Section 1.3By BayLynn Ellsworth

Becky FallkMorgan Palmiter

Explanation of Content

Section 1.3: Collinearity, Betweenness, and

Assumptions

Collinearity

Collinear points are points that lie on the same line.

Noncollinear points are points that do not lie on the same line.

A B

M

Q

S

NONCOLLINEAR

the points don’t lie on the same line

COLLINEAR

the points lie on the same line

Betweenness of Points

All points must be collinear in order to say that a point is between two other points.

B L X

Triangle Inequality

There are only two possibilities for any three points: The points are collinear, which means

that one point is between the other two and two of the distances add up to the third.

They are noncollinear, which means that the three points determine a triangle.

*Something to Remember*

In a triangle, the sum of the lengths of any two sides is always greater than the length of the third side.

2015

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Assumptions from Diagrams

Should Be Assumed Should Not Be Assumed

Straight lines and angles Right Angles

Collinearity of points Congruent segments

Betweenness of points Congruent angles

Relative positions of points Relative sizes of segments and angles

Sample Problems

x6

11

_____ < x < ______

Identify range of values for x in each triangle.

Difference < x < Sum

5 17

Sample Problems (cont.)For each diagram, tell whether M is between B and R.

RM

B

1.) No

2.)

R

B

M

X

No

3.)

B RM

YES

Sample Problems (cont.)

F R

O

G

Should we assume that GF is a straight line?

↔ YES

Should we assume that <O is congruent to <ORG?

straight lines CAN be assumed

No

congruent angles can NOT be assumed

Practice Problem #1F

U

O

R

<>

O

O

FO is twice as long as

FR.

The perimeter of FOUR is 156.

Find: OU

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Practice Problem #1 Answer

Let FR = x Let FO = 2x

2x + 2x + x + x = 156

6x = 156x = 26

Since FR is congruent to OU, and FR = x, OU = x. Since x = 26, OU is 26.

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‾‾‾‾OU = 26

Practice Problem #2

W

D

O

R<WOR = 81˚ 19’ 10”

<WOD = 119˚ 31’ 43”

<ROD = ?

Practice Problem #2 Answer

119˚ 31’ 43”- 81˚ 19’ 10”38˚ 12’ 33”

<ROD = 38˚ 12’ 33”

Since <WOD = 119˚ 31’ 43” , and <WOR = 81˚ 19’ 10”, we need to subtract <WOR from <WOD to find <ROD.

- =38 ˚ 12’ 33” - cool81 ˚ 19’ 10” - warm119 ˚ 31’ 43” - HOT

Practice Problem #3

32

1

If <1, <2, and <3 are in a ratio of 2 : 4 : 6

Find m <1, m <2, m <3

Practice Problem #3 Answer

2x + 4x + 6x = 18012x = 180x = 152x = 30 = <14x = 60 = <26x = 90 = <3

Since we can assume straight angles, the sum of <1, <2, and <3 = 180. When solving for x, <1 = 30, <2 = 60, and <3 = 90.

Practice Problem #4

R

O

A

D

S

Given:<ROD is a right angle<AOS is a right angle<ROA is (3x + 2y)°<AOD is (3x + 3y)°<DOS is (6x)

Find: x and y

Practice Problem #4 Answer

R

O

A

D

S

3x + 2y +3x + 3y = 906x + 5y = 90

3x + 2y + 6x =909x + 3y = 903x + y = 30y = -3x + 30

6x + 5(-3x + 30) = 906x – 15x + 150 =90-9x + 150 = 90-9x = -60x = 6 2/3

6x + 5y =906 (6 2/3) + 5y = 9040 + 5y = 905y = 50y = 10

Since <ROD and <AOS are right angles, the two angles that make up the right angles are complementary and have a sum of 90°. Therefore, < ROA (3x + 2y) and <AOD (3x + 3y) = 90. The equation formed from that is 6x + 5y = 90. Then the sum of <AOD (3x + 3y) and <DOS (60) = 90. That equation simplifies to y = -3x + 30.When substituting y into the first equation, x = 6 2/3. After putting x back into the original equation, y = 10.

Works Cited Page

Rhoad, Richard, George Milauskas, and

Robert Whipple. Geometry for

Enjoyment and Challenge. Boston:

McDougal Little/Houghton Mifflin, 1991.