PLANES, LINES AND POINTS

A B

A

A B C

A B Cm

A

A B

C

D

Points A, B and D are collinear Points A, B and C are not collinear

BDBA and are opposite rays

B is between A and D C is not between A and D

Any two points are collinear. Two points define a line.

M

A B

C

D

E

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

R

XY

Z

A

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.

9

A B

3Find AB:AB= 3-9 or 9-3 AB=6

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

B

A C

BA C5 7

12

B is between A and C

B is not between A and C

ExEx: if DE=2, EF=5, and DE=FG, : if DE=2, EF=5, and DE=FG, find FG, DF, DG, & EG.find FG, DF, DG, & EG.

DD EE

FFGG

FG=2FG=2

DF=7DF=7

DG=9DG=9

EG=7EG=7

Ex. Y is between X and Z. Find XY, YZ and XZ if:

XY= 3x + 4

YZ= 2x + 5

XZ= 9x - 3

(3x + 4) + (2x +5) = 9x - 3

5x + 9 = 9x - 3

12 = 4x

x = 3

XY = 3(3) + 4XY = 13

YZ = 2(3) + 5YZ = 11

XZ = 9(3) -3XZ = 24

If two line segments have the samelengths they are said to be congruent.The symbol for congruence is ≅

If AB = 5 and CD = 5 then, AB = CD, distances are equalAB ≅ CD, segments are congruent

The Distance Formula

A

B

1 1( , )x y

2 2( , )x y

If A ( x , y ) and B ( x , y ) are points in a ₁ ₁ ₂ ₂coordinate planethen the distance between A and B is: AB = √ ( x - x ) + ( y - y )₂ ₁ ² ₂ ₁ ²

G

H

(3, 2)

(11, 6)Using the Distance Formula

GH = √ (11 -3) + (6 - ²2)²GH = √ 8 + 4² ²GH = √ 64 + 16GH = √ 80GH = 4√ 5

The Distance Formula comes from the Pythagorean Theorem a + b = c ² ² ² where a and b are the legs of a right triangle and c is the hypotenuse.AB is the hypotenuse of a right triangle (x - x ) = a, the length of the horizontal leg₂ ₁ ²

(y - y ) = b, the length of the vertical leg₂ ₁ ²

A

B

b

a

c

(x , y ) ₂ ₂

(x , y )₁ ₁

G

H

(3, 2)

(11, 6)Using the Pythagorean Theorem

(11,2)

a

cb

a + b = c ² ² ²√a + b = c² ²√∣ 11-3 ∣ + ∣ 6-2 ∣ = c² ²√8 + 4 = c² ²√64 + 16 = c√80 = c4√5 = c

J (2,5), K (7,11) Find JK

a is the distance between 2 and 7, a=5

b is the distance between 5 and 11, b=6

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

JK = √5 + ²6²JK = √61

RS = √4 + ²10²RS = √116RS = 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

B

side

side

vertex

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.

HJ

I

∠JHI or ∠IHJ or ∠H

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.

H

J

I

K

∠JHI or ∠IHJ ∠JHK or ∠KHJ∠IHK or ∠KHI

The measure of ∠ A is denoted by m∠ A. The measure of an angle can be approximated using a protractor, using units called degrees(°).

For instance, ∠ BAC has a measure of 50°, which can be written as m∠ BAC = 50°.CA

B

Angles that have the same measure are called congruent anglesIf ∠BAC and ∠DEC both measure 75˚ thenm ∠BAC = m∠DEC, measures are equal ∠BAC ≅ ∠DEC , angles are congruent

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.

CA

B

A

D C

B

Applying the Protractor Postulate

m∠ CAB = 60˚ , m∠ BAD = 130˚m∠ CAD = ∣ 130˚ - 60˚ ∣ or ∣ 130˚ - 60˚ ∣ = 70˚

A

A point is interior of an angle if it is between the sides of the angleA point is exterior of an angle if it is not on the angle or in its interior

C

B

Point C is interior of angle A Point B is exterior of angle A

H

J

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 ∠ABCC is interior of ∠DBEm ∠ABC = 75˚m ∠DBC = 45˚m ∠ABE = 100˚Find the m ∠ABD , m ∠CBE and m ∠DBE

Ans→

D is interior of ∠ABCC is interior of ∠DBEm ∠ABC = 75˚m ∠DBC = 45˚m ∠ABE = 100˚ A

B

C

D

E

75˚ 45˚

100˚

Find the m ∠ABD , m ∠CBE and m ∠DBE m∠ABC - m∠DBC = m∠ABD 75˚ - 45˚ = 30˚ m∠ABE - m∠ABC = m∠CBE100˚ - 75˚ = 25˚ m∠ABD = 30˚ m∠CBE = 25˚

m∠DBC + m∠CBE = m∠DBE45˚ + 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.

H

J

M

K

∠JHM and ∠MHK are adjacent angles∠JHM and ∠JHK are not adjacent angles

Bisecting a SegmentThe midpoint of a segment is the point that bisects the segment, dividing the segment into two congruent segments.Bisect: to divide into two equal parts.

A segment bisector is a segment, ray, line or plane that intersects a segment at its midpoint.

A BC

Congruent segments are indicated using marks through the segments.

A BC

X

Y

XY is a bisector of AB

If C is the midpoint of AB, then AC ≅ CB

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.

2,

22121 yyxx

The midpoint of AB =

B (10, 6)

A ( 3,2 )

11, yxA

B 22 , yx

The midpoint of AB =

4,2

13

2

62,

2

103

Finding the coordinates of an endpointThe midpoint of GH is M ( 7, 5 ). One endpoint is H ( 15, -1 ). Find the coordinates of point G.

2

2

2

1

1514

2

157

x

x

x

The x coordinate of G The y coordinate of G

2

2

2

11

110

2

15

y

y

y

G is at ( -1, 11 )

H ( 15 , -1 )

M ( 7 , 5 )

G ( x₂ , y₂ )

Practice Problems

JK has endpoints J ( -1, 7 ) and K ( 3, -3 ) find the coordinates of the midpoint

NP has midpoint M ( -8, -2 ) and endpoint N ( -5, 9 ). Find the coordinates of P

x

x

x

11

516

2

58

P ( -11, -13 )

2

37,

2

31

( 1, 2 )

y

y

y

13

94

2

92

Angle BisectorAn angle bisector is a ray that divides an angle into two congruent adjacent angles.

C

A

B

D

Congruent angles are indicated by separate arcs on each angleIf BD is a bisector of ∠ ABC, then ∠ ABD ≅ ∠ DBC

Practice Problems

KL is a bisector of ∠ JKM. Find the two angle measures not given in the diagram

J

M

L

K85⁰

J

M

L

K

m ∠ JKL = 85/2 = 42 ½⁰m ∠ LKM = 85/2 = 42 ½⁰

37⁰

m ∠ JKL = m ∠ LKM = 37⁰m ∠ JKM = 2 x 37 = 74 ⁰

J

M

L

K

KL is a bisector of ∠ JKM. Find the value of xPractice Problem

( 10x – 51 )⁰

( 6x – 11 )⁰ 6x – 11 = 10x – 51 -11 = 4x -51 40 = 4x x = 10

Check6 (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.

9

A B

3Find AB:AB= 3-9 or 9-3 AB=6

Vertical Angles: angles whose sides form opposite rays

12

34

∠ 1 and ∠ 3 are vertical angles ∠ 2 and ∠ 4 are vertical angles

Linear Pair: two adjacent angles whose noncommon sides are opposite rays

∠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

12

34

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?

43

2

1110˚

110˚

70˚ 70˚

Practice Problem

( 4x + 15 )°

( 3y + 15 )°( 3y - 15 )°

( 5x + 30 )°

Find the value of x and y( 4x + 15 ) + ( 5x + 30 ) = 180

9x + 45 = 180

9x = 135

X = 15

( 3y + 15 ) + ( 3y - 15 ) = 180 6y = 180

Y = 60

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°67°

67°

23°

Adjacent complementary angles

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° 130°

50°

Adjacent supplementary angles Nonadjacent supplementary angles

Practice Problems

T and S are supplementary. The measure of T is half the measure of S. Find the measure of S

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 = 180m∠S + m∠S =180½(3/2)∠S =180m∠S = 120˚m∠S = m∠T½

m∠T + m∠S = 90 4m∠S = m∠T4m∠S + m∠S = 90 5m∠S = 90m∠S = 18˚

Rectangle

( I = length, w = width )

Perimeter: P = 2l + 2w

l

w

Area: A = l w

12 in

5 in

P = 2(12) + 2(5)

P = 24 + 10

P = 34 in

A = 12(5)

A = 60 in²

Find the perimeter and area of each rectangle

9ft

4ft 15ft

A rectangle has a perimeter of 32 in and a length of 9 in. Find its area.

9ft

P = 2l + 2w32 = 2(9) + 2w32 = 18 + 2w14 = 2ww = 7 in

P = 2(9) + 2(4)

P = 18 + 8

P = 26 ft

A = 12(9)

A = 108 ft²

A = 9(7)A = 63 in²

Practice Problems

A = 9(4)

A = 36 ft²

P = 2(12) + 2(9)

P = 24 + 18

P = 42 ft

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

Find the perimeter of a square with an area of 81 ft²

a = ba² + a² = c²2a² = 4²2a² = 16 a² = 8 a = √ 8

s² = 81 s = 9

P = 4(8)P = 32 ft

A = 8²A = 64 ft²

P = 4√ 8P = 4(2)√2P = 8√2 m

A = (√8)²A = 8m²

P = 4(9)P = 36 ft²

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:

4m5m

15m

14m ft277 ft

Find the perimeter and area of each triangle:

4m5m

15m

14m 277 ft

ft

P = 5 + 15 + 14P = 34 m

A = ½ 15(4)A = 30 m²

P = 7 + 7 + P = 14 + A = ½ 7 (7)A = 24½ ft²

2727

Circle

( r = radius)

Find the circumference and area of each circle

6in8 ft

Circumference: C = 2 π r Area: A = πr²

R = 8/2 = 4

C = 2 π 4C = 8 πC ≈ 25.13 ft

A = π 4²A = 16A ≈ 50.27 ft²

Find the circumference and area of each circle

6in 8 ft

C = 2 π 6C = 12 πC ≈ 37.70 in

A = π 6²A = π36A ≈ 113.10 in²

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