Geometry
Analytic Geometry
2018-08-13
www.njctl.org
Table of Contents
Click on a topic to go to that section
Origin of Analytic Geometry
The Distance Formula
The Midpoint Formula
Partitions of a Segment
Slopes of Parallel & Perpendicular Lines
Equations of Parallel & Perpendicular Lines
Triangle Coordinate Proofs
Equation of a Circle & Completing the Square
Throughout this unit, the Standards for Mathematical Practice
are used.
Math Practice
MP1: Making sense of problems & persevere in solving them.
MP2: Reason abstractly & quantitatively.
MP3: Construct viable arguments and critique the reasoning of
others.
MP4: Model with mathematics.
MP5: Use appropriate tools strategically.
MP6: Attend to precision.
MP7: Look for & make use of structure.
MP8: Look for & express regularity in repeated reasoning.
Additional questions are included on the slides using the "Math
Practice" Pull-tabs (e.g. a blank one is shown to the right on this
slide) with a reference to the standards used.
If questions already exist on a slide, then the specific MPs that
the questions address are listed in the Pull-tab.
Origin of Analytic
Geometry
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of Contents
https://njctl.org/video/?v=qbBOX8rE8nY
The Origin of Analytic Geometry
Analytic Geometry is a powerful combination of geometry
and algebra.
Many jobs that are looking for employees now, and will
be in the future, rely on the process or results of analytic
geometry.
This includes jobs in medicine, veterinary science,
biology, chemistry, physics, mathematics, engineering,
financial analysis, economics, technology, biotechnology,
etc.
The Origin of Analytic Geometry
Euclidean Geometry
Was developed in Greece about
2500 years ago.
Was lost to Europe for more than a
thousand years.
Was maintained and refined during
that time in the Islamic world.
Its rediscovery was a critical part of
the European Renaissance.
http://www.christies.com/lotfinder/book
-manuscripts/euclid-milliet-dechalesclaude-francois-milliet-5541389details.aspx
The Origin of Analytic Geometry
Algebra
Started by Diophantus in Alexandria
about 1700 years ago.
Ongoing contributions from Babylon,
Syria, Greece and Indians.
Named from the Arabic word al-jabr
which was used by al-Khwarizmi in the
title of his 7th century book.
http://en.wikipedia.org/wiki/Mathe
matics_in_medieval_Islam
The Origin of Analytic Geometry
Analytic Geometry
A powerful combination of
algebra and geometry.
Independently developed, and
published in 1637, by Rene
Descartes and Pierre de Fermat in
France.
The Cartesian Plane is named for
Descartes.
The Origin of Analytic Geometry
Math Practice
How would you
describe to someone
Thethe
question
on this slide
location of these
addresses
& MP3.
five points MP2
so they
could draw them on
another piece of
paper without seeing
your drawing?
Discuss.
The Origin of Analytic Geometry
y
x
Adding this Cartesian
coordinate plane
makes that task
simple since the
location of each point
can be given by just
two numbers: an xand y-coordinate,
written as the ordered
pair (x, y).
The Origin of Analytic Geometry
y
x
With the Cartesian
Plane providing a
numerical description
of locations on the
plane, geometric
figures can be
analyzed using
algebra.
The Distance Formula
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https://njctl.org/video/?v=rC3_pwx6OLU
The Distance Formula
y
Did you get this?
(x2,
y2)
c
(x1,
y1 )
2
2
c =a +b
b
a (x ,
2
y1) x
2
The next step is to write
expressions for the lengths of
sides a and b based on the x
and y subscript coordinates.
Use the distance along each
axis to find those.
The coordinates that we just
added for the third vertex
should help.
The Distance Formula
y
(x2,
y2)
The Distance Formula
d
or
2
d=
(x1,
y1 )
(x2,
y1)
( x2 − x1) + ( y2 −
2
y1)
x
Example
What is the distance between (8, 10) and (3, 3)?
y
(x2, y2) =
2
d=
( x2 − x1) + ( y2 −
2
y1)
d
(x1, y1) =
(x2,
y1)
x
Example
What is the distance between (8, 10) and (3, 3)?
y
(x2, y2) =
2
d=
( x2 − x1) + ( y2 −
2
y1)
2
d
(x1, y1) =
d = (8 − 3) + (10 − 3)
(x2,
y1)
d=
2
(5) +
2
(7)
x
d = 74
2
3 What is the distance between (4, 8) and (7, 3)?
5 What is the distance between (−4, 8) and (7, −3)?
A 11 __
B
_
C 2√11
___
D
11√2
E I need help
√232
https://njctl.org/video/?v=t_ZWOkz0fHE
6 What is the distance between (−2, −5) and (7, −3)?
A
9.3__
B
_
C 2√15
__
D
9√2
E I need help
√85
https://njctl.org/video/?v=BeKsYZHdk7U
The Midpoint Formula
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https://njctl.org/video/?v=kEBcxfnhNfw
The Midpoint Formula
(x2,
y2 )
(x1,
y1)
Another question which is
easily solved using
analytic geometry is to
find the midpoint of a
segment.
Once again, we make use
of the fact that it's easy to
determine distance
parallel to an axis, so let's
include a coordinate
system.
The Midpoint Formula
(x2,
y2)
(x1,
y1)
The x-coordinate for the
midpoint between (x1, y1) and
(x2, y2) is halfway between
x1 and x2.
Similarly, the y-coordinate of
that midpoint will be halfway
between y1 and y2.
If you're provided a graph of a
line, and asked to mark the
midpoint, you can often do that
without much calculating.
The Midpoint Formula
(x2,
y2)
→→
→→
(x1,
y1)
In this example,
x1 = 1 and x2 = 9,
coordinates are 8 units
apart. Half of 8 is 4, so go
horizontally 4 units, turn
and go up until you reach
the segment.
That will give you the
midpoint.
In this case, the midpoint
is (5, 7), which can be
read from the graph.
The Midpoint Formula
(x2,
y2)
→→
→→
(x1,
y1)
Similarly,
y1 = 3 and y2 = 11,
coordinates 8 units apart.
Half of 8 is 4 so go
vertically up 4 units, turn
right and go until you
reach the segment.
This method will give you
the midpoint.
Of course, again, the
midpoint is (5, 7), which
can be read from the
graph.
The Midpoint Formula
Calculate the x-coordinate
midway between the two
endpoints by finding the
average of x1 and x2.
The same is true for the
midpoint's y-coordinate.
Just add the two values and
divide by two.
The x-coordinate of the
midpoint is (1 + 9)/2 = 5
The y-coordinate of the
midpoint is (3 + 11)/2 = 7
The Midpoint Formula
y
The Midpoint Formula
(x2,
y2)
The more general solution is
given below for any points:
(x1, y1) and (x2, y2).
The x-coordinate of the
midpoint is given by
xM = (x1 + x2) / 2
(x1,
y1,)
x
The y-coordinate of the
midpoint is given by
yM = (y1 + y2) / 2
8 What is the midpoint between the indicated points?
y
A (4, 9)
C (5, 6)
Answer
B (−5, −4)
D
D (5, 7)
E I need help
x
https://njctl.org/video/?v=VvOqGyRdKiA
9 What is the midpoint between the indicated points?
y
A (1, 1)
C (0.5, 1)
D (0, 0)
E I need help
https://njctl.org/video/?v=DN2I3lXi0Go
Answer
B (−1, −1)
D
x
10 What is the midpoint between
the indicated points?
A (3, 3)
C (4, 3)
Answer
B (3, 4)
A
D (5, 3)
E I need help
x
https://njctl.org/video/?v=_0w2LMdHP6g
11 What is the midpoint between (4, 8) and (7, 3)?
A (8, 2)
C (5.5, 5.5)
D (6, 5)
E I need help
https://njctl.org/video/?v=Pyf4s9yMsSQ
Answer
B (4, 7)
C
12 What is the midpoint between (−4, 8) and (4, −8)?
B (0, 0)
C (12, 12)
D (4, 4)
E I need help
https://njctl.org/video/?v=CC9XTwfw-UY
Answer
A (8, 2)
B
13 What is the midpoint between (−4, −8) and (−6, −4)?
A (−5, −6)
C (5, 6)
D (5, −6)
E I need help
https://njctl.org/video/?v=66QUOVMya5c
Answer
B (−10, −12)
A
14 What is the midpoint between (k, 6k) and (5k, −4k)?
A
C
(6k, k)
D (6k, 5k)
E I need help
https://njctl.org/video/?v=t8HYwni4Tg8
Answer
(3k, k)
B (3k, 5k)
A
Finding the Coordinates of an Endpoint
__
The midpoint of AB is M(1, 10).
One endpoint is A(4, 6). What
are the coordinates of the other
endpoint B(x, y)?
You can use the midpoint
formula to write equations
using x and y. Then, set up
two equations and solve
each one.
https://njctl.org/video/?v=IhMnQDXSFLo
M(1, 10)
A(4, 6)
Finding the Coordinates of an Endpoint
B(x, y)
M(1, 10) = 4 + x, 6 + y
2
2
(
M(1, 10)
Math Practice
A(4, 6)
)
10 = 6 + y
1= 4+x
2
2part of this example
The
first
20 = 6 + y
2=4+x
addresses
−2 = x MP2.14 = y
Therefore, the coordinates of
The questions at the bottom of
B(−2, 14).
the slide address MP8.
- Answer is found on the next 3
Can you
findusing
a shortcut
slides
(first
the to
graph &
solveusing
this problem?
How w/ the
then
the pattern
would your shortcut make the
points/numbers).
problem easier?
Finding the Coordinates of an Endpoint
B(?, ?)
→
→
→
→
4
−3
M(1, 10)
4
−3
A(4, 6)
Another way of approaching
this problem is to look for the
pattern that occurs between
the endpoint A(4, 6) and
midpoint M(1, 10).
Looking only at our points, we
can see that we traveled left
−3 units and up 4 units to get
from A to M. If we repeat the
pattern starting at M, we will
get to B(−2, 14).
Finding the Coordinates of an Endpoint
__
Consider AB with A(4, 6) and its midpoint M(1, 10). What are
the coordinates of B ?
→
→
A(4, 6)
∆ x = −3
∆y = 4
M(1, 10)
∆y = 4
∆ x = −3
B(??, ?? )
We can do this without a
graph. Identify the change
in x and the change in y
and follow the pattern.
Finding the Coordinates of an Endpoint
__
Consider AB with A(4, 6) and its midpoint M(1, 10). What are
the coordinates of B ?
→
→
A(4, 6)
∆ x = −3
∆y = 4
M(1, 10)
∆y = 4
∆ x = −3
B(−2, 14)
We can do this without a
graph. Identify the change
in x and the change in y
and follow the pattern.
__
15 Consider AB with A(1, −6) and its midpoint M(3, −2).
What are the coordinates of B?
y
D
(4, 2)
(4, 3)
(5, 2)
(5, 3)
E
I need help
B
C
Answer
A
x
C
M
A
https://njctl.org/video/?v=o60q3VzffLM
B
16 One endpoint of a segment is (7, 2) and the midpoint
is (3, 0). What are the coordinates of the other
endpoint?
(−1, −2)
B
(−2, −1)
C
D
(4, 2)
(2, 4)
E
I need help
Answer
A
A
M(x, y) = (
M(3, 0) = (
3=
,0=
(−1, −2)
https://njctl.org/video/?v=KaTJQTpYd5U
)
,
,
)
__
17 Consider AB with A(1, 4) and its midpoint M(5, −2).
What are the coordinates of B?
A
C
D
Answer
B
(11, −8)
(9, 0)
(9, −8)
(3, 1)
C
M(x, y) =(
M(5, −2) = (
E I need help
5=
,
, −2 =
(9, −8)
https://njctl.org/video/?v=I3uT80uqNls
)
,
)
__
18 Consider ST with S(−4, −1) and its midpoint M(−2, 3).
What are the coordinates of T ?
(−6, −5)
B
(−3, −2)
(0, 7)
(1, 9)
C
D
E I need help
https://njctl.org/video/?v=i1FD5p050xU
Answer
A
C
19 One endpoint of a segment is (−2k, 5k) and the
midpoint is (0, 2k). What are the coordinates of the
other endpoint?
E
I need help
B
C
https://njctl.org/video/?v=pHZLJShkxSA
Answer
D
(−k, −3.5k)
(−4k, 8k)
(k, 0.5k)
(2k, −k)
A
D
20 One endpoint of a segment is (k − 2, k + 5) and the
midpoint is (0, 2k). What are the coordinates of the
other endpoint?
(−k, −3.5k)
B
(2 − k, 3k − 5)
C
(2k − 4, 2k + 10)
D
(−2, 2k + 10)
E
I need help
https://njctl.org/video/?v=JH_tlEp0pd8
Answer
A
B
Partitions of a Segment
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https://njctl.org/video/?v=15XYnaWZCVg
Partitioning a Ruler
Partitioning a ruler simply means to divide the ruler into two
(or more) parts strategically to achieve a desired ratio.
A midpoint partition would dissect the ruler into two six-inch
rulers, each of equal length, with a ratio 6" to 6" = 1:1.
6"
six-inch partition
6"
six-inch partition
Partitioning a Ruler
The ruler can be partitioned into three pieces with a threeway ratio 4":4":4" = 1:1:1 like this.
4"
4"
4"
Or partitioned into three unequal pieces with a three-way left
to right ratio 6":4":2" = 3:2:1 as shown here.
6"
4"
2"
Partitioning a Ruler
Math Practice
Exercise A Where would you partition a twelve-inch ruler to
create two pieces in the left to right ratio of 5:1?
This slide & the next 5 slides
address MP1, MP4, MP5 &
MP7.
Exercise B Where would you partition a twelve-inch ruler to
create four pieces in the left to right ratio of 6:3:2:1?
Partitioning a Ruler
Exercise A Where would you partition a twelve-inch ruler to
create two pieces in the left to right ratio of 5:1?
10"
5k
Here
k
To start, we want to partition
5k + k = 6k = 12"
the ruler into 5 + 1 congruent
k = 2"
sub-sections.
5:1 = 5k:k = 10":2"
Exercise B Where would you partition a twelve-inch ruler to
create four pieces in the left to right ratio of 6:3:2:1?
6k
6"
Here
3k
9"
Here
2k 11"
Here
k
To start, we want to partition the 6k + 3k + 2k + k = 12k = 12"
ruler into 6 + 3 + 2 + 1
k = 1"
congruent sub-sections.
6": 3": 2": 1"
Partitioning a Ruler
Exercise C Where would you partition a twelve-inch ruler to
create two pieces in the left to right ratio of 17:7?
Partitioning a Ruler
Exercise C Where would you partition a twelve-inch ruler to
create two pieces in the left to right ratio of 17:7?
8½
17k
Here
7k
To start, we want to partition the ruler into
17 + 7 = 24 congruent sub-sections.
17k + 7k = 24k
The length of the ruler 12" divided by the
24k = 12"
k = 12/24 =
number of partitions 24 gives k =
½
Finally compute the desired ratio 17k:7k
½"
17k:7k = 17/2:7/2
= 8.5":3.5"
"
Answer The ruler should be partitioned at the 8½ mark.
Partitioning a Ruler
Exercise D Where would you partition a 30-centimeter ruler
to create three pieces in the left to right ratio of 5:3:2?
Partitioning a Ruler
Exercise D Where would you partition a 30-centimeter ruler to
create three pieces in the left to right ratio of 5:3:2?
5k
Here
To start, we want to partition the
ruler into 5 + 3 + 2 = 10 congruent
sub-sections.
The length of the ruler 30 cm
divided by the number of partitions
10 gives k = 3 cm
3k
Here
2k
5k + 3k + 2k = 10k
10k = 30
k = 3 cm
5k:3k:2k = 15 cm:9 cm:6 cm
Finally compute the desired ratio
Answer The ruler should be partitioned at 15 cm and 24 cm.
5k:3k:2k
21 Where would you partition a twelve-inch ruler to
create two pieces in the left to right ratio of 3:1?
Answer
A
at the 4" mark
B
at the 8" mark
C
at the 8½" mark
D
at the 9" mark
E
I need help
https://njctl.org/video/?v=FUKquRq4LT4
3k + k = 4k
4k = 12"
k = 3"
3k:k = 9":3"
The ruler should be
partitioned at the 9" mark.
D
22 Where would you partition a twelve-inch ruler to
create two pieces in the left to right ratio of 5:3?
at the 6½" mark
B
at the 7½" mark
C
at the 8" mark
D
at the 9" mark
E
I need help
Answer
A
5k + 3k = 8k
8k = 12"
k = 12/8 = 1.5"
5k:3k = 15/2":9/2"
The ruler should be partitioned at
the 7½" mark.
B
https://njctl.org/video/?v=2c-b0KwQYsY
23 Where would you partition a twelve-inch ruler to create
three pieces in the left to right ratio of 13:9:2?
D
at the 6½" at the 11½"
Themarks
ruler should be partitioned
E
I need help
A
B
Answer
C
13k + 9k + 2k = 24k
at the 4½" at the 6½" marks 24k = 12"
k = 12/24 = ½"
at the 6½" at the 10½" marks
13k:9k:2k = 13/2":9/2":1"
at the 6½" at the 11" marks
at the 6½" and 11" marks.
C
https://njctl.org/video/?v=ZINXrOxEXZ4
24 Where would you partition a twelve-inch ruler to create
three pieces in the left to right ratio of 10:5:3?
E
I need help
A
B
C
Answer
D
10k + 5k + k = 16k
16k = 12"
at the 7½" at the 11¼" marks k = 12/16 = ¾"
10k:5k:k = 30/4":15/4":3/4"
at the 7½" at the 11½" marks
The ruler should be partitioned
at the 6½" at the 11¼"
marks
at 7½"
and 11¼"
at the 6½" at the 11" marks
https://njctl.org/video/?v=MUbiwTXnaSM
A
25 Where would you partition a thirty centimeter ruler to
create three pieces in the left to right ratio of 4:1:1?
4k + k + k = 6k
E
I need help
A
B
C
Answer
D
6k = 30 cm
at the 5 cm at the 15 cm marks
k = 30/6 = 5 cm
at the 5 cm at the 25 4k:k:k
cm marks
= 20 cm: 5 cm: 5 cm
The
should be partitioned
at the 20 cm at the 25
cmruler
marks
at 20 cm and 25 cm.
at the 20 cm at the 26 cm marks
https://njctl.org/video/?v=GzcoSuhi724
C
Partitions of Segments in the Plane
A(1, 0)
B(13, 0)
Let's now take the idea of partitioning and overlay it on the plane.
It is effectively the same problem with a wrinkle.
__
Given AB where A = (1, 0) and B = (13, 0).
__
Find the points on the plane that would partition AB into three
congruent segments.
https://njctl.org/video/?v=5j6ESFqNQKc
Partitions of Segments in the Plane
A(1, 0)
__
Given AB where A = (1, 0) and B = (13, 0).
B(13, 0)
__
Find the points on the plane that would divide AB into three
__
congruent segments.
Solution The length of AB is ∆x = 13 − 1 = 12 units. To create
three congruent segments, we want a 1:1:1 = k:k:k ratio.
Therefore k + k + k = 3k = 12 making k = 4.
Now comes the wrinkle: each partition is 4 units long but there is a
starting position (1,0) that we must take into account. The
26 Where would you partition the segment joining (−3, 0)
and (7, 0) to create two pieces in the ratio of 3:1?
A(−3,0)
at (3.5, 0)
B
at (4,0)
C
at (4.5, 0)
D
at (5,0)
E
I need help
https://njctl.org/video/?v=WyuJ2TTt46I
Answer
A
3k + k = 4k B(7,0)
4k = 10
k = 10/4 = 2.5
3k:k = 7.5: 2.5
The segment gets
partitioned at (4.5, 0)
C
27 Where would you partition the segment joining (−5, 0)
and (10, 0) to create three pieces in the ratio of 3:1:1?
A(−5, 0)
3k + k + k = 5k B(10, 0)
Answer
5k = 15
k = 15/5 = 3
3k:k:k = 9:3:3
A
at (4, 0) and (7, 0)
B
C
at (4, 0) and (8, 0) The segment gets partitioned
at (3, 0) and (6, 0)
at (4, 0) and (7,0).
D
at (3, 0) and (7, 0)
E
I need help
https://njctl.org/video/?v=ufe2hgOgdhA
A
28 Where would you partition the segment joining (−2, 3)
and (6, 3) to create three pieces in the ratio of 2:1:1?
A
C
D
E
Answer
B
2k + k + k = 4k
B(6, 3)
at (0, 3) and (3, 3) A(−2, 3)
4k = 8
at (0, 3) and (4, 3)
k=2
at (2, 3) and (4, 3)
2k:k:k = 4:2:2
at (2, 3) and (5, 3) The segment gets partitioned at
(2, 3) and (4, 3).
I need help
C
https://njctl.org/video/?v=1uInbJEzWcc
29 Where would you partition the
B(0, 8)
segment joining (0, −4) and (0, 8) to
create three pieces in the ratio
3k of
+ 2k + k = 6k
3:2:1?
Answer
A
at (0, 0) and (0, 4)
B
at (0, 0) and (0, 6)
C
at (0, 2) and (0, 4)
D
at (0, 2) and (0, 6)
E
I need help
6k = 12
k=2
3k:2k:k = 6:4:2
The segment is partitioned at
(0, 2) and (0, 6).
D
A(0, −4)
https://njctl.org/video/?v=_Jmskvc5x_E
Partitions of a Sloped Segment
B(6, 8)
Math Practice
Let's now take the idea of
partitioning a sloped
segment and overlay it on
the &
plane.
How does
this
This slide
the next
2 slides
to the MP5
horizontal
addresscompare
MP1, MP4,
&
and vertical segment
MP7. problems?
__
Given AB where
A = (0, 0) and B = (6, 8).
A(0, 0)
https://njctl.org/video/?v=IqGGJ69CtRY
__ the points on the plane
Find
that would partition AB into
three congruent segments.
Partitions of a Sloped Segment
B(6, 8)
__
Given AB where
A = (0, 0) and B = (6, 8).
Find
__ the points on the plane
that would partition AB into
three congruent segments.
This is two problems in one:
A(0, 0)
a) trisect the horizontal (green)
segment to get the x-partition
and then
b) trisect the vertical (purple)
segment to get the y-partition.
Partitions of a Sloped Segment
(4, 5⅓)
2
that would partition AB into
three congruent segments.
∆ y/3 2⅔
(2, 2⅔)
A(0, 0) ∆ x/3
__
B(6, 8)
Given AB where
A = (0, 0) and B = (6, 8).
__
∆ y/3 2⅔ Find the points on the
plane
∆ x/3 = 6/3 = 2
∆ y/3 = 8/3 = 2⅔
∆ x/3
2
∆ y/3 2⅔ The trisection points are
(2, 2⅔) and (4, 5⅓).
∆ x/3
2
30 Consider the segment joining (0, 0) and (8, 6). To trisect
the segment, which of the following statements is
consistent with the figure below?
A ∆x/3 = 7/3
∆y/3 = 2
B ∆x/3 = 8/3
∆y/3 = 2
C ∆x/3 = 2
∆y/3 = 7/3
D ∆x/3 = 2
∆y/3 = 8/3
Answer
E I need help
(0, 0)
https://njctl.org/video/?v=LBFEj3sKsSo
(8, 6)
B
31 Consider the segment joining (0, 0) and (10, 6). To trisect
the segment, which of the following statements is
consistent with the figure below?
(10, 6)
A ∆ x/3 = 3
E I need help
https://njctl.org/video/?v=cs5szuZXRqw
Answer
∆ y/3 = 2
B ∆ x/3 = 3
∆ y/3 = 3
C ∆ x/3 = 8/3
∆ y/3 = 2
D ∆ x/3 = 10/3
∆ y/3 = 2
(0, 0)
D
32 Consider the segment joining (−5, 0) and (5, 12). What
points do you choose to dissect the segment into three
pieces with a left to right ratio of 2:1:1?
(5, 12)
Answer
A (0, 0) and (2.5, 0)
B (−2.5, 3) and (0, 6)
C (0, 6) and (2.5, 9)
D (0, 0) and (2.5, 0)
E I need help
(−5, 0)
https://njctl.org/video/?v=349oakJunWM
C
33 Consider the segment joining (−5, 7) and (10, −2). What
point do you choose to dissect the segment into two
pieces with a left to right ratio of 2:1?
(−5,7)
Answer
B
(10,−2)
A (0, 4)
B (5, 1)
https://njctl.org/video/?v=C15o17z0y5o
C (1, 3.4)
D (6, 0.4)
E I need help
34 Consider the segment joining (−5,7) and (10, −2). What
points do you choose to dissect the segment into three
pieces with a left to right ratio of 2:2:1?
Answer
(−5,7)
D
(10,−2)
A (0, 4) and (5, 1)
C (1, 6) and (7, −⅓)
B (0, 4) and (7, 0)
D (1, 3.4) and (7, −0.2)
E I need help
https://njctl.org/video/?v=uvqCF1rStUE
35 Point Q lies on ST, where point S is located at (−2, −6)
and point T is located at (5, 8). If SQ:QT = 5:2, where is
point Q on ST?
y
B (0, −2)
Answer
A (−1, −4)
T(3, 4)
C (1, 0)
x
D (3, 4)
E I need
help
D
S
From PARCC Sample Test - Non-Calculator
https://njctl.org/video/?v=z9hGzCenVJ4
Slopes of Parallel &
Perpendicular Lines
Return to Table
of Contents
https://njctl.org/video/?v=6GT3vx3HVe8
Slope of a Line
The slope of a line indicates the steepness of the line with
respect to the x- and y-axes.
y
run
rise
run
rise
x
The slope m equals a number — a ratio — of rise to run where
the rise is the change in y and the run is the change in x between
any two points on a given line.
Slope of a Line
The slope m equals a number — a ratio — of rise to run where
the rise is the change in y and the run is the change in x between
any two points on a given line.
y
→
→
run = 8
→
rise = 4
run = 8
rise = 4
→
x
Vertical Lines
A vertical line through (h, k) has the equation x = h.
MP6: Attend to precision
x = −5
x=2
Math Practice
A vertical line through (2, 0)
Sometimes
students
confuse
has the
equation
x = 2x. = _ and
y = _ for
vertical
lines.
A horizontal
vertical lineand
through
(−5,
3) A
way to make sure that they get the
has the equation x = −5.
equation correct is to have them select
two to three points on each line and find
the similarity between the points.
> When it's x = _ on a vertical line, all of
the x-coordinates are the same.
The slope of a vertical line is> undefined
run is zero
When it's ybecause
= _ on athe
horizontal
line, all
and division by zero is undefined.
of the y-coordinates are the same.
Horizontal Lines
A horizontal line through (h, k) has the equation y = k.
MP6: Attend to precision
A horizontal line through (−5, 3)
Sometimes students confuse x = _ and
has the equation y = 3.
y = _ for horizontal and vertical lines. A
A horizontal line through (0, −2)
way to make sure that they get the
has correct
the equation
y = them
−2. select
equation
is to have
Math Practice
y = 3.
two to three points on each line and find
the similarity between the points.
y = −2
> When it's x = _ on a vertical line, all of
the x-coordinates are the same.
> When it's y = _ on a horizontal line, all
The slope of a horizontal line
zero
(m = 0) because
rise is zero
ofisthe
y-coordinates
are thethe
same.
and zero divided by any non-zero number is zero.
Positive vs. Negative Slopes
A line with a positive slope is uphill from left to right.
A line with a negative slope is downhill from left to right.
If we are careful to compute slope making sure that the run is
always positive by looking from left to right, then it is the rise (or fall)
that determines the sign of the slope.
36 Match the line with the correct equation.
C y=4
Answer
y
Easy way to remember horizontal and
A vertical
y = 4xequations:
B y=x+4
Horizontal
C lines
y = 4"cut" the y-axis at the
number indicated. Therefore, it is
D x = y4= b.
x E I need help
Vertical lines "cut" the x-axis at the
number indicated. Therefore, it is
x = a.
https://njctl.org/video/?v=8MUGo5RKSWc
37 What is the slope of the sloped line?
y
Answer
A m = 2/3
B m = 3/2
A
C m=2
x
https://njctl.org/video/?v=w4TKv8SkHNw
D m=3
E I need help
38 The slope of the line below is .
y
A negative
C zero
D undefined
E I need help
https://njctl.org/video/?v=Rsjh-7zo47M
Answer
B positive
B
x
39 The graph of x = 6 is a .
A vertical line
C diagonal line
D wavy line
E
I need help
https://njctl.org/video/?v=Tl70Js4BnLM
Answer
B horizontal line
A
40 The slope of the line below is .
y
B positive
Answer
A negative
A
C zero
D undefined
E I need help
https://njctl.org/video/?v=yK12rGaOMSI
x
41 What is the slope of the red line?
y
A m = −3/4
Answer
B m = −4/3
C m = 3/4
B
D m = 4/3
x
https://njctl.org/video/?v=l_ocPqvNTbQ
E I need help
42 The slope of the line below is .
y
A negative
C zero
D undefined
E I need help
https://njctl.org/video/?v=JyRnmYX4K2g
Answer
B positive
D
x
43 The slope of the line below is .
y
A negative
C zero
Answer
B positive
C
D undefined
E I need help
https://njctl.org/video/?v=bxsNpXiQ9e4
x
Slope of a Line Again
The slope m equals a number — a ratio — of the change in y to
the change in x between any two points on a line and is given by
∆ y : ∆ x.
y
→
→
∆
x2
∆
x1
∆
y2
∆
y1
→
—
→
x
https://njctl.org/video/?v=QNr9t0o_-WA
Slope of a Line Again
y
→
→
→
—
→
∆ y1 =
2
Answer
∆ y2 =
6
∆x2 =
9
∆ x1 =
3
The slope m equals a
number — a ratio —
of the change in y to
the change in x
between any two
points on a line and is
given by ∆ y : ∆ x.
x0369 12 15 18
y2468 1012 14
x
x0369???
y2468???
Can you identify more
points on the line?
Determining Other Points on a Line
y
Answer
(10, ?) is 3
1) the y-coordinate
x is 7
2) the y-coordinate
1)The line passes through a point where x = 10. What is its
y-coordinate?
2)The line passes through a point where x = 20. What is its
y-coordinate?
44 Three of the points lie on the line shown below; one
does not. Identify which point is not on the line.
y
A (10, 8)
D (45, 30)
Answer
B (15, 11)
C (30, 20)
D
E I need help
x
https://njctl.org/video/?v=TK1I_R0B88U
45 Three of the points lie on the line shown below; one
does not. Identify which point is not on the line.
y
A (−6, 15)
C (6, −10)
Answer
B (−3, 10)
C
D (15, −20)
E I need help
x
https://njctl.org/video/?v=i6uQXOgKx1E
46 The table gives linear data that relates Fahrenheit to
Celsius degrees. Five of the data points (°F, °C) are
correct; one is incorrect. Which point is not correct in the
table?
B (120, 50)
C (158, 70)
D (194, 90)
E I need help
https://njctl.org/video/?v=DYcwzVDJ004
Answer
A (68, 20)
℉
32
68
120
158
194
212
℃
0B
20
50
70
90
100
Slopes of Parallel Lines
Definition In a plane, two lines are parallel if and only if they
have thesame slope.
Math Practice
ℓ1
With what we've learned
about parallel lines this
is easy to understand.
This example (and the next 3
ℓ2 The slope
of a line
is & MP7.
questions)
addresses
MP3
related to the angle it
makes with the x-axis.
Now imagine the x-axis
is a transversal.
https://njctl.org/video/?v=E293q2WtFkM
47 Two lines are cut by a transversal below. What is the
name of this pair of angles labeled 1 and 2?
A Alternate interior angles
ℓ1
Answer
B Corresponding angles
C Alternate exterior angles
ℓ2
B
D Same-side interior angles
E I need help
https://njctl.org/video/?v=0peAEPsbvQM
1
2
48 If the sloped lines are parallel, then the labeled angles
must be .
A supplementary
C congruent
D adjacent
E I need help
https://njctl.org/video/?v=gUddxZVBTv4
Answer
B complementary
ℓ1
ℓ2
C
1
2
49 If ∠1 ≅ ∠2, then the slopes of ℓ1 and ℓ2 must be
.
B additive inverses
ℓ1
Answer
A reciprocals
C
ℓ2
C equal
D nothing special
E I need help
https://njctl.org/video/?v=2RsAjD-poOc
1
2
50 Given ℓ1 || ℓ2 . The line ℓ1 passes through the points
(−4, −1) and (0, 5). The line ℓ2 passes through the
point (2, 0). Three of the
ℓ1
following points lie on ℓ2;
one does not. Which
point is not on ℓ2?
ℓ2
D (−2, −3)
E I need help
https://njctl.org/video/?v=UWYvtMJlIGI
Answer
A (4, 3)
B (0, −3)
C (6, 6)
1
D
2
51 Given ℓ1 || ℓ2. The vertical distance between the
lines remains constant. What is the y-coordinate on
ℓ2 where x = 5?
C −4
D −3
Answer
A −6
B −5
ℓ1
E I need help
ℓ2
https://njctl.org/video/?v=jxfgaYXeD2c
C
52 Given ℓ1 || ℓ2 . The points (2, 2) and (5, 5) lie on ℓ1. The
point (1, 5) lies on ℓ2. Which of the following points also
lies on the line ℓ2?
B (4, 4)
C (5, 6)
D (−1, 3)
E I need help
https://njctl.org/video/?v=xUS7whPAHz8
Answer
A (2, 2)
D
53 Given ℓ1 || ℓ2 . The points (−3, 4) and (0, 10) lie on ℓ1.
The point (−1, −4) lies on ℓ2. Which of the following points
also lies on the line ℓ2?
B (2, −5)
C (3, 1)
D (−1, 3)
E I need help
https://njctl.org/video/?v=q-6kiGxcyUo
Answer
A (0, −2)
A
54 Given ℓ1 || ℓ2 . The points (3, 5) and (5, −1) lie on ℓ1.
The point (−1, −1) lies on ℓ2. Which of the following
points also lies on the line ℓ2?
B (−2, 2)
C (4, 8)
D (−4, −4)
E I need help
https://njctl.org/video/?v=Yi8GkOcg11k
Answer
A (0, 1)
B
Slopes of Perpendicular Lines
Theorem In a plane, two lines are perpendicular if and only if the
product of their slopes is −1.
Let m1 be the slope of ℓ1 and m2 be the slope of ℓ2.
ℓ
1
ℓ1 ⊥ ℓ2 if and only if the
following statements are true.
m1m2 = −1
m1 = −1/m2
m2 = −1/m1
ℓ
https://njctl.org/video/?v=wfVlIMNViBk
2
All are mathematically
equivalent provided the slopes
are non-zero numbers.
Slopes of Perpendicular Lines
Theorem In a plane, two lines are perpendicular if and only if the
product of their slopes is −1.
Another way of stating this idea is to say that the slopes of
perpendicular lines are negative reciprocals.
ℓ
1
This will be useful in cases when
you need to prove lines
perpendicular, including proving
that certain triangles are right
triangles.
First, let's prove the theorem.
ℓ
2
Slopes of Perpendicular Lines
ℓ
ℓ
1
3
Given ℓ1
|| ℓ3 and ℓ2 || ℓ4.
ℓ
To prove the theorem, we can
focus our attention on lines
that pass through the origin.
4
Because parallel lines have
equal slopes, proving ℓ1 ⊥ ℓ2
is enough to prove ℓ3 ⊥ ℓ4.
ℓ
2
So let's prove ℓ1 ⊥ ℓ2 by
demonstrating the product of
their slopes equals −1.
Slopes of Perpendicular Lines
Now, let's zoom in and focus
on the lines between x = 0
and x = 1.
(1,
m1 )
∆y1 =
(0, 0)
∆x = 1 m1
∆y2 =
(1,
m2 )
m2
At x = 1, the y-coordinates of
ℓ1 and ℓ2 are given by m1
and m2 as shown.
This makes sense as the
slope is the change in y over
the change in x. As x goes
from 0 to 1, the change in x
= 1. Therefore, the y-values
for ℓ1 and ℓ2 are m1 and m2.
Slopes of Perpendicular Lines
(1,
m1)
a
c = m1 − m2
(0, 0)
b
(1,
m2)
Let's now use the Pythagorean
Theorem
2
2
c = a +b
2
to insist on a right angle at the
origin.
Side c is hypotenuse and is the
distance between the points
along the vertical line x = 1. It
has length m1 − m2.
Next use the distance formula to
find the length of each leg.
Slopes of Perpendicular Lines
(1,
m1)
The lengths of legs a and b can
be found using the distance
formula.
2
2
2
a
c = m1 − m2
(0, 0)
2
d = (x2 − x1) + ( y2 − y1)
2
a = (1 − 0) + (m1 − 0)
= 1 + m1
2
2
2
2
b = (1 − 0) + (m2 − 0)
b
(1,
m2)
= 1 + m2
2
2
2
And, we get c by squaring our
result for c from the prior slide.
2
2
c = (m1− m2)
2
2
Slopes of Perpendicular Lines
From the previous slide
2
(1, m1) 2
a = 1 + m1
2
a
c
2
2
2
b = 1 + m2 c = m1 − 2m1m2 +
22
2
a2 + b
c
m
=
2
2
2
1 + m1 + 1 + m2 = m1 − 2m1m2 +
2
(0, 0)
b
m2 2 + m 2 + m 2 = m 2 − 2m m + m 2
1
2
1
1 2
2
2 = − 2m1m2
(1, m2)
m1m2= −1 Q.E.D.
m1m2 = −1 ⇒ m1 = −1/m2 ⇒ m2 = −1/m1
Theorem In a plane, two lines are perpendicular if and only if the
product of their slopes is −1.
55 If ℓ1 has a slope of 4, what must be the slope of any
line perpendicular to ℓ1?
C −4
D 4
E I need help
https://njctl.org/video/?v=ahtaDkba4tw
Answer
A −1/4
B 1/4
A
56 Given ℓ1 ⊥ ℓ2 and ℓ1 and ℓ2 intersect at the origin.
What is the slope of ℓ2?
ℓ
C −8/3
D 8/3
Answer
A −3/8
B −1/3
1
A
E I need help
ℓ
2
https://njctl.org/video/?v=aj0Aw4dxs7Y
57 Given ℓ1 ⊥ ℓ2. The points (2, 2) and (5, 5) lie on ℓ1. The
point (1, 5) lies on ℓ2. Which of the following points also
lies on the line ℓ2?
B (2, 4)
C (3, 2)
D (−1, 6)
E I need help
https://njctl.org/video/?v=fdGixCIp87I
Answer
A (0, 4)
B
58 If ℓ1 passes through the points (0, 0) and (4, 2) what
must be the slope of any line perpendicular to ℓ1?
A −1/2
C −2
D 2
E I need help
https://njctl.org/video/?v=55T9NrBA_H8
Answer
B 1/2
C
59 Given ℓ1 ⊥ ℓ2. The points (1, 2) and (5, −1) lie on ℓ1. The
point (1, 5) lies on ℓ2. Which of the following points also
lies on the line ℓ2?
B (5, 8)
C (3, 9)
D (4, 8)
E I need help
https://njctl.org/video/?v=GW-_DYMprSI
Answer
A (4, 9)
A
60 If ℓ1 passes through the points (−5, 9) and (5, 8) what
must be the slope of any line perpendicular to ℓ1?
B 1/10
C −10
D 10
E I need help
https://njctl.org/video/?v=8n34WFuYCgc
Answer
A −1/10
D
61 Given ℓ1 ⊥ ℓ2. The points (−3, 4) and (0, 10) lie on ℓ1. The
point (−1, −4) lies on ℓ2. Which of the following points also
lies on the line ℓ2?
B (2, −5)
C (3, 1)
D (−3, −3)
E I need help
https://njctl.org/video/?v=V3X64VXo9Nc
Answer
A (0, −2)
D
Equations of Parallel &
Perpendicular Lines
Return to Table
of Contents
https://njctl.org/video/?v=ECwHyLdvNG8
Sloped Lines (Review)
Lines in the plane can be
1) vertical (no slope),
2) horizontal (slope is zero), or
3) sloped (slope is a non-zero real number).
The slope m of a line connecting
the points (x1, y1) and (x2, y2) is
(x2,
y2)
y2 −
(x1,
y1)
y1
x2 −
(0, b)
x1
The sloped line
y = mx + b
The y-intercept of a line is the
point where the line crosses the yaxis where x = 0. It is standard to
use the letter b to represent the yintercept.
Equation of a Sloped Line
To determine an equation of a sloped line, find the coordinates of
two points. (Ideally one is the y-intercept.)
Q1What is the value of the yintercept, b?
Q2What are the coordinates of a
third point on the line?
Q3What is the slope m?
Q4What is the equation of the
line in y = mx + b form?
Equation of a Sloped Line
To determine an equation of a sloped line, find the coordinates of two
points. (Ideally one is the y-intercept.)
Ans1What is the value of the yintercept, b?
b = 1; (0, 1)
Ans2What are the coordinates
of a second point?
(3, 3) or (−3, −1)
Ans3What is the slope m?
m = 2/3
Ans4What is the equation of the
line in y = mx + b form?
y=⅔x+1
Forms of Lines from Algebra 1
Teacher Note
The previous problem asked for the equation of a line in slopeintercept form. Slope-Intercept is one of the three standard ways to
Function
Form
with
= f is
(x)
write the equation of a line. Getting
comfortable
with
all ythree
important and is the goal of the next set of exercises.
Slope-Intercept Form
Slope-Intercept Form
y = mx + b
Standard Form
f (x) = mx + b
Standard Form
Ax + B f (x) = C
Ax + By = C
Point-Slope Form
Point-Slope Form
f (x) = m(x − x1) + y1
y − y1 = m(x − x1)
Slope-Intercept Form
Slope-Intercept Form
y = mx + b or f (x) = mx + b
m is the slope and b is the y-intercept
Exercise A
Write the equation of the line in slope-intercept form that includes
the points (0, 5) and (2, 9).
Slope-Intercept Form
Slope-Intercept Form
y = mx + b or f (x) = mx + b
m is the slope and b is the y-intercept
Exercise A
Write the equation of the line in slope-intercept form that includes
the points (0, 5) and (2, 9).
Solution (slope) and b = 5 (intercept).
y = 2x + 5 or f (x) = 2x + 5
Slope-Intercept Form
Slope-Intercept Form
y = mx + b or f (x) = mx + b
m is the slope and b is the y-intercept
Exercise B
Write the equation of the line AB in slope-intercept form given A(0,
k) and B(3k, −5k).
Slope-Intercept Form
Slope-Intercept Form
y = mx + b or f (x) = mx + b
m is the slope and b is the y-intercept
Exercise B
Write the equation of the line AB in slope-intercept form givenA(0, k)
and B(3k, −5k).
Solution (slope) and b = k (intercept) .
y = −2x + k
or
f (x) = −2x + k
Standard Form of the Line
Standard Form
Ax + By = C
A, B, and C are integers and A > 0
Exercise C
What are the x- and y-intercepts of the line 5x + 4y = 20?
Standard Form of the Line
Standard Form
Ax + By = C
A, B, and C are integers and A > 0
Exercise C
What are the x- and y-intercepts of the line 5x + 4y = 20?
SolutionThe x-intercept is the value of x
when y = 0.
5x + 4(0) = 20
5x + 4y = 20
5x = 20
x=4
The y-intercept is the value of y
when x = 0.
5(0) + 4y = 20
Standard Form of the Line
Exercise D
Write the equation of the line in Standard Form Ax + By = C.
Standard Form of the Line
Exercise D
Write the equation of the line in Standard Form Ax + By = C.
Solution
The slope is m = 3/2.
The y-intercept is (0, −3).
In slope-intercept form,
Multiply by 2
and use algebra
to rewrite in
standard form
Ax + By = C.
Point-Slope Form of the Line
Point-Slope Form
y − y1 = m(x − x1) or f (x) − y1 = m(x − x1)
m is the slope and (x1, y1) is a point on the line
Exercise E
Write the equation of the line in point-slope form that passes through
the point (6, −7) with slope 5.
Exercise F
Write the equation of the line AB in point-slope form that passes
though the points A(k, 2k) and B(−3k, 10k).
Point-Slope Form of the Line
Point-Slope Form
y − y1 = m(x − x1) or f (x) − y1 = m(x − x1)
m is the slope and (x1, y1) is a point on the line
Answer
Exercise E
Write the equation of the line in point-slope form that passes through
the point (6, −7) with slope 5.
Solutionm = 5 and (x1, y1) = (6, −7).
y + 7 = 5(x − 6) or f (x) + 7 = 5(x − 6)
Exercise F
y + 7form
= 5(x
6)
Write the equation of the line AB in point-slope
that −
passes
though the points A(k, 2k) and B(−3k, 10k).
Solutionm = (10k − 2k)/(−3k −k) = −2 and (x1, y1) = (k, 2k).
y − 2k = −2(x − k) or f (x) − 2k = −2(x − k)
62 Which standard form equation is equivalent to the line
in slope-intercept form?
B 9x + 15y = 35
C 15x − 9y = 35
D 5x − 3y = 15
E I need help
https://njctl.org/video/?v=B2uWh7PHphQ
Answer
A 3x + 5y = 15
B
Multiply each term by LCD: 15
63 Which line passes through (3, 0) and (0, −5)?
A 3x + 5y = 15
C 15x − 9y = 35
D 5x − 3y = 15
E I need help
https://njctl.org/video/?v=hHrqhcDR00k
Answer
B 9x + 15y = 45
D
64 Which equation is equivalent to 5x − 4y = 24 in slopeintercept form?
A
Answer
B
C
D
E
I need help
https://njctl.org/video/?v=YpVAtc59LNU
D
65 What is the sum of the x- and y-intercepts of the line
6x − 3y = 18?
−18
B
−9
C
−3
D
3
E
I need help
https://njctl.org/video/?v=k3nXc4_2ti8
Answer
A
C
66 Which of the lines in standard form is equivalent to
the line in point-slope form?
B x + 2y = 9
C x − 2y = −12
D 2x − y = 6
E
I need help
https://njctl.org/video/?v=DC6S8VoqUtM
Answer
A 2x + y = 12
C
67 Which of the lines in standard form is equivalent to
the line in point-slope form?
B x − 2y = −8k
C x − 2y = 8k
D 2x − y = 8k
E
I need help
https://njctl.org/video/?v=dZI1ADdkuA8
Answer
A x + 2y = −8k
B
Linear Functions
Exercise A Sketch the graph of the line 2x − 6y = 10.
Remember - Two points determine a line.
https://njctl.org/video/?v=0TWwfhvZov4
Linear Functions
Exercise A Sketch the graph of the line 2x − 6y = 10.
Remember- Two points determines
a line.
It is in
Standard Form. Graph
Answer
findxthe
intercepts
of
usingTothe
- and
y-intercepts.
2x − 6y = 10.
x-int: (5, 0)
Set x = 0 and solve for y.
−6y = 10
y-int:
y = −5/3 ⇒ (0, −5/3)
Set y = 0 and solve for x.
2x = 10,
x = 5 ⇒ (5, 0)
Linear Functions
Exercise B Sketch the graph of y + 3 = −5(x + 1) using the pointslope form.
Linear Functions
Exercise B Sketch the graph of y + 3 = −5(x + 1) using the pointslope form.
To sketch y + 3 = −5(x + 1) find
two points.
Answer
It is in Point-Slope Form.
Graph
the slope and
(−1, −3)itisusing
the obvious point.
the given point.
Use slope to find a second point.
point: (−1, −3)
Go up five units and left one unit to
slope: −5
(−2, 2) and you have a second
point.
68 Match the graph with the correct function.
A f(x) = −6x − 6
Answer
https://njctl.org/video/?v=R5Uk6SMeA_w
B g(x) = −x − 6
D L(x) = x − 6
C h(x) = 6x − 6
Ask
D students
L(x) = xwhich
− 6 form it is in:
Slope-intercept
E
I need help
69 Match the graph with the equation 3x + y = −6.
A
Answer
B
C
D
https://njctl.org/video/?v=tSnTB_8uLnE
A
Line A
B
Line B
C A Line C
D
Line D
E
I need help
70 Match the graph with the function L(x) = 3x − 2.
B
https://njctl.org/video/?v=rEN6FSLCcNg
C
Answer
A
D
A
Line A
B
Line B
C
Line C
C
D
Line D
E
I need help
71 Match the graph with the correct function.
A f (x) − 3 = −3x
Answer
B g(x) − 3 = −2x
C h(x) − 3B= 2x
Dg(x)
L(x)− −3 3= =−2(x
3x − 0)
E
https://njctl.org/video/?v=szcM7z7nWbg
I need help
Two Lines in the Plane
In a plane, two lines intersect unless they are parallel. Two distinct
lines that intersect have different slopes. Parallel lines have the
same slope. Perpendicular lines have slopes whose product is −1.
Given Two distinct lines
L1(x) = m1x + b1 and L2(x) = m2x + b2
Intersecting lines
L1 and L2 intersect if m1 ≠ m2.
Parallel lines
L1 and L2 are parallel if m1 = m2
Perpendicular lines
L1 and L2 are perpendicular if m1m2 = −1
https://njctl.org/video/?v=GwsE85HxZ10
Intersecting Lines
Definition of Intersecting lines
GivenL1(x) = m1x + b1 and L2(x) = m2x + b2
L1 and L2 intersect if m1 ≠ m2
Exercise A
If f (x) = 2x − k intersects g(x) = k − x at x = 4, what is k?
Intersecting Lines
Definition of Intersecting lines
g(x) = 6 − x
Teacher Note
GivenL1(x) = m1x + b1 and L2(x) = m2x + b2
L1 and L2 intersect if m1 ≠ m2
Exercise A
If f (x) = 2x − k intersects g(x) = k − x at x = 4, what is k?
f
Solutionf (4) = 2(4) − k and g(4) = k − 4.
f (4) = 8 − k and g(4) = k − 4.
8 − k = k − 4.
12 = 2k
k=6
(x) = 2x − 6
Parallel Lines
Definition of Parallel Lines
GivenL1(x) = m1x + b1 and L2(x) = m2x + b2
L1 and L2 are parallel if m1 = m2
Exercise B
Are the lines L1(x) + 3 = −5(x + 1) and 5x + y = 15 parallel?
Parallel Lines
Definition of Parallel Lines
5x + y = 15
Teacher Note
GivenL1(x) = m1x + b1 and L2(x) = m2x + b2
L1 and L2 are parallel if m1 = m2
Exercise B
Are the lines L1(x) + 3 = −5(x + 1) and 5x + y = 15 parallel?
SolutionL1(x) + 3 = −5(x + 1) is in point-slope form, m1 = −5.
5x + y = 15 can be rewritten y = L(x)
−5x ++15.
3 = −5(x + 1)
This is slope-intercept form with m2 = −5.
The lines are parallel. m = −5.
Perpendicular Lines
Definition of Perpendicular Lines
GivenL1(x) = m1x + b1 and L2(x) = m2x + b2
L1 and L2 are perpendicular if m1m2 = −1
Exercise C
If kx + y = 3 and y =
−0.5x − 2 are perpendicular, what is k ?
Perpendicular Lines
Definition of Perpendicular Lines
GivenL1(x) = m1x + b1 and L2(x) = m2x + b2
L1 and L2 are perpendicular if m1m2 = −1
If kx + y = 3 and y =
Teacher Note
Exercise C
−0.5x − 2 are perpendicular, what is k ?
SolutionFind the slopes of kx + y = 3 and y =
−0.5x − 2
72 If kx − 2y = 4 and 2x + 3y = k intersect to (2, −1) as
shown, what is the value of k?
A
E
I need help
B
C
kx − 2y = 4
Answer
D
1
2
3
4
A
2x + 3y = k
https://njctl.org/video/?v=P9r9JiVhGjw
73 What is the equation of the line passing through (6, −2)
and parallel to y = 2x − 3?
y = 2x + 2
B
y = −2x + 10
C
y = 1/2 x − 5
D
y = 2x − 14
E
I need help
https://njctl.org/video/?v=xW3YGTw3Qvk
Answer
A
D
74 Which is the equation of a line parallel to y = −x − 22?
x − y = 22
B
y − x = 22
C
y + x = −17
D
2y + x = −22
E
I need help
https://njctl.org/video/?v=RnrUiNarfVw
Answer
A
C
75 Two lines are given by the equations
−3y = 12x − 14 and y = kx + 14
For what value of k will the lines be parallel?
B −14
C 3
D −4
E I need help
https://njctl.org/video/?v=W2fNyKwibJk
Answer
A 12
D
76 Which line is parallel 3y + 4x = 21?
12y + 16x = 12
B
3y − 4x = 22
C
3y = 4x + 21
D
4y + 3x = 21
E
I need help
https://njctl.org/video/?v=WiRbFhxllZQ
Answer
A
A
77 What is the equation of a line that passes through
(9, 3) and is perpendicular to the line given by the
equation 4x − 5y = 20?
y − 3 = −5/4(x − 9)
B
y − 3 = 4/5(x − 9)
C
y − 3 = 4(x − 9)
D
y − 3 = −5(x − 9)
E
I need help
Answer
A
https://njctl.org/video/?v=e8BmQovgeJw
A
78 What is an equation of the line that passes through
the point (5, −2) and is parallel to the line given by
9x − 3y = 12?
B
y=x
C
y = −x + 17
D
y = −3x + 15
E
I need help
https://njctl.org/video/?v=tBeCoZHtvNY
answer
y = 3x − 17
Answer
A
A
A
79 What is an equation of the line that contains the
point (−4, 1) and is perpendicular to the line given by
y = −2x − 3?
y = 2x + 1
B
y= x+3
C
y = −2x − 1
D
y=− x+3
E
I need help
https://njctl.org/video/?v=t7akXsCck5Q
Answer
A
B
80 Two lines are represented by the given equations.
What would be the best statement to describe
these two lines?
5(x + 1) = −2y + 20
2x + 5y =15
Answer
A
The lines are parallel.
B
The lines are the same line.
C
The lines are perpendicular.
D
The lines intersect at an angle other than 90°.
E
I need help
https://njctl.org/video/?v=ydcDssvJAi0
D
Application Problems
We can also relate equations of parallel and perpendicular lines to
other concepts that we've studied. For example, the
radius/diameter of a circle is perpendicular to a tangent line, and the
point of tangency.
In the graph shown to the right,
the slope of the radius of a circle
is given, as well as the equation
of the tangent line.
The radius is perpendicular to
the tangent line.
https://njctl.org/video/?v=RANk8hCW3fM
m= 4
3
Application Problems
Example
Write the equation of a line, in slope-intercept form, which has a
point of tangency at (3, 6) with a circle whose center is at the origin.
We start off by calculating the slope of the radius.
Since the radius and tangent are perpendicular, we know that the
slope of the tangent line will be −1/2.
Application Problems
Example
Write the equation of a line, in slope-intercept form, which has a
point of tangency at (3, 6) with a circle whose center is at the origin.
The tangent line passes through the point of tangency, (3, 6), so
use the point and its slope to determine the equation. You can
approach it in two ways:
Point-slopeORSlope-intercept
(y − y1) = m(x − x1)y = mx + b
(y − 6) = −(1/2)(x − 3)6 = −0.5(3) + b
y = −0.5x + 1.5 + 66 = −1.5 + b
b = 7.5
y = 0.5x + 7.5
Application Problems
Example
Write the equation of a line, in slope-intercept form, which has a
point of tangency at (2, −2) with a circle whose center is (−1, 2).
Answer
OR
81 What is the slope of a line tangent at (7, 2) to a circle
whose center is at (2, 3)?
B
5
C
−5
D
E
I need help
https://njctl.org/video/?v=lVy26Ooi8Tw
Answer
A
B
82 What is the slope of a line tangent at (−4, −5) to a circle
whose center is at the origin?
Answer
A
B
C
D
E
I need help
https://njctl.org/video/?v=Dge8U-4OEIc
D
83 What is the equation of a line, in slope-intercept form, that
is tangent at the point (−2, 3) to a circle whose center is at
the origin?
Answer
A
B
C
D
E
I need help
https://njctl.org/video/?v=OCUZo0uMb84
C
84 What is the equation of a line, in slope-intercept form, that
is tangent at the point (5, −1) to a circle whose center is
(2, 4)?
Answer
A
B
C
D
E
I need help
https://njctl.org/video/?v=gHIsKM9VcY4
A
Triangle Coordinate
Proofs
Return to Table
of Contents
https://njctl.org/video/?v=Tp9yBljaueo
Triangle Coordinate Proof
Math Practice
Coordinate Proofs place figures on the Cartesian Plane to
This lesson addresses MP1, MP2, MP3 &
make use of the coordinates ofMP6.
key features of the figure,
combined with formulae (e.g. distance formula, midpoint
formula and slope formula) to prove
something.
Additional
Q's to address MP standards
What information are you given? (MP1)
What do
need to (MP1)
The use of the coordinates is simply
anyou
alternative
approach in
Can you represent any side lengths with
conducting the proof.
symbols and numbers? (MP2)
Whatthen
Mathhave
language
willsome
help you prove
We'll provide a few examples and
you do
your answer? (MP3)
proofs.
What is the reason for this step? (MP6)
Triangle Coordinate Proof
GivenThe points A(0, 4), B(3, 0), C(−3, 0), and
Q(0, 0) are
__the vertices of ΔABC and Δ AQB
Prove QA bisects ∠CAB
Sketch the triangles on some graph paper and then
discuss a strategy to accomplish the proof.
Example
GivenThe points A(0, 4), B(3, 0), C(−3, 0), and
Q(0, 0) are
__the vertices of ΔABC and Δ AQB
Prove QA bisects ∠CAB
A(0, 4)
C(−3, 0)
Q(0, 0)
B(3, 0)
Does this sketch look like yours?
If not, take a moment to see if
this is correct. Looking at this, our
strategy becomes clear.
If we can prove that ΔAQC ≅
ΔAQB, then ∠CAQ ≅ ∠BAQ
which would allow us to conclude
AQ bisects__∠CAB.
Let's get to work.
Example
GivenThe points A(0, 4), B(3, 0), C(−3, 0), and
Q(0, 0) are
__the vertices of ΔABC and Δ AQB
Prove QA bisects ∠CAB
A(0, 4)
From the graph,
CQ = BQ = 3 and AQ = 4.
C(−3, 0)
Q(0, 0)
B(3, 0)
Example
GivenThe points A(0, 4), B(3, 0), C(−3, 0), and
Q(0, 0) are
__the vertices of ΔABC and Δ AQB
Prove QA bisects ∠CAB
A
You should be able to see that
these are two right triangles
with identical length sides, so
they must be congruent.
4
C
3
3
Q
B
Let's work through the proof as
good practice.
Example
Given The points A(0, 4), B(3, 0), C(−3,
0), and Q(0, 0) are the vertices of Δ ABC
and Δ AQB
__
Prove QA bisects ∠CAB
C
A
4
3
StatementsReasons
1. CQ
and BQ = 31. Given in graph
__ = 3__
2. QC
__ ≅ QB2.
__ ≅ segments have equal measure
__ ≅ QA3.
___ Reflexive Property of Congruence
3. AQ
4. AQ ⊥ CQB4. x- and y-axes are perpendicular
5. ∠CQA ≅ ∠BQA5. ⊥ lines form ≅ adjacent angles
6. ΔAQC ≅ ΔAQB6. SAS triangle congruence
__
7. ∠CAQ
≅ ∠BAQ7. CPCTC
8. QA bisects ∠CAB8. Definition of angle bisector
3
Q
B
Triangle Coordinate Proof
Given A(4, −1), B(5, 6), and C(1, 3)
Prove Δ ABC is an isosceles right triangle
Make a sketch first and think of a strategy for this proof.
Triangle Coordinate Proof
Given A(4, −1), B(5, 6), and C(1, 3)
Prove Δ ABC is an isosceles right triangle
B
Plan
To start,
__ let's__show that the slopes
of BC and AC are negative (or
opposite) reciprocals.
Then use the Pythagorean
Theorem to determine if the
__
__
lengths of BC and AC are equal.
C
A
Write a paragraph proof.
Triangle Coordinate Proof
Given A(4, −1), B(5, 6), and C(1, 3)
Prove Δ ABC is an isosceles right triangle
B
C
Paragraph Proof
Given A(4, −1), B(5, 6), and C(1, 3), the
__
slope of BC is mBC = (6 − 3)/(5 − 1) = 3/4
A
__
and the slope of AC is mAC = (3 + 1)/(1__
− 4) __
= −4/3.
The slopes are negative reciprocals so AC ⊥ BC which makes Δ
ABC is a right triangle. By the Pythagorean
Theorem,
__
2
2
AC = √(4 − 1) + (−1 − 3) = __
√25 = 5
2
2
BC = √(5 − 1)
__ + (6
__− 3) = √25 = 5
which makes AC ≅ BC. Thus Δ ABC is an isosceles right
triangle. Q.E.D.
Triangle Coordinate Proof
Given A(1, 1), B(4, 4), and C(6, 2)
Prove Δ ABC is a right triangle
Make a sketch first and think of a strategy for this proof.
Triangle Coordinate Proof
B
Given A(1, 1), B(4, 4),
and C(6, 2)
Prove Δ ABC is a right triangle
A
Come up with a plan that would allow you
to prove that Δ ABC is a right triangle.
Write a paragraph proof.
C
Triangle Coordinate Proof
B
Given A(1, 1), B(4, 4),
and C(6, 2)
Prove Δ ABC is a right triangle
A
Paragraph Proof
Given A(1, 1), B(4, 4), and C(6, 2), the slope of AB is
mAB = (4 − 1)/(4 − 1) = 3/3 = 1 and the slope of BC is
mBC = (2 − 4)/(6 − 4) = −2/2 = −1.
__ __
The product of the slopes is (1)(−1) = −1 so AB ⊥ BC which
makes Δ ABC is a right triangle. Q.E.D.
C
Equation of a Circle &
Completing the Square
Return to Table
of Contents
https://njctl.org/video/?v=DLtUjkcqNP8
The Circle as a Locus of Points
Definition
r
center
A circle is a set of points in a
plane that are equidistant
from a fixed point called the
center of the circle. The
distance from the center to
any point on the circle is the
radius r.
Note
The center is not a part of
the circle but a reference
point for constructing a
circle.
The Circle Centered at the Origin
r
(0, 0)
Math Practice
(4, 8) This derivation
(this
anda the
The circleexample
to the left
is on
next 3 coordinate
slides) addresses
MP1
& MP2.
plane. Its
center
is (0, 0) and the circle
Additional
Q's through
to address
passes
theMPs:
point
What information
are you given? (MP1)
(4, 8) as shown.
What do you need to find? (MP1)
Let's
determine
radius
of the
Can you
represent
anythe
side
lengths
circle. and numbers? (MP2)
with symbols
2
d = ((x2 − x1) + (y2 −
2 1/2
)
2
2 1/2
r = ((4_− 0) + (8 − 0) )
1/2
1/2
y1)
= (16 + 64)
= 4√5
= (80)
The Circle Centered at the Origin
y
(x, y)
r
y
x
We can generalize a formula for all
points (x, y) which lies on the circle
whose center is located at the origin
(0, 0).
x This will give us the equation of a
circle with its center at the origin.
No matter where we place the point
on the circle labeled (x, y) the
Pythagorean Theorem applies.
2
2
x +y =r
2
The Circle Centered at the Origin
y
x
Definition
x
A circle centered at the origin
with the radius r is the set of
all points (x, y) which
satisfies the equation
2
y
(x, y)
2
x +y =r
r
2
The Circle Centered at the Origin
(3, 4)
Example
A circle centered at the origin with the
radius r = 5 is the set of all points (x, y)
which satisfies the equation
2
2
x + y = 25.
2
2
3 +4 =
25
(4, 3)
Can you name more points that satisfy
the equation?
(−4, 3)
(−3, −4)
2
2
4 +3 =
25
2
2
(−3) + (−4) = 25
2
2
(−4) + 3 =
25
2
2
85 What is the radius of the circle x + y = 25?
B −25
C 5
D −5
E I need help
https://njctl.org/video/?v=uxmEHpD2eD4
Answer
A 25
C
2
2
86 If the y-coordinate of a point on the circle x + y = 25 is 5,
what is the x-coordinate?
B −5
C 5
D 25
E I need help
https://njctl.org/video/?v=y1WbMYfBnP8
Answer
A 0
A
2
2
87 How many points on the circle x + y = 25 have an
x-coordinate of 3?
Let x = 3 and solve for y:
2
2
(3 ) + y = 25
B One
y = 25__− 9
y = ±√16
y=±4
C Two
Answer
A None
2
D Not enough information
E
I need help
https://njctl.org/video/?v=fu1rmzYzZCo
There are TWO points,
(3, 4) and (3, −4)
C
2
2
88 How many points on the circle x + y = 25 have an
x-coordinate of 6?
B One
Answer
A None
None, because it is outside of
the radius of 5.
C Two
D
Not enough information
E
I need help
https://njctl.org/video/?v=x2Pdgi7pC8k
A
The Circle Centered at the Origin
y
Definition
x
y
r
(x, y)
https://njctl.org/video/?v=zuuS_tDBVow
A circle centered at the origin
with the radius r is the set of
all points (x, y) which
x satisfies the equation
2
2
2
xHow
+ about
y = rcircles whose
center is not the origin (0, 0)?
The Circle Centered at (h, k)
y
(x, y)
r
If a circle is centered at
the point (h, k), the
Pythagorean Theorem
produces a new
equation.
|y − k|
(h, k)
|x − h|
x
With the length as
indicated, what is the
new equation?
The Circle Centered at (h, k)
y
(x, y)
r
Definition
A circle centered at (h, k)
with the radius r is the set of
all points (x, y) which
satisfies the equation
|y − k|
(h, k)
|x − h|
2
x
2
(x − h) + ( y − k) = r
2
2
2
2
Note ( |x − h| ) + ( | y − k| ) = r
is equivalent to
2
2
(x − h) + ( y − k) = r
2
Circle Centered at (h, k)
y
r
(2, 3)
(x, y)
Consider a circle with center
(2, 3) as shown.
|y − 3|
The distance from the center
to each point on the circle is
the constant r. The equation
that describes this is
|x − 2|
x
2
2
(x − 2) + ( y − 3) = r
2
To determine the value of r
requires more information.
Circle Centered at (h, k)
y
The circle passes through the
point (6, 11). This information
allows us to determine the
radius.
r
Substitute x = 6 and y = 11
(2, 3)
x
2
2
2
(6 − 2) + (11 − 3) = r
2
2
2
(4) + (8) = r
2
16 + 64 = r
2
r = 80
2equation is2
2
The(x
circle's
− 2) + ( y − 3) = r
2
2
(x − 2) + ( y − 3) = 80
Circle Centered at (h, k)
Exercise
Write the equation of
y a circle with center (−2, 3) and radius 3.8.
(x, y)
r
x
Circle Centered at (h, k)
Exercise
Write the equation of
y a circle with center (−2, 3) and radius 3.8.
Solution
(x, y)
To start, write the equation with center
r
(−2, 3).
2
2
2
(x − h) + ( y − k) = r
2
2
2
(x + 2) + ( y − 3) = r
x
Now substitute r = 3.8.
2
2
2
(x − h) + ( y − k) = 3.8
2
2
(x + 2) + ( y − 3) = 14.44
89 What is the radius of the circle given by
2
2
(x − 5) + ( y − 3) = 36?
B −5
C 6
D 36
E
I need help
https://njctl.org/video/?v=C93tUzZVMuA
Answer
A 3
C
90 What is the equation of the circle?
y
2
2
A (x + 5) + (y − 3) = 25
2
2
2
2
2
Answer
2
x
B (x − 5) + (y + 3) = 20
C (x − 5) + (y + 3) = 25
D (x + 5) + (y − 3) = 20
E
I need help
https://njctl.org/video/?v=pSa9j-T6YFg
B
91 What is the x-coordinate of the center of the circle
2
2
whose equation is (x − 5) + ( y − 3) = 49?
B 5
C 7
D
49
E
I need help
https://njctl.org/video/?v=M2fTEsTasWk
Answer
A 3
B
92 What is the center and radius of the circle whose equation
2
2
is (x + 3) + ( y − 4) = 144?
C(−3, 4); r = 12
B
C(3, −4); r = 12
Answer
A
C
C(3, −4); r = 144
D
C(−3, 4); r = 144
E
I need help
https://njctl.org/video/?v=5fAIbeQ2Uag
A
93 What is the center and the radius of the circle whose
2
2
equation is (x − 5) + ( y − 3) = 169?
C(−5, −3); r = 12
B
C(−5, −3); r = 12
Answer
A
C
C(5, 3); r = 13
D
C(5, 3); r = 169
E
I need help
https://njctl.org/video/?v=EPHSwUJou18
C
Example
The point (−5, 6) is on a circle with center (−1, 3).
First, we need to find the length of the
Write the standard equation
of the circle.
radius:
Answer
Then substitute the center and radius
into the standard equation of a circle:
https://njctl.org/video/?v=5xAHT9JztS4
Example
y
Math Practice
The equation of a circle is
2
2
(x −Q's
4) to
+ (y
+ 2) = MPs:
36.
Additional
address
What information are you given?
(MP1) Graph the circle in the
coordinate plane.
What do
you need to find?
(MP1)
x
How do you graph the circle?
(MP4 & MP5)
Example
y
The equation of a circle is
2
2
(x − 4) + (y + 2) = 36.
Graph the circle in the
coordinate plane.
x Solution
(4, −2)
Center: (4, −2)
Radius = √36 = 6
94 Which is the standard equation of the circle below?
y
Answer
2
2
A
2
x
2
A x + y = 400
C (x + 10) + (y − 10) = 400
2
2
2
2
B (x − 10) + (y − 10) = 400 D (x − 10) + (y + 10) = 400
E I need help
https://njctl.org/video/?v=pW-0rWy9tGE
95 Which is the standard equation of the circle?
y
Answer
2
2
D
2
x
2
A (x − 4) + (y − 3) = 81
C (x + 4) + (y + 3) = 81
2
2
2
2
B (x − 4) + (y − 3) = 9
D (x + 4) + (y + 3) = 9
E I need help
https://njctl.org/video/?v=CBxnzjYxerY
96 What is the center and radius of
2
2
(x − 4) + (y − 2) = 64?
A C(0, 0); r = 64
C C(−4, −2); r = 8
D C(4, −2); r = 64
E I need help
https://njctl.org/video/?v=KRjNJGlY46Y
Answer
B C(4, 2); r = 8
B
97 What is the diameter of a circle whose equation is
2
2
(x − 2) + ( y + 1) = 16?
B 4
C 8
D 16
E I need help
https://njctl.org/video/?v=hU4F8pnshCw
Answer
A 2
C
98 Which point does not lie on the circle described by the
2
2
equation (x + 2) + ( y − 4) = 25?
B (1, 8)
C (3, 4)
D (0, 5)
E
I need help
https://njctl.org/video/?v=LAGOHUCl-3g
Answer
A (−2, −1)
D
Completing the Square
Math Practice
example
(this slide
the next 4)
Sometimes you're going to beThis
given
the equation
of a and
circle
which is not in standard form.addresses MP2 & MP7.
Additional
Q's to address
MPs:
You need to be able to transform
the equation
to standard
form
information
you given? (MP1)
in order to find the location of What
the center
and theare
radius.
What do you need to find? (MP1)
What
the numbers
represent
in the
For instance, it's not clear what
the do
radius
and center
are of the
problem? (MP2)
circle described by this equation.
What do you know about perfect square
2
2
trinomials that can apply to this
x − 4x + y − 12 = 0
situation? (MP7)
https://njctl.org/video/?v=QvgqHhLTuuA
Completing the Square
2
2
x − 4x + y − 12 = 0
To find the radius and the coordinates of the center, we need to
transform the equation above into the form
2
2
(x − h) + (y − k) = r
2
The first step is to separate groups of terms which have x, which
have y, and are constants.
Just moving those around makes this equation:
2
[x
2
− 4x] + y = 12
Take a moment to confirm that this is true.
Completing the Square
2
2
x − 4x + y − 12 = 0
2
[x
2
− 4x] + y = 12
We already see that the y-coordinate of the center is 0 (k = 0),
2
since y is by itself.
2
But what to do with the expression (x − 4x)?
2
We have to convert that into the form (x − h) to find the xcoordinate of the center...and then the radius.
Completing the Square
If you recall, when completing the square, you divide the coefficient of
x by 2, square it, and add that number to both sides of the equation.
2
2
(x
− 4x) + y = 12
Dividing the coefficient −4 by 2 yields −2, so h = 2
2
Then h = 4
2
2
(x
− 4x + 4) + y = 12 + 4
2
2
(x − 2) + y = 16
The center is at (2, 0) and the radius is 4.
The same steps are used to find k, when needed, as in the next
example.
Completing the Square
In general, the steps given below are used to determine the equation
of a circle. Note that the coefficients of x and y are negative.
2
2
x − 2hx + y − 2ky = C
Divide the coefficients of the x and y terms by −2 and square them.
Add these terms to both sides.
2
2
2
2
2
2
(x
− 2hx + h
) + (y
− 2ky + k
)=C+h
+k
Factor the perfect square trinomials on the left side of the equation.
2
2
2
2
(x − h) + (y − k) = C + h + k
Example of Completing the Square
Determine the radius and center of this circle.
2
2
example
MP2 & MP7.
x + y − 2xThis
+ 6y
+ 6 = addresses
0
Math Practice
Additional Q's to address MP standards
What information are you given? (MP1)
What do you need to find? (MP1)
What do the numbers represent in the
problem? (MP2)
What do you know about perfect square
trinomials that can apply to this situation?
(MP7)
Example of Completing the Square
Determine the radius and center of this circle.
2
2
x + y − 2x + 6y + 6 = 0
2
[x
2
[x
2
− 2x] + [y + 6y] = −6
2
− 2x][y + 6y]
2h = −22k = 6
2
2
h = −2/2 = −1 ⇒ h = 1k = 6/2 = 3 ⇒ k = 9
2
2
(x − 2x + 1) + (y + 6y + 9) = −6 + 1 + 9
2
2
(x − 1) + (y + 3) = 4
The center is (1, −3) and the radius is 2.
99 What are the center and radius of the circle described by
this equation?
2
2
x 2+ y
Answer
− 8x + 6y − 11= 0
2
x + y − 8x + 6y = 11
2
2
[x − 8x] + [y + 6y + 9] = 11
A C(4, 3); r2 = 6
2
[x − 8x + 16] + [y + 6y + 9]
B C(4, −3); r = 6
= 11 + 16 + 9
2
2
(x − 4)
C C(−4, −3);
r=+
36(y + 3) = 36
C(4, −3); r = 6
D C(−4, 3); r = 36
E
I need help
https://njctl.org/video/?v=KKp-kUaAgXo
B
100 What are the center and radius of the circle described by
this equation?
2
2
x + y − 8x + 4y − 5 = 0
2
A
C
D
E
Answer
B
2
x + y − 8x + 4y − 5 = 0
2
C(4, −2); 2r = 5
[x − 8x] + [y + 4y] = 5
C(4, 2);[rx2=−58x + 16] + [y2 + 4y + 4]
= 5 + 16 + 4
C(−4, −2); r =225
2
(x − 4) + (y + 2) = 25
C(−4, 2); r = 25
C(4, −2); r = 5
I need help
A
https://njctl.org/video/?v=VqO8IwJeuTg
101 What are the center and radius of the circle described by
this equation?
2
A
C
D
E
x + y + 16x − 22y + 174 = 0
2
2
x + y + 16x − 22y + 174 = 0
C(−8, −11);
2 r = 11
2
[x + 16x] + [y − 22y] = −174
2
C(8, −11);
√11+ 64] + [y2 − 22y + 121]
[x r+=16x
= −174 + 64 + 121
C(−8, 11); r = 2√11
2
(x + 8) + (y − 11) = 11
C(8, 11);
r = 11
C(−8,
11); r = √11
Answer
B
2
I need help
C
https://njctl.org/video/?v=JGfYPT6_Hp8
102 Part A
2
2
The equation x + y − 4x + 2y = b describes a circle.
2y-coordinate
2
Determine the
of the
x + y − 4x + 2y
= bcenter of the circle.
2
Answer
A
−2
=b+4+1
2
2
(x − 2) + (y + 1) = b + 5
C(2, −1)
B
−1
C
1
D
2
E
I need help
https://njctl.org/video/?v=reyKNxN7DJ8
2
[x − 4x] + [y + 2y] = b
2
2
[x − 4x + 4] + [y + 2y + 1]
B
From PARCC Sample Test - Non-calculator - Response Format
103 Part B
2
2
The equation x + y − 4x + 2y = b describes a circle.
The radius of the circle is 7 units. What is the value of b
in the equation?
2
Answer
A
2
B
12
C
44
D
54
E
I need help
https://njctl.org/video/?v=apeyZkC--6w
2
(x − 2) + (y + 1) = b + 5
2
r =b+5
2
(7) = b + 5
49 = b + 5
44 = b
C
From PARCC Sample Test - Non-calculator - Response Format
2
2
104 The equation x − 8x + y = 9 defines a circle in the xycoordinate plane. To find the radius of the circle, the
2
2
equation can be rewritten as (________) + y = ___.
Answer
A
I and VIII
B
II and VI
C
IV and VII
D
II and V
E
D II and V
1/2
I need help
https://njctl.org/video/?v=PMIODy_QS8I
I.x +x24− 8x + y2V.=25
9
2 −4
2
II.x
[x
− 8x] + [y −VI.
0] 13
=9
2
2
[x III.x
− 8x++1616] + yVII.
= 99 + 16
2
2
IV.x
(x−−16
4) + y VIII.
= 255
C(4, 0); r = (25)
=5
From PARCC Sample Test - Calculator - Response Format