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PLANES, LINES AND POINTS
Point A
A
Ray AB, AB , AC
B
A
Line Segment AB, AB or BA
A
B
C
A is the initial point
Line AB, AB or BA or AC, CA or line m
A
B
C
m
C
A
D
B
Collinear Points: points that line on the same line
Points A, B and D are collinear
Points A, B and C are not collinear
Any two points are collinear. Two points define a line.
Opposite rays: Collinear rays with the same initial point, extending in
opposite directions .
BA and
BD are opposite rays
Between two points: In order to say that a point is between two
other points all three points must be collinear.
B is between A and D
C is not between A and D
M
D
C
Plane M or plane ABC or plane BCD
A
B
E
Coplanar points: points that line on the same plane.
A, B, C, D are all coplanar. A, B, C, E are not coplanar
Any three points are coplanar. Any three points define a plane.
Plane ABE exists although it is not shown.
Two or more figures intersect if they have one or more points in common.
Two lines intersect as a point
Two planes intersect at a line
Draw and label one diagram that includes all of the
following:
a. Plane R, points X, Y and Z coplanar, but not collinear.
b. XY and ZX
c. AB non-coplanar to plane R, but intersecting plane R
at point X
A
R
Z
Y
X
B
1.3 Segments and Their Measures
Ruler Postulate
The points on a line can be matched one to one with real
numbers. The real number that corresponds to a point is
the coordinate of the point.
The distance between points A and B, written as AB, is
the absolute value of the difference between the
coordinates of A and B.
A
3
B
9
Find AB:
AB= 3-9
AB=6
or
9-3
Segment Addition Postulate:
If B is Between A and C, then AB + BC = AC
If AB + BC = AC then B is between A and C
12
A
5
B
7
C
B is between A and C
C
A
B
B is not between A and C
Ex: if DE=2, EF=5, and DE=FG,
find FG, DF, DG, & EG.
D
E
F
FG=2
DF=7
DG=9
EG=7
G
Ex. Y is between X and Z. Find XY, YZ and XZ if:
XY= 3x + 4
YZ= 2x + 5
(3x + 4) + (2x +5) = 9x - 3
5x + 9 = 9x - 3
XZ= 9x - 3
12 = 4x
x=3
XY = 3(3) + 4
XY = 13
YZ = 2(3) + 5
YZ = 11
XZ = 9(3) -3
XZ = 24
If two line segments have the same
lengths they are said to be congruent.
The symbol for congruence is ≅
If AB = 5 and CD = 5 then,
AB = CD, distances are equal
AB ≅ CD, segments are congruent
The Distance Formula
If A ( x₁, y₁) and B ( x₂, y₂) are points in a coordinate plane
then the distance between A and B is:
AB = √ ( x₂ - x₁ )² + ( y₂ - y₁ )²
B ( x2 , y2 )
A ( x1, y1 )
Using the Distance Formula
G (3, 2)
GH = √ (11 -3)² + (6 - 2)²
GH = √ 8² + 4²
GH = √ 64 + 16
GH = √ 80
GH = 4√ 5
H (11, 6)
The Distance Formula comes from the Pythagorean
Theorem a² + b² = c² where a and b are the legs of a
right triangle and c is the hypotenuse.
AB is the hypotenuse of a right triangle
(x₂ - x₁)² = a, the length of the horizontal leg
(y₂ - y₁)² = b, the length of the vertical leg
B (x₂ , y₂)
c
b
A (x₁ , y₁)
a
Using the Pythagorean Theorem
H (11, 6)
c
b
a
(11,2)
G (3, 2)
a² + b² = c²
√a² + b² = c
√∣ 11-3 ∣² + ∣ 6-2 ∣² = c
√8² + 4² = c
√64 + 16 = c
√80 = c
4√5 = c
J (2,5), K (7,11) Find JK
a is the distance between 2 and 7, a=5
JK = √5² + 6²
b is the distance between 5 and 11, b=6
JK = √61
R (5,12), S (9, 2) Find RS
a is the distance between 5 and 9, a=4
b in the distance between 12 and 2, b=10
RS = √4² + 10²
RS = √116
RS = 2√29
An angle consists of two
different rays that have
the same initial point.
The rays are the sides of
the angle. The initial
point is the vertex of the
angle.
A
C
vertex
B
Angles are named using three points , the vertex and a point on
each ray which makes up the angle. The vertex is always the
second point listed. If an angle stands alone it can be named by
its vertex only.
J
H
∠JHI or ∠IHJ or ∠H
I
J
H
∠JHI or ∠IHJ
I
∠JHK or
∠KHJ
∠IHK or ∠KHI
K
You should not name any of these angles as H
because all three angles have H as their vertex.
The name H would not distinguish one angle
from the others.
The measure of ∠ A is denoted by m∠ A.
The measure of an angle can be approximated using a
protractor, using units called degrees(°).
B
A
C
For instance, ∠ BAC has a measure of 50°, which can be
written as m∠ BAC = 50°.
Angles that have the same measure are called congruent angles
If ∠BAC and ∠DEC both measure 75˚ then
m ∠BAC = m∠DEC, measures are equal
∠BAC ≅ ∠DEC , angles are congruent
B
Consider a point B on one side of AC. The rays
of the form AB can be matched one to one
with the real numbers from 1-180.
The measure of BAC is equal to the absolute
value of the difference between the real
numbers for AC and AB.
A
C
Applying the Protractor Postulate
C
D
A
m∠ CAB = 60˚ , m∠ BAD = 130˚
m∠ CAD = ∣ 130˚ - 60˚ ∣ or ∣ 130˚ - 60˚ ∣ = 70˚
B
B
C
A
A point is interior of an angle if it is between the sides of the angle
Point C is interior of angle A
A point is exterior of an angle if it is not on the angle or in its interior
Point B is exterior of angle A
J
H
M
K
If M is in the interior of ∠ JHK, then
m∠ JHM + m∠ MHK = m∠ JHK
Draw a sketch using the following information
D is interior of ∠ABC
C is interior of ∠DBE
m ∠ABC = 75˚
m ∠DBC = 45˚
m ∠ABE = 100˚
Find the m ∠ABD , m ∠CBE and m ∠DBE
Ans→
D is interior of ∠ABC
C is interior of ∠DBE
m ∠ABC = 75˚
m ∠DBC = 45˚
m ∠ABE = 100˚
D
A
75˚
45˚
100˚
B
Find the m ∠ABD , m ∠CBE and m ∠DBE
m∠ABC - m∠DBC = m∠ABD
75˚ - 45˚ = 30˚
C
E
m∠ABE - m∠ABC = m∠CBE
100˚ - 75˚ = 25˚
m∠CBE = 25˚
m∠ABD = 30˚
m∠DBC + m∠CBE = m∠DBE
45˚ + 25˚ = 70˚
m∠DBE = 70˚
Angles are classified according to their measures.
Angles have measures greater than 0°
and less than or equal to 180°
Two angles are adjacent if they share a common
vertex and side, but no common interior points.
J
H
M
K
∠JHM and ∠MHK are adjacent angles
∠JHM and ∠JHK are not adjacent angles
Bisecting a Segment
Bisect: to divide into two equal parts.
The midpoint of a segment is the point that bisects the segment,
dividing the segment into two congruent segments.
A
B
C
If C is the midpoint of AB, then AC ≅ CB
Congruent segments are indicated using marks through the
segments.
A segment bisector is a segment, ray, line or plane that
intersects a segment at its midpoint.
Y
C
A
X
B
XY is a bisector of AB
Midpoint Formula
To find the coordinates of the midpoint of a segment you find the mean of the
x coordinates and the y coordinates of the endpoints.
B x2 , y2 
A x1 , y1 
The midpoint of AB =
 x1  x2 y1  y2 
,


2 
 2
B (10, 6)
The midpoint of AB =
A ( 3,2 )
 3  10 2  6   13 
,

   ,4 
2  2 
 2
Finding the coordinates of an endpoint
The midpoint of GH is M ( 7, 5 ). One endpoint is H ( 15, -1 ). Find the coordinates
of point G.
G ( x₂ , y₂ )
M(7,5)
H ( 15 , -1 )
The x coordinate of G
The y coordinate of G
 15  x2 
7

 2 
14  15  x2
  1  y2 
5

2


10  1  y2
11  y2
 1  x2
G is at ( -1, 11 )
Practice Problems
JK has endpoints J ( -1, 7 ) and K ( 3, -3 ) find the coordinates of the midpoint
  1  3   7  3 

, 

 2  2 
( 1, 2 )
NP has midpoint M ( -8, -2 ) and endpoint N ( -5, 9 ). Find the coordinates of P
9 y
2

2


4 9 y
 5 x 
8  

 2 
 16  5  x
 11  x
 13  y
P ( -11, -13 )
Angle Bisector
An angle bisector is a ray that divides an angle into two
congruent adjacent angles.
D
A
B
C
If BD is a bisector of ∠ ABC, then ∠ ABD ≅ ∠ DBC
Congruent angles are indicated by separate arcs on each
angle
Practice Problems
KL is a bisector of ∠ JKM. Find the two angle measures not given
in the diagram
J
K
J
L
85⁰
K
M
L
37⁰
M
m ∠ JKL = 85/2 = 42½⁰
m ∠ JKL = m ∠ LKM = 37⁰
m ∠ LKM = 85/2 = 42½⁰
m ∠ JKM = 2 x 37 = 74⁰
Practice Problem
KL is a bisector of ∠ JKM. Find the value of x
J
6x – 11 = 10x – 51
-11 = 4x -51
40 = 4x
x = 10
L
K
( 10x – 51 )⁰
M
Check
6 (10) – 11 = 10 (10) -51
60 - 11 = 100 – 51
49 = 49
END
Ruler Postulate
The points on a line can be matched one to one with real
numbers. The real number that corresponds to a point is
the coordinate of the point.
The distance between points A and B, written as AB, is
the absolute value of the difference between the
coordinates of A and B.
A
3
B
9
Find AB:
AB= 3-9
AB=6
or
9-3
Vertical Angles: angles whose sides form opposite rays
1
2
4
3
∠ 1 and ∠ 3 are vertical angles
∠ 2 and ∠ 4 are vertical angles
Linear Pair: two adjacent angles whose noncommon sides are
opposite rays
1
2
4
3
∠1 and ∠ 2 are a linear pair
∠ 2 and ∠ 3 are a linear pair
∠ 3 and ∠ 4 are a linear pair
∠ 4 and ∠ 1 are a linear pair
110˚
1
270˚
70˚ 4
3
110˚
If m ∠ 1 = 110° , what is the m ∠ 2 and m ∠ 3 and m ∠ 4?
What conclusions can you draw about vertical angles?
What conclusions can you draw about linear pairs?
Practice Problem
( 4x + 15
)°
( 5x + 30 )°
( 3y + 15 )°
( 3y - 15 )°
Find the value of x and y
( 4x + 15 ) + ( 5x + 30 ) = 180
( 3y + 15 ) + ( 3y - 15 ) = 180
9x + 45 = 180
6y = 180
9x = 135
Y = 60
X = 15
Complementary angles: two angles whose sum is 90°.
Each of the angles is called the complement of the other.
Complementary angles can be adjacent or nonadjacent
23°
Adjacent
complementary angles
23°
Nonadjacent complementary angles
Supplementary angles: two angles whose sum is 180°.
Each of the angles is called the supplement of the other.
Supplementary angles can be adjacent or nonadjacent
130°
50°
Adjacent supplementary angles
130°
Nonadjacent supplementary angles
Practice Problems
T and S are supplementary. The measure of
measure of S. Find the measure of S
m∠T + m∠S = 180
T is half the
½m∠S = m∠T
½m∠S + m∠S =180
(3/2)∠S =180
m∠S = 120˚
T and S are complementary. The measure of T is four
times the measure of S. Find the measure of S
m∠T + m∠S = 90
4m∠S + m∠S = 90
5m∠S = 90
m∠S = 18˚
4m∠S = m∠T
Rectangle
l
( I = length, w = width )
w
Perimeter: P = 2l + 2w
P = 2(12) + 2(5)
12 in
P = 24 + 10
P = 34 in
Area: A = l w
A = 12(5)
A = 60 in²
5 in
Practice Problems
Find the perimeter and area of each rectangle
4ft
9ft
9ft
P = 2(9) + 2(4)
A = 9(4)
P = 18 + 8
A = 36 ft²
P = 2(12) + 2(9)
A = 12(9)
P = 24 + 18
A = 108 ft²
P = 42 ft
P = 26 ft
A rectangle has a perimeter of 32 in and a length of 9 in. Find its area.
P = 2l + 2w
32 = 2(9) + 2w
32 = 18 + 2w
14 = 2w
w = 7 in
A = 9(7)
A = 63 in²
Square
(s = side length)
Perimeter: P = 4s
Area: s²
Find the perimeter and area of each square
8ft
4m
Find the perimeter of a square with an area of 64 ft²
Find the perimeter and area of each square
8ft
4m
a=b
a² + a² = c²
2a² = 4²
2a² = 16
a² = 8
a =√8
P = 4(8)
P = 32 ft
A = 8²
A = 64 ft²
Find the perimeter of a square with an area of 81 ft²
s² = 81
s=9
P = 4(9)
P = 36 ft²
P = 4√ 8
P = 4(2)√2
P = 8√2 m
A = (√8)²
A = 8m²
Triangle
( a, b, c = side lengths, b = base, h = height)
Perimeter: P = a + b + c
Area: A = ½ bh
Find the perimeter and area of each triangle:
14m
5m
4m
15m
7 2ft
7 ft
Find the perimeter and area of each triangle:
14m
5m
4m
7 2 ft
7 ft
15m
P = 5 + 15 + 14
P = 34 m
P = 7 + 7 +7 2
P = 14 + 7 2
A = ½ 15(4)
A = 30 m²
A = ½ 7 (7)
A = 24½ ft²
Circle
( r = radius)
Circumference: C = 2 π r
Area: A = πr²
Find the circumference and area of each circle
6in
8 ft
Find the circumference and area of each circle
6in
8 ft
R = 8/2 = 4
C=2π6
C = 12 π
C ≈ 37.70 in
A = π 6²
A = π36
A ≈ 113.10 in²
C=2π4
C= 8π
C ≈ 25.13 ft
A = π 4²
A = 16
A ≈ 50.27 ft²
Problem Solving Steps
1. Define the variable(s)
2. Write an equation using the variable
3. Solve the equation
4. Answer the question
Practice Problem
A painter is painting one side of a wooden fence along a highway. The
fence is 926 ft long and 12 ft tall. Each five gallon can of paint can cover
2000 square feet. How many cans of paint will be needed to paint the
fence.
1. Define the variable: x = number of 5 gallon cans needed
2. Write an equation using the variable:
2000x = 926(12)
3. Solve the equation
2000x = 926(12)
2000x = 11112
x = 5.56
4. Answer the question: 6 cans of paint will be needed to paint the fence