1 PROBLEM 2A PROBLEM 3A PROBLEM 2B PROBLEM 3B PROBLEM 4 PROBLEM 9BPROBLEM 9A PROBLEM 5 LINES STANDARDS 13, 17 SEGMENT ADDITION POSTULATE MIDPOINT PROBLEM

1

PROBLEM 2A

PROBLEM 3A

PROBLEM 2B

PROBLEM 3B

PROBLEM 4

PROBLEM 9BPROBLEM 9A

PROBLEM 5

LINESSTANDARDS 13, 17

SEGMENT ADDITION POSTULATE

MIDPOINT

PROBLEM 10BPROBLEM 10A

PLANES

PROBLEM 1A PROBLEM 1B

DISTANCE FORMULA

PROBLEM 6

MIDPOINT FORMULA

COORDINATE GEOMETRY

END SHOW

PROBLEM 7 PROBLEM 8

PYTHAGOREAN THEOREM

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

2

STANDARD 13:

Students prove relationships between angles in polygons using properties of complementary, supplementary, vertical and exterior angles.

STANDARD 15:

Students use the Pythagorean Theorem to determine distance and find missing lengths of sides of right triangles.

STANDARD 17:

Students prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles.



3

ESTÁNDAR 13:

Los estudiantes prueban relaciones entre ángulos en polígonos usando propiedades de ángulos complementarios, suplementarios, verticales y ángulos exteriores.

ESTÁNDAR 15:

Los estudiantes usan el Teorema de Pitágoras para determinar distancias y encontrar las longitudes de los lados de triángulos.

ESTÁNDAR 17:

Los estudiantes prueban teoremas usando geometría coordenada, incluyendo el punto medio de un segmento, la fórmula de la distancia y varias formas de ecuaciones de líneas y círculos.



4

AB

Line ABp

or

Line p

H

I

Line HI

M N

Line MN

is VERTICAL

is HORIZONTAL

How do we call this line?

What about this other?

LINES STANDARD 13



5

CF

Line CD

m

Line m?

LINES

D E

How many different ways can we call

Line CE

Line CF

Line DE

Line DF

Line EF

Line DC

Line EC

Line FC

Can you figure out other names?

OR

IN GENERAL A LINE IS NAMED BY TWO POINTS.

STANDARD 13



6

D E

SEGMENTS

If we have line DE and we take one part of the line

STANDARD 13



7

D E

SEGMENTS

If we have line DE

D E

and we take one part of the line

then this part is called:

LINE SEGMENT DE

Can you name the different line segments in the following line:

S T U

ST

SU

TU

TS

US

UT

SEGMENTS:OR

STANDARD 13



8

RAYS

AB

This is RAY AB

What are the differences between a Line, a Line Segment and a Ray?

The line is infinite and never ends at either side.

The line segment has two endpoints.

The ray has on one side one endpoint at the other side it never ends it goes on to infinite.

M N

D E

A

B

What do they have in common?

They are named using TWO POINTS.

STANDARD 13



9

CF

D E

Points C, D, E and F: are they COLLINEAR?

Yes, they are COLLINEAR because they lie in the same line

No, they are NONCOLLINEAR because they don’t lie in the same line.

Are points A, B, C, and D collinear?

A

C

BD

COLLINEAR VS NONCOLLINEAR STANDARD 13



10

Can you explain where the following two lines intersect?

l

m

A

BC

D

E

They INTERSECT at point E.

Can they intersect at other point? NO

STANDARD 13



11

PLANES

MA

C

DB

The figure above represents a PLANE, which is a flat surface that has no end at any of the sides.

What examples can you give of objects lying in a PLANE?

This is:

PLANE M

PLANE ADC

PLANE BDC

PLANE ABD

In general a Plane can be named using three non-collinear points.

• The wall of a house• The lid of a shoebox.• The ground of a football field.

STANDARD 13



12

P

M

A

B

l

m

n

Can you describe the INTERSECTION of planes M and P?

Planes M and P intersect at line AB. PLANES ALWAYS INTERSECT AT A LINE.

Where do lines m and l intersect? They intersect at point B. LINES ALWAYS INTERSECT AT A POINT.

Where do lines n and l intersect? At point A.

STANDARD 13



13

Are Points L, K, and M COPLANAR?

Yes, they are COPLANAR because they LIE ON THE SAME PLANE P.

Is point H, coplanar with points L, K, and M?

P

Q

A

B

LK

M

H

C

No, because it lies on plane Q and points L, K, and M are in different plane, on plane P.NON-COPLANAR points are points that lie in different planes.

D

STANDARD 13



14

P

Q

A

B

C

D

On what planes does point C lie?

STANDARD 13



15

P

Q

A

B

C

On what planes does point C lies? On planes P and Q.

D

STANDARD 13



16

On what planes does point D lie?

P

Q

A

B

C

D

STANDARD 13



17

On what planes does point D lie? It only lies on plane Q.

P

Q

A

C

B D

STANDARD 13



18

P

Q

A

C

B D

On what plane is line k lying?

Since points B and D lie on plane Q

then line k lies on its entirety on plane Q

k

STANDARD 13



19

SUMMARIZING FINDINGS:

• Through any two points there is exactly one line.

• Through any three points not on the same line there is exactly one plane or through any three points non-collinear there is one plane.

• A line contains at least two points.

• A plane contains at least three points not on the same line.

• A plane contains at least three non-collinear points.

• If two points lie in a plane, the entire line containing those points lies in that plane.

STANDARD 13



20

A B C

+ =

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.

AB BC AC

STANDARD 17



21

A B C

+ =

STANDARD 17Find the length for AB and BC if AC = 60 and AB = 4x + 6 and BC= 6x + 14. (B is between A and C)

+ =AB BC AC

AB BC AC

4x +6 + 6x + 14 = 60

4x + 6x + 6 + 14 = 60

10x + 20 = 60

-20 -20

10x = 4010 10

x = 4

Applying Segment Addition Postulate:

Now finding AB and BC:

AB = 4x + 6 BC = 6x + 14

= 4( ) + 6 = 6( ) + 144 4

= 16 + 6

= 22

= 24 + 14

= 38

Verifying the solution:

22 + 38 = 60

60 = 60



22

E F G

+ =

STANDARD 17Find the length of EF and FG if EG = 80 and EF = 3x + 8 and FG= 7x + 12. (F is between E and G)

+ =EF FG EG

EF FG EG

3x +8 + 7x + 12 = 80

3x + 7x + 8 + 12 = 80

10x + 20 = 80

-20 -20

10x = 6010 10

x = 6


Now finding EF and FG:

EF = 3x + 8 FG = 7x + 12

= 3( ) + 8 = 7( ) + 126 6

= 18 + 8

= 26

= 42 + 12

= 54

Verifying the solution:

26 + 54 = 80

80 = 80



23

A C

AB BC

STANDARD 17

and we place point B at the same

distance from point A than from Point C, then:

MIDPOINT OF A SEGMENT:

B

If we have segment AC,

and then point B is THE MIDPOINT OF SEGMENT AC.

Point B is also BISECTING segment AC, because it is dividing it into two halves.

AB = BC

Means congruent



24

STANDARD 17

A

B

C

D

E

AC is bisected by ED. AB= 6X + 8 and BC=4X + 18. Find the length for AC.

AB BC

AB = BC

6X + 8 = 4X + 18

-8 -8

6X = 4X + 10

-4X -4X

2X = 102 2

X = 5

Now finding AB:

AB = 6X + 8

= 6( ) + 85

= 30 + 8

= 38

Since AB = BC

BC = 38


AC = AB + BC

AC = AB + BC

AC = 38 + 38

AC = 76



25

STANDARD 17

R

S

T

V

U

RT is bisected by VU. RS= 8X + 4 and ST=4X + 28. Find the length for RT.

RS ST

RS = ST

8X + 4 = 4X + 28

-4 -4

8X = 4X + 24

-4X -4X

4X = 244 4

X = 6

Now finding RS:

RS = 8X + 4

= 8( ) + 46

= 48 + 4

= 52

Since RS = ST

ST = 52


RT = RS + ST

RT = RS + ST

RT = 52 + 52

RT = 104



26

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x-axis

y-axis

CARTESIAN COORDINATE PLANE

O

Origin

Quadrant III

Quadrant II Quadrant I

Quadrant IV

STANDARD 17



27

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x-axis

y-axis


(9, 4)

x-coordinate

y-coordinate

O

STANDARD 17



28

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x-axis

y-axis


(10,-8)

x-coordinate

y-coordinate

O

STANDARD 17



29

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x-axis

y-axis


(-9,-3)

x-coordinate

y-coordinate O

STANDARD 17



30

STANDARD 17DISTANCE FORMULA in a number line is given by:

|a – b|

ED H

-6 -4 -2 0 2 4 6 8 10 12

Find measure of DE, and EH:

DE = |-2 – (-6)|

= |-2 + 6|

= |4|

= 4

EH = |12 – (-2)|

= |12 + 2|

= |14|

= 14



31

STANDARD 17DISTANCE FORMULA in a number line is given by:

|a – b|

QR T

-6 -4 -2 0 2 4 6 8 10 12

Find measure of RT, and QT:

RT = |12 – (-6)|

= |12 + 6|

= |18|

= 18

QT = |12 – 6|

= |6|

= 6



32

Distance Formula between two points in a plane:

d = (x –x ) + (y –y )2 2

1 12 2

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

y1x

1

y2x2

,

,

STANDARD 17



33

Find the distance between points at A(2, 1) and B(6,4).y1

y2 x1

x2

AB= ( - ) + ( - )2 2

AB= ( -4 ) + ( -3 )2 2

= 16 + 9

= 25

AB=5

2 6 1 4

1

2

3

4

5

6

7

8

9

21 3 4 5 76 8 9 10 x

y

B

A

STANDARD 17

d = (x –x ) + (y –y )2 2

1 12 2

Remember:



34

d = ( - ) + ( - )2 2

= ( 5 ) + ( -1 )2 2

= 25 + 1

2 -3 -6 -5

d = (x –x ) + (y –y )2 2

1 12 2

y1x

1

y2x2

=(-3,-5)

=(2,-6)

d= 26

= ( + ) + ( + )2 22 3 -6 5

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Find the distance between (-3,-5) and (2,-6).

STANDARD 17



35

Find the value of a, so that the distance between (-6,2) and (a,10) be 10 units.

We use the distance formula:

d = (x –x ) + (y –y )2 2

1 12 2

10 = ( - ) + ( - )2 210

y1

2

y2

a

x1

-6

x2

10 = (-6-a) + (-8)22

100 = (-6-a) + 642

22

-64 -64

36 = (-6-a)2

6 = |-6-a|

6 = -(-6-a) 6 = -6-a

6 = 6 + a

-6 -6

a = 0

+6 +6

a = -12

6 = |-6-a|

Check:

6 =|-6- ( )| 6 =|-6- ( )|0 -12

6 = |-6|

6 = 6

6 =|-6+12|

6 =|6|6 = 6

6 = |-6-a|

Solving this absolute value equation:

12 = -a(-1) (-1)

STANDARD 17



36

STANDARD 15

Right Angle = 90°LEG

LEG

hypotenuse

Right triangle parts:



37

STANDARD 15Pythagorean Theorem:

xy

z

y + z = x2 2 2

The square of the hypotenuse is equal to the sum of the square of the legs.

The Pythagorean Theorem applies ONLY to RIGHT TRIANGLES!



38

STANDARD 15Find the value for x:

x= ?y= 4

z= 3

4 + 3 = x2 2 2

16 + 9 = x 2

25 = x 2

25 = x 2

|x|= 5

x= 5 x= -5

y + z = x2 2 2



39

STANDARD 15

x= 10

y= 8

z= ?

8 + z = 102 2 2

264 + z = 100-64 -64

z = 362

|z|= 6

z= 6 z= -6

z = 362

Find the value for z:

y + z = x2 2 2


40

STANDARD 17

R T

-6 -4 -2 0 2 4 6 8 10 12

Find Midpoint Q of RT:

MIDPOINT FORMULA in a number line is given by:

a + b2

-6 + 122

= 6 2

= 3

Q


41

STANDARD 17

K L

-4 -2 0 2 4 6 8 10 12 14

Find Midpoint R of KL:

MIDPOINT FORMULA in a number line is given by:

a + b2

-4 + 142

=10 2

= 5

R


42

Midpoint of a Line Segment:

If a line segment has endpoints at and , then the midpoint of the line segment has coordinates:

y1x1 y2x2

yx, =x1 x2 ,

2+ y1 y2

2+

21 3-1-2-3

1

2

3

-1

-2

-3

4 5-4-5

4

-4

5

x

y

y1x

1,

y2x2 ,

(x,y)

STANDARD 17


43

Find the midpoint of the line segment that connects points (1,6) and (9,8). Show it graphically.

1

2

3

4

5

6

7

8

9

21 3 4 5 76 8 9 10 x

y

(1,6)

(9,8)(5,7)

yx, = ,2+

2+

x1

1

x2

9

y1

6

y2

8

=yx, 142

102

,

=yx, 75,

yx, =x1 x2 ,

2+ y1 y2

2+Using:

STANDARD 17


44

Find the midpoint of the line segment that connects points (4,7) and (8,9). Show it graphically.

1

2

3

4

5

6

7

8

9

21 3 4 5 76 8 9 10 x

y

(4,7)

(8,9)(6,8)yx, = ,

2+

2+

x1

4

x2

8

y1

7

y2

9

=yx, 162

122

,

=yx, 86,

yx, =x1 x2 ,

2+ y1 y2

2+Using:

STANDARD 17


45

yx, =x1 x2 ,

2+ y1 y2

2+

= ,2+

2+

x1

-2

y1

-65

y

8,

x,

Using the Midpoint Formula:

x2y2

5=2+-6 y28=

2+-2 x

2(2) (2) (2) (2)

10 =-6 + y216 =-2 + x2

+2 +2 +6 +6

x2 =18 y2 =16

= 1618,y2x2 ,

y

84 12-4-8-12

4

8

12

-4

-8

-12

16 20-16-20

16

-16

20

xK

M

L

Given the coordinates of one endpoint of KL are K(-2,-6) and its midpoint M(8, 5). What are the coordinates of the other endpoint L. Graph them.

STANDARD 17


46

yx, =x1 x2 ,

2+ y1 y2

2+

= ,2+

2+

x1

-1

y1

-54

y

7,

x,

Using the Midpoint Formula:

x2y2

4=2+-5 y27=

2+-1 x

2(2) (2) (2) (2)

8 =-5 + y214 =-1 + x2

+1 +1 +5 +5

x2 =15 y2 =13

= 1315,y2x2 ,

y

84 12-4-8-12

4

8

12

-4

-8

-12

16 20-16-20

16

-16

20

xK

M

L

Given the coordinates of one endpoint of KL are K(-1,-5) and its midpoint M(7, 4). What are the coordinates of the other endpoint L. Graph them.

STANDARD 17


Documents

1 PROBLEM 2A PROBLEM 3A PROBLEM 2B PROBLEM 3B PROBLEM 4 PROBLEM 9BPROBLEM 9A PROBLEM 5 LINES STANDARDS 13, 17 SEGMENT ADDITION POSTULATE MIDPOINT PROBLEM