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Segments and their Measure
The Ruler Postulate
The points on a line can be matched one to one with the real numbers.
·A
·B
The real number that corresponds to a point is the coordinate of the point
What is the coordinate of:
1) Point A
2) Point B
= 1
= 4.5
Segments and their Measure
Distance between point A and B in symbol is AB
AB = | coordinate of A - coordinate of B|
Example: Find the distance between the given points
·A
·B1.)
Solution: AB = | coordinate of A - coordinate of B|
AB = | |1 - 4.5 = |-3.5 | = 3.5
Segments and their Measure
Distance between point A and B in symbol is AB
AB = | coordinate of A - coordinate of B|
Example: Find the distance between the given points
2.)
Find: AG
GR
AE
=| |2 - -7 = |9| = 9
876543210-1-2-3-4-5-6-7-8
● ● ●● ●G R A C E
=| |-7 - -3 = |-4| = 4
=| |2 - 7 = |-5| = 5
Segments and their Measure
Distance between point A and B in symbol is AB
AB = | coordinate of A - coordinate of B|
Try it yourself: Find the distance between the given points
3.)
Find: CG
ER
EG
=| |6 - -7 = |13| = 13
876543210-1-2-3-4-5-6-7-8
● ● ●● ●G R A C E
=| |7 - -3 = |11| = 11
=| |7 - -7 = |14| = 14
Segments and their Measure
Segment Addition Postulate
·A ·C·B
If B is between A and C
Then
AB + BC = AC
•EXAMLE 3
Example 1: Use the diagram to find GH.
Use the Segment Addition Postulate to write an equation. Then solve the equation to find GH.
SOLUTION
Segment Addition Postulate.
Substitute 36 for FH and 21 for FG.
Subtract 21 from each side.
21 + GH=36
FG + GH=FH
=15 GH
Segments and their Measure
Segment Addition Postulate
SOLUTION
Example 2: Maps
The cities shown on the map lie approximately in a straight line. Use the given distances to find the distance from Lubbock, Texas, to St. Louis, Missouri.
Because Tulsa, Oklahoma, lies between Lubbock and St. Louis, you can apply the Segment Addition Postulate.
LS = LT + TS = 380 + 360 = 740
The distance from Lubbock to St. Louis is about 740 miles.
ANSWER
Segments and their Measure Segment Addition Postulate
Segments and their Measure
Segment Addition Postulate
Example 3:
In the figure: AD = 20, BD = 15, CD = 4
A CB· ·· D·Find: AB
BC
AC
Segments and their Measure
Segment Addition Postulate
Example 4:
RS = 2x + 10, ST = x - 4, RT= 21
·R
·T
·S
Find x
RS
Connecting with Algebra
ST
Midpoint and BisectorLesson 1.2 continued
Congruent Segment
Midpoint
Bisector
Friday, January 31, 2020
Congruent Segment• Congruent segments are segments with
equal length.
Congruence Symbol:
EXAMPLE:
1. If AB = ED THEN AB ED
2. Use the number line to determine whether the statement is true or false. Explain why.
876543210-1-2-3-4-5-6-7-8
● ● ●● ●G R A C E
a) AR AE b) GR AEtrue false
Midpoint of Segment and Bisector
• The midpoint of a segment is a point that divides the segment into two congruent segments.
ADE● ● ●
In the Figure Point D is the midpoint of EA if
1. D is between E and A
2. ED = DA
Midpoint of Segment and Bisector
Bisect means to separate into two
congruent parts.
A segment bisector can be a point,
line, ray, segment or plane.
A B● ●●
P
Midpoint of Segment and Bisector
Solution: 5x – 6 = 2x
5x – 2x = 6
3x = 6
x = 2
AB = 8
BCA● ● ●
5x – 6 2x
Example 1: Algebra Problem
In the Figure Point C is the midpoint of AB
Find: x , AB
Friday, January 31, 2020
In the skateboard design, VW bisects XY at point T,
and XT = 39.9 cm. Find XY.
Example 2: Skateboard
SOLUTION
EXAMPLE 1
Point T is the midpoint of XY .
So, XT = TY = 39.9 cm.
XY = XT + TY
= 39.9 + 39.9
= 79.8 cm
Segment Addition Postulate
Substitute.
Add.
Midpoint of Segment and Bisector
Midpoint in the Coordinate Plane
Midpoint of AB =
y
x
·A
(x1, y1)
·B
(x2, y2)
•Midpoint
2,
2
1212 yyxxMidpoint Formula
In the coordinate plane,
B with coordinates (x2, y2)
whose endpoints are A with
coordinate (x1, y1)
the midpoint of a segment
is given by the
and
Midpoint in the Coordinate Plane
FIND MIDPOINT The endpoints of RS are R(1,–3)
and S(4, 2). Find the coordinates of the midpoint
M.
Example 1 Using the Midpoint Formula
M =
2,
2
1212 yyxx
SOLUTION
M =
R( 1, -3 ) S( 4, 2 ) x1
y1 x2y2
4 + 12
2 + -3
2,
M = 52
-1
2,
Midpoint in the Coordinate Plane
Example 2
Missing one Endpoint given a Midpoint
Given one endpoint (x1, y1) and a midpoint (x, y)
Missing Endpoint= ( 2x - x1, 2y - y1)
Find the endpoint K of JK if J is (1,
4) and the midpoint is M(2, 1)
SOLUTION J( 1, 4 ) M( 2, 1 ) x1
y1 x y
K = ( 2x- x1, 2y - y1)
K = (2)- 1 (1) - 4,2 2
K = 3 -2,
Midpoint in the Coordinate Plane
Try it yourself!
Midpoint Formula
Missing Endpoint = ( 2x - x1, 2y - y1)
2
yy,
2
xx 1212
Guided Practice
M = 4 5,
V = -6 -8,
Midpoint in the Coordinate Plane
Try it yourself!
Midpoint Formula
Missing Endpoint = ( 2x - x1, 2y - y1)
2
yy,
2
xx 1212
More Practice
Find the midpoint between
Find the missing endpoint with the given
Distance in the Coordinate Plane
·A
(x1, y1)
·B(x2, y2)
In the coordinate plane
B with coordinates (x2, y2)
whose endpoints are A with coordinate (x1, y1) and
the distance of a segment
is denoted by
2
12
2
12 )()( yyxx AB =
Distance AB
Distance Formula
Friday, January 31, 2020
Distance in the Coordinate Plane
1. What is the approximate length of RS with
endpoints R (2, 3) and S (4, -1).
Example Using the Distance Formula
RS =
SOLUTION
RS =
R( 2, 3 ) S( 4, -1 ) x1
y1 x2y2
4 – 22
-1 – 32
+
2
12
2
12 )()( yyxx
RS = 22
-4 2+
RS = 20 = 2 5
Distance in the Coordinate Plane
Practice Work
Distance Formula 2
12
2
12 )yy()xx(
Find the distance between the 2 given points.
1. (2, 4) and (-1, 3) 2. (-3, 4 ) and (2, -1)
10D 2D 5
Pythagorean TheoremLesson 1.5
Solving for One missing side of a right triangle
Friday, January 31, 2020
The Pythagorean TheoremThe Pythagorean Theorem
In a right triangle, the square of the length of the
hypotenuse is equal to the sum of the squares of
the lengths of the two sides.
c
a
b
c2= a2 + b2
zx
y
z2 = x2 + y2
Note: Pythagorean Theorem is only use in RIGHT TRIANGLES to find one missing side.
Missing Hypotenuse
12
x5
SOLUTION
x = x = 13
In the right triangle, find x
22 12 5
x = 144 25
x = 169
THE PYTHAGOREAN THEOREM
52 + 122 = x2
52 + 122=x2
Missing side
Find the missing side of the right triangle.
x
14
7SOLUTION
x = 147
x =
THE PYTHAGOREAN THEOREM
37
x2 + 72 = 142
x2 + 𝟒𝟗 = 196
x2 = 196 - 49
x2 = 147
THE PYTHAGOREAN THEOREM
Practice Problems
x
3
6
14x
Solve the missing side of each right triangle
1) 2)
37