematics I Solutions Manual
− 52 − 12
−3
=
=3
−1 − 0
−1
9
2 − (− 52 )
s2 = 1
= 23 = 3
− (−1)
2
2
s1 =
s1 = s2 so the three points are collinear.
• Chapter 8, page 328
“000200010271723958_CH08_p305-331”
4. The slope of line 2x + 3y = 0 is − 23 , so the slope of any
line perpendicular to it must be 32 . Here is a general
equation for a line perpendicular to 2x + 3y = 0:
2013/2/27
1:46
page 329
3.
y
3x − 2y = q, where q is any real number,
4
so in this case, q = 0 because the coordinates of the
origin must satisfy the equation.
5. If the x-coordinates of a pair of points are the same then
the points lie on a vertical line and the line is parallel to
the y-axis. If the y-coordinates of a pair of points are the
same then the points lie on a horizontal line and the line
is parallel to the x-axis.
6. The distance formula states that the distance between two
points (x1 , y1 ) and (x2 , y2 ) is
(x2 − x1 )2 + (y2 − y1 )2
CHAPTER REVIEW
1. Refer to this figure:
D
2
x
O
4
Here is the graph of 2x + y = 4. One way to find an
equation for the image is to find two points that must lie
on it. For example, (0, −4) and (2, 0). So an equation is
2x − y = 4.
Another way of solving this problem is to substitute
y = −y , x = x in the equation of the line:
2x − y = 4
4. The reflection line, r, will be the perpendicular bisector
of segment QQ . It
contain the midpoint
must, therefore,
of QQ , which is −12+ 3 , −32+ 1 = (1, −1). The slope of
r is the negative reciprocal of the slope of QQ . The slope
of QQ is −3−1
= 1, so the slope of r must be −1. We can
−1−3
=
write the equation for r in point-slope form as y−(−1)
x−1
y+1
=
−1.
Solving
for
y,
we
get
y
=
−x.
x−1
5. There are two lines of symmetry.
From the construction, you know that ACB ∼
= DCB
and ABC ∼
= DBC. You also know that BC ∼
= BC. So,
ABC ∼
= DBC by ASA.
2. Refer to this figure:
D
6. A B C is the image of this rotation.
C'
O
B'
B
A'
In Exercise 1, you showed that ABC ∼
= DBC, so by
CPCTC you know that AC ∼
= DC. You also know that
ACB ∼
= DCB from the original construction of the
figure. CP ∼
= CP, so you can conclude that ACP ∼
=
DCP. So, by CPCTC, AP ∼
= DP.
A
C
7. The length of AB is
√
√
(3 − (−1))2 + (4 − 1)2 = 16 + 9 = 25 = 5
Mathematics I Solutions Manual
• Chapter 8, page 329
“000200010271723958_CH08_p305-331”
page 330
3 − (−2)
5
=
7
1
3
−2
2
−1 − 4
5
mHI =
=
−2 − 1
3
mFG =
mHG =
4−3
1− 72
= − 25 , mIF =
−1−(−2)
−2− 12
= − 25
11. The line with equation 4x − 6y = 24 has slope 23 . The
slope of the line with equation 4x + 6y = 24 has slope
−2
. The lines are therefore not parallel.
3
12. Here is an equation of the line through P = (5, 6) and
perpendicular to the line through A = (−3, 4) and
B = (4, −2):
− 32
1
slope first, second =
=−
2
2−0
1
1− 2
1
slope first, third =
=−
1−2
2
1
2
These points are collinear.
(b)
1:46
To prove that FGHI is a parallelogram, you can prove
that its opposite sides are parallel.
The coordinates of the midpoint of AB are
5
3−1 4+1
,
= 1,
2
2
2
8. The midpoint is −2.52+ 6.7 , 3.1 + 2(−3.3) = (2.1, −0.1).
9. To check the collinearity of three points, you can
calculate the slope of two pairs of points.
(a)
2013/2/27
7x − 6y = −1
12 − 5
=7
1−0
1−5
slope first, third =
=4
−1 − 0
slope first, second =
. Any
13. The line with equation 3x + 7y = 42 has slope −3
7
line perpendicular to this line must have a slope that is
the negative reciprocal of −3
, which is 73 .
7
These points are not collinear.
(c)
6 − 20
= −1
7 − (−7)
10 − 20
slope first, third =
= −1
3 − (−7)
slope first, second =
CHAPTER TEST
These points are collinear.
10. Use the lettering in the figure below.
C
6
4
D
–10
H
G
2
I
–5
5
–2
A
B
F
–4
The coordinates of the midpoints are
1
4 − 3 −1 − 3
,
=
, −2
F=
2
2
2
4+3 7−1
7
G=
,
=
,3
2
2
2
3−1 7+1
H=
,
= (1, 4)
2
2
−1 − 3 1 − 3)
I=
,
= (−2, −1)
2
2
10
1. When reflecting a line over the y-axis, replace x with −x
in the original equation. Doing this, we get 2(−x) + y =
−4. Solving for y, we get y = 2x − 4. The correct answer
choice is A.
2. Reflecting first over the y-axis, we get F (1, 0) and
G (−2, 3). Reflecting over the x-axis, we end up
with F (1, 0) and G (−2, −3). The correct answer
choice is B.
3. The translation has subtracted 5 from the x-coordinate
and 7 from the y-coordinate. The correct rule to apply is
therefore answer choice B.
4. Find
averages
−4+2the−2+3
of the
corresponding endpoints to get
1
=
−1,
. The correct answer choice is A.
,
2
2
2
5. To find GH
,
plug
the
endpoints
into the distance formula
2
to
(−1)) + (5 − (−3))2 =
obtain (−3 − √
√
2
2
(−2) + (8) = 4 + 64 = 68 ≈ 8.2. So, the
correct answer choice is A.
6. The correct answer is C because is it the only congruence
statement involving corresponding parts of the triangles.
7. To check if three points are collinear, you could calculate
the slopes between the points and the third point. If the
slopes are the same, then the points are collinear.
Alternatively, notice that x = 3y for the given points. The
only one that doesn’t satisfy that is (0,2). The correct
answer choice is C.
8. (a) Not congruent; Take length or angle measurements
will show that the butterfly on the right is slightly
smaller and slightly distorted.
(b) Congruent; One way to check would be to find a
radius or diameter and measure its length.
(c) Congruent; The angles have the same measure.
9. The coordinates of the end points of the image of AB
after the reflection are
Mathematics I Solutions Manual
A = (1, −2) and B = (6, −8).
• Chapter 8, page 330
“000200010271723958_CH08_p305-331”
2013/2/27
1:46
page 331
10. The slope of s is − 34 , so the slope of any line
perpendicular to it must be 43 . Now you have to check
whether the slope of r is 43 or not. The slope of r is
12
= 43 , so s⊥r.
16
11. The coordinates for B must be C − A =
(3 − 4, 6 − 5) = (−1, 1).
12. The x-coordinate of C must satisfy the equation
−2 = 6+x
, so x = −10. The y-coordinate must satisfy
2
,
so y = 4.
8 = 12+y
2
13. (a) The triangles are congruent.
Since BC AD, then BCA ∼
= DAC, and since
AB DC, then BAC ∼
= DCA, both because of the
PAI Theorem. AC is congruent to itself, so the
triangles are congruent because of ASA.
(b) They are not congruent. The two triangles have three
pairs of congruent angles since angle A is congruent
to itself, but AAA is not a valid congruence test.
Mathematics I Solutions Manual
• Chapter 8, page 331