Set up and solve linear equations using segment and midpoint properties. Use correct notation for distance and segments.

1. Segment Addition Postulate
Objectives:
• Calculate the distance between two points on a number
line and on a graph
• Set up and solve linear equations using midpoint
properties
• Correctly use notation for distance and segments

2. distancedistancedistancedistance – the absolute value of the difference of the
coordinates. Also called the lengthlengthlengthlength.
Example:
The distance from R to S is written RS:
Distance is alwaysalwaysalwaysalways positive. If you come up with a
negative answer, you’ve done something wrong!
Notation: Notice the different notations:
AB line AB
segment AB
AB length AB
AB
2 3 5 5RS = − − = − =
R S

3. congruent segmentscongruent segmentscongruent segmentscongruent segments – segments that have the same length.
Notation: “Tick marks” indicate congruent segments. To
match more than one set of congruent segments, match
up the tick marks (2 tick marks go with 2 tick marks, etc.)
YX
A B
Since XY , (and vice versa)AB XY AB= ≅
• •
t

4. betweenbetweenbetweenbetween – point B is between two points A and C if all three
points are collinearcollinearcollinearcollinear and
AB + BC = AC.
(part + part = whole)
Note: This is also called the Segment Addition PostulateSegment Addition PostulateSegment Addition PostulateSegment Addition Postulate.
●
A B C

5. bisectbisectbisectbisect – to cut or divide into two congruent pieces.
Example:
Point B bisectsbisectsbisectsbisects FI ⇒ FB = BI
midpointmidpointmidpointmidpoint – the point that bisects a segment.
Example: Point B is the midpointmidpointmidpointmidpoint of FI.
●
F B I

6. Examples 1. O is the midpoint of and DO = 16.
Find DG.
2. K is the midpoint of and SY = 24.
Find SK.
3. E is the midpoint of ; SE = 2x + 7 and
EA = 5x – 2. Find SA.
DG
SY
SA
●
D O G
16
●
S K Y
24
●
S E A
2x+7 5x–2

7. Examples 1. O is the midpoint of and DO = 16.
Find DG.
2. K is the midpoint of and SY = 24.
Find SK.
3. E is the midpoint of ; SE = 2x + 7 and
EA = 5x – 2. Find SA.
DG
SY
SA
●
D O G
16
●
S K Y
24
●
S E A
2x+7 5x–2
DO + OG = DG
16 + 16 = 32
SK = ½ SY = ½(24) = 12
SE = EA
2x + 7 = 5x – 2
9 = 3x
x = 3
SA = SE + EA
= 2(3)+7+5(3)-2
= 26