## Presentation on theme: "SEGMENT ADDITION POSTULATE"— Presentation transcript:

LINES PLANES STANDARDS 13, 17 SEGMENT ADDITION POSTULATE PROBLEM 1A PROBLEM 1B MIDPOINT PROBLEM 2A PROBLEM 2B COORDINATE GEOMETRY PROBLEM 3A PROBLEM 3B DISTANCE FORMULA PROBLEM 4 PROBLEM 5 PROBLEM 6 PYTHAGOREAN THEOREM PROBLEM 7 PROBLEM 8 MIDPOINT FORMULA PROBLEM 9A PROBLEM 9B PROBLEM 10A PROBLEM 10B END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

LINES STANDARD 13 Line AB B p A or Line p How do we call this line? H

LINES STANDARD 13 F E m D C How many different ways can we call
Line m? Line CD Line DC Line CE Line EC OR Line CF Line FC Line DE Can you figure out other names? Line DF Line EF IN GENERAL A LINE IS NAMED BY TWO POINTS. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SEGMENTS STANDARD 13 If we have line DE
and we take one part of the line D E PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SEGMENTS STANDARD 13 If we have line DE
and we take one part of the line D E D E then this part is called: LINE SEGMENT DE Can you name the different line segments in the following line: S T U ST TS SEGMENTS: OR SU US TU UT PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

RAYS STANDARD 13 B A This is RAY AB
What are the differences between a Line, a Line Segment and a Ray? M N The line is infinite and never ends at either side. D E 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. A B What do they have in common? They are named using TWO POINTS. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

COLLINEAR VS NONCOLLINEAR
STANDARD 13 COLLINEAR VS NONCOLLINEAR Points C, D, E and F: are they COLLINEAR? C F D E Yes, they are COLLINEAR because they lie in the same line Are points A, B, C, and D collinear? A C B D No, they are NONCOLLINEAR because they don’t lie in the same line. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 13 Can you explain where the following two lines intersect? l m A B C D E They INTERSECT at point E. NO Can they intersect at other point? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

PLANES STANDARD 13 A M This is: D B PLANE M PLANE ADC C PLANE BDC
PLANE ABD In general a Plane can be named using three non-collinear points. 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? The wall of a house The lid of a shoebox. The ground of a football field. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 13 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

P Q STANDARD 13 A L D K M B H C 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? 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

P Q STANDARD 13 A D B C On what planes does point C lie?

P Q STANDARD 13 A D B C On what planes does point C lies?

P Q STANDARD 13 A D B C On what planes does point D lie?

P Q STANDARD 13 A D B C On what planes does point D lie?

P Q STANDARD 13 A D B C k 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SUMMARIZING FINDINGS:
STANDARD 13 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SEGMENT ADDITION POSTULATE. If B is between A and C then AB + BC = AC.
STANDARD 17 A B C AC AB BC + = 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 17 Find the length for AB and BC if AC = 60 and AB = 4x + 6 and BC= 6x (B is between A and C) AC AB BC Applying Segment Addition Postulate: A B C + = Now finding AB and BC: + = AB BC AC 4x +6 + 6x + 14 = 60 AB = 4x + 6 BC = 6x + 14 4x + 6x = 60 = 4( ) + 6 4 = 6( ) + 14 4 10x + 20 = 60 = = = 22 = 38 10x = 40 Verifying the solution: x = 4 = 60 60 = 60 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 17 Find the length of EF and FG if EG = 80 and EF = 3x + 8 and FG= 7x (F is between E and G) EG EF FG Applying Segment Addition Postulate: E F G + = Now finding EF and FG: + = EF FG EG 3x +8 + 7x + 12 = 80 EF = 3x + 8 FG = 7x + 12 3x + 7x = 80 = 3( ) + 8 6 = 7( ) + 12 6 10x + 20 = 80 = = = 26 = 54 10x = 60 Verifying the solution: x = 6 = 80 80 = 80 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

and we place point B at the same
STANDARD 17 MIDPOINT OF A SEGMENT: A C B and we place point B at the same distance from point A than from Point C, then: If we have segment AC, Means congruent AB BC AB = BC 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 17 AC is bisected by ED. AB= 6X + 8 and BC=4X Find the length for AC. A B C D E AB BC Now finding AB: Applying Segment Addition Postulate: AB = BC AB = 6X + 8 AC = AB + BC 6X + 8 = 4X + 18 = 6( ) + 8 5 AC = AB + BC AC = = 6X = 4X + 10 AC = 76 -4X -4X = 38 2X = 10 Since AB = BC BC = 38 X = 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 17 RT is bisected by VU. RS= 8X + 4 and ST=4X Find the length for RT. R S T V U RS ST Now finding RS: Applying Segment Addition Postulate: RS = ST RS = 8X + 4 RT = RS + ST 8X + 4 = 4X + 28 = 8( ) + 4 6 RT = RS + ST RT = = 8X = 4X + 24 RT = 104 -4X -4X = 52 4X = 24 Since RS = ST ST = 52 X = 6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

CARTESIAN COORDINATE PLANE
STANDARD 17 CARTESIAN COORDINATE PLANE y-axis 10 Quadrant II 8 Quadrant I 6 4 Origin 2 -10 -8 -6 -4 -2 O 2 4 6 8 10 x-axis -2 -4 Quadrant III Quadrant IV -6 -8 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

CARTESIAN COORDINATE PLANE
STANDARD 17 CARTESIAN COORDINATE PLANE y-axis 10 8 y-coordinate 6 4 (9, 4) 2 x-coordinate -10 -8 -6 -4 -2 O 2 4 6 8 10 x-axis -2 -4 -6 -8 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

CARTESIAN COORDINATE PLANE
STANDARD 17 CARTESIAN COORDINATE PLANE y-axis 10 8 6 4 2 -10 -8 -6 -4 -2 O 2 4 6 8 10 x-axis -2 -4 y-coordinate -6 -8 (10,-8) x-coordinate PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

CARTESIAN COORDINATE PLANE
STANDARD 17 CARTESIAN COORDINATE PLANE y-axis 10 8 6 4 2 y-coordinate -10 -8 -6 -4 -2 O 2 4 6 8 10 x-axis -2 (-9,-3) -4 x-coordinate -6 -8 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

DISTANCE FORMULA in a number line is given by:
STANDARD 17 DISTANCE FORMULA in a number line is given by: |a – b| E D H -6 -4 -2 2 4 6 8 10 12 Find measure of DE, and EH: DE = |-2 – (-6)| EH = |12 – (-2)| = |-2 + 6| = |12 + 2| = |4| = |14| = 4 = 14 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

DISTANCE FORMULA in a number line is given by:
STANDARD 17 DISTANCE FORMULA in a number line is given by: |a – b| Q R T -6 -4 -2 2 4 6 8 10 12 Find measure of RT, and QT: RT = |12 – (-6)| QT = |12 – 6| = |12 + 6| = |6| = |18| = 6 = 18 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Distance Formula between two points in a plane:
STANDARD 17 Distance Formula between two points in a plane: d = (x –x ) + (y –y ) 2 1 4 2 6 -2 -4 -6 8 10 -8 -10 x y y 1 x , y 2 x , PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Find the distance between points at A(2, 1) and B(6,4).
STANDARD 17 d = (x –x ) + (y –y ) 2 1 Remember: x 2 y 2 x 1 y 1 Find the distance between points at A(2, 1) and B(6,4). 1 2 3 4 5 6 7 8 9 10 x y AB= ( - ) + ( - ) 2 2 6 1 4 AB= ( -4 ) + ( -3 ) 2 = B = 25 A AB=5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Find the distance between (-3,-5) and (2,-6). y x =(-3,-5)
STANDARD 17 Find the distance between (-3,-5) and (2,-6). y 1 x =(-3,-5) 4 2 6 -2 -4 -6 8 10 -8 -10 x y y 2 x =(2,-6) d = (x –x ) + (y –y ) 2 1 d = ( ) + ( ) 2 2 -3 -6 -5 = ( ) + ( ) 2 3 -6 5 = ( 5 ) + ( -1 ) 2 = d= 26 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

We use the distance formula: Solving this absolute value equation:
STANDARD 17 -6 x 2 a x 1 2 y 10 y 1 Find the value of a, so that the distance between (-6,2) and (a,10) be 10 units. We use the distance formula: Solving this absolute value equation: d = (x –x ) + (y –y ) 2 1 6 = |-6-a| 10 = ( ) + ( ) 2 6 = -(-6-a) 6 = -6-a 6 = 6 + a 2 10 = (-6-a) + (-8) 2 (-1) (-1) 12 = -a 100 = (-6-a) + 64 2 a = -12 a = 0 Check: 6 = |-6-a| 36 = (-6-a) 2 6 =|-6- ( )| 6 =|-6- ( )| -12 6 = |-6-a| 6 =|-6+12| 6 = |-6| 6 =|6| 6 = 6 6 = 6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Right triangle parts: STANDARD 15 hypotenuse LEG Right Angle = 90° LEG

Pythagorean Theorem: x y y + z = x z STANDARD 15
2 z The square of the hypotenuse is equal to the sum of the square of the legs. The Pythagorean Theorem applies ONLY to RIGHT TRIANGLES! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Find the value for x: y + z = x x= ? y= 4 4 + 3 = x 16 + 9 = x 25 = x
STANDARD 15 Find the value for x: y + z = x 2 x= ? y= 4 = x 2 = x 2 25 = x 2 z= 3 25 = x 2 |x|= 5 x= 5 x= -5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Find the value for z: y= 8 y + z = x z= ? x= 10 8 + z = 10
STANDARD 15 Find the value for z: y= 8 y + z = x 2 z= ? x= 10 z = 10 2 2 64 + z = 100 z = 36 2 z = 36 2 |z|= 6 z= 6 z= -6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

MIDPOINT FORMULA in a number line is given by:
STANDARD 17 MIDPOINT FORMULA in a number line is given by: a + b 2 R T -6 -4 -2 2 4 6 8 10 12 Find Midpoint Q of RT: Q 2 = 6 2 = 3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

MIDPOINT FORMULA in a number line is given by:
STANDARD 17 MIDPOINT FORMULA in a number line is given by: a + b 2 K L -4 -2 2 4 6 8 10 12 14 Find Midpoint R of KL: R 2 = 10 2 = 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Midpoint of a Line Segment:
STANDARD 17 Midpoint of a Line Segment: If a line segment has endpoints at and , then the midpoint of the line segment has coordinates: y 1 x 2 x 1 2 , + y 1 2 + = x, y 2 1 3 -1 -2 -3 4 5 -4 -5 x y , (x,y) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 17 x , + y + = Using: x, y x x 9 y 6 y 8
2 , + y 1 2 + = Using: x, y x 1 x 2 9 y 1 6 y 2 8 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 10 x y = , 2 + (9,8) x, y (5,7) (1,6) = y x, 14 2 10 , = y x, 7 5, PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARD 17 x , + y + = Using: x, y x 4 x 8 y 7 y 9
2 , + y 1 2 + = Using: x, y x 1 4 x 2 8 y 1 7 y 2 9 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 10 x y (8,9) = , 2 + (6,8) x, y (4,7) = y x, 16 2 12 , = y x, 8 6, PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Using the Midpoint Formula: L
STANDARD 17 x 1 -2 y 1 -6 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. 8, x, 5 y y 8 4 12 -4 -8 -12 16 20 -16 -20 x Using the Midpoint Formula: L y x, = x 1 2 , + M = , 2 + x 2 y 2 K 8= 2 + -2 x 5= 2 + -6 y (2) (2) (2) (2) 16 =-2 + x 2 10 =-6 + y 2 y 2 =16 x 2 =18 = 16 18, y 2 x , PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Using the Midpoint Formula:
STANDARD 17 x 1 -1 y 1 -5 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. 7, x, 4 y y 8 4 12 -4 -8 -12 16 20 -16 -20 x Using the Midpoint Formula: y x, = x 1 2 , + L M = , 2 + x 2 y 2 K 7= 2 + -1 x 4= 2 + -5 y (2) (2) (2) (2) 14 =-1 + x 2 8 =-5 + y 2 y 2 =13 x 2 =15 = 13 15, y 2 x , PRESENTATION CREATED BY SIMON PEREZ. All rights reserved