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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 10BPROBLEM 10A PLANES PROBLEM 1A PROBLEM 1B DISTANCE FORMULA PROBLEM 6 MIDPOINT FORMULA COORDINATE GEOMETRY END SHOW PROBLEM 7PROBLEM 8 PYTHAGOREAN THEOREM PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 4 A B Line AB p 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 5 C F 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 6 D E SEGMENTS If we have line DEand we take one part of the line STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 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 TU ST SU TU TS US UT SEGMENTS: OR STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 8 RAYS A B 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 9 C F 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 B D COLLINEAR VS NONCOLLINEAR STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 10 Can you explain where the following two lines intersect? l m A B C D E They INTERSECT at point E. Can they intersect at other point? NO STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 11 PLANES M A C D B 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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 L K 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14 14 P Q A B C D On what planes does point C lie? STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15 15 P Q A B C On what planes does point C lies? On planes P and Q. D STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16 16 On what planes does point D lie? P Q A B C D STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17 17 On what planes does point D lie? It only lies on plane Q. P Q A C B D STANDARD 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21 21 A B C + = 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) + = AB BC AC AB BC AC 4x x + 14 = 60 4x + 6x = 60 10x + 20 = x = x = 4 Applying Segment Addition Postulate: Now finding AB and BC: AB = 4x + 6 BC = 6x + 14 = 4( ) + 6 = 6( ) = = 22 = = 38 Verifying the solution: = = 60 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

22 22 E F G + = 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) + = EF FG EG EF FG EG 3x x + 12 = 80 3x + 7x = 80 10x + 20 = x = x = 6 Applying Segment Addition Postulate: Now finding EF and FG: EF = 3x + 8 FG = 7x + 12 = 3( ) + 8 = 7( ) = = 26 = = 54 Verifying the solution: = = 80 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

23 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

24 24 STANDARD 17 A B C D E AC is bisected by ED. AB= 6X + 8 and BC=4X Find the length for AC. AB BC AB = BC 6X + 8 = 4X X = 4X X 2X = 10 2 X = 5 Now finding AB: AB = 6X + 8 = 6( ) = = 38 Since AB = BC BC = 38 Applying Segment Addition Postulate: AC = AB + BC AC = AC = 76 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

25 25 STANDARD 17 R S T V U RT is bisected by VU. RS= 8X + 4 and ST=4X Find the length for RT. RS ST RS = ST 8X + 4 = 4X X = 4X X 4X = 24 4 X = 6 Now finding RS: RS = 8X + 4 = 8( ) = = 52 Since RS = ST ST = 52 Applying Segment Addition Postulate: RT = RS + ST RT = RT = 104 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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

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

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

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

30 30 STANDARD 17 DISTANCE FORMULA in a number line is given by: |a – b| E D H Find measure of DE, and EH: DE =|-2 – (-6)| = |-2 + 6| = |4| = 4 EH =|12 – (-2)| = |12 + 2| = |14| = 14 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

31 31 STANDARD 17 DISTANCE FORMULA in a number line is given by: |a – b| Q R T Find measure of RT, and QT: RT =|12 – (-6)| = |12 + 6| = |18| = 18 QT =|12 – 6| = |6| = 6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

32 32 Distance Formula between two points in a plane: d = (x –x ) + (y –y ) x y y 1 x 1 y 2 x 2,, STANDARD 17 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

33 33 Find the distance between points at A(2, 1) and B(6,4). y 1 y 2 x 1 x 2 AB= ( - ) + ( - ) 22 AB= ( -4 ) + ( -3 ) 22 = = 25 AB= x y B A STANDARD 17 d = (x –x ) + (y –y ) Remember: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

34 34 d = ( - ) + ( - ) 22 = ( 5 ) + ( -1 ) 22 = d = (x –x ) + (y –y ) y 1 x 1 y 2 x 2 =(-3,-5) =(2,-6) d= 26 = ( + ) + ( + ) x y Find the distance between (-3,-5) and (2,-6). STANDARD 17 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

35 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 ) = ( - ) + ( - ) y 1 2 y 2 a x 1 -6 x 2 10 = (-6-a) + (-8) = (-6-a) = (-6-a) 2 6 = |-6-a| 6 = -(-6-a) 6 = -6-a 6 = 6 + a -6 a = 0 +6 a = = |-6-a| Check: 6 =|-6- ( )| = |-6| 6 = 6 6 =|-6+12| 6 =|6| 6 = 6 6 = |-6-a| Solving this absolute value equation: 12 = -a (-1) STANDARD 17 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

36 36 STANDARD 15 Right Angle = 90° LEG hypotenuse Right triangle parts: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

37 37 STANDARD 15 Pythagorean Theorem: x y z y + z = x 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

38 38 STANDARD 15 Find the value for x: x= ? y= 4 z= = x = x 2 25 = x 2 2 |x|= 5 x= 5 x= -5 y + z = x PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

39 39 STANDARD 15 x= 10 y= 8 z= ? 8 + z = z = z = 36 2 |z|= 6 z= 6 z= -6 z = 36 2 Find the value for z: y + z = x PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

40 40 STANDARD 17 R T Find Midpoint Q of RT: MIDPOINT FORMULA in a number line is given by: a + b = 6 2 = 3 Q PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

41 41 STANDARD 17 K L Find Midpoint R of KL: MIDPOINT FORMULA in a number line is given by: a + b = 10 2 = 5 R PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

42 42 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 1 y 2 x 2 y x, = x 1 x 2, 2 + y 1 y x y y 1 x 1, y 2 x 2, (x,y)(x,y) STANDARD 17 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

43 43 Find the midpoint of the line segment that connects points (1,6) and (9,8). Show it graphically x y (1,6)(1,6) (9,8)(9,8) (5,7)(5,7) y x, =, x 1 1 x 2 9 y 1 6 y 2 8 = y , = y x, 7 5,5, y = x 1 x 2, 2 + y 1 y Using: STANDARD 17 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

44 44 Find the midpoint of the line segment that connects points (4,7) and (8,9). Show it graphically x y (4,7)(4,7) (8,9)(8,9) (6,8)(6,8) y x, =, x 1 4 x 2 8 y 1 7 y 2 9 = y , = y x, 8 6,6, y = x 1 x 2, 2 + y 1 y Using: STANDARD 17 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

45 45 y x, = x 1 x 2, 2 + y 1 y =, x 1 -2 y y 8, x, Using the Midpoint Formula: x 2 y 2 5= y 2 8= x 2 (2) 10 =-6 + y 2 16 =-2 + x x 2 =18 y 2 =16 = 16 18, y 2 x 2, y x K 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

46 46 y x, = x 1 x 2, 2 + y 1 y =, x 1 y y 7, x, Using the Midpoint Formula: x 2 y 2 4= y 2 7= 2 + x 2 (2) 8 =-5 + y 2 14 =-1 + x x 2 =15 y 2 =13 = 13 15, y 2 x 2, y x K 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


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