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1 Standards 2, 5, 25 SLOPE a conceptual approach SLOPE FORMULA & examples SLOPE FORMULA VS. run and rise variation problems Falling to the right or to the left? SLOPE FORMULA problems SUMMARIZING SLOPE CASES PERPERDICULAR OR PARALLEL LINES? PERPENDICULAR SEGMENTS END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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2 Standard 2: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. Estándar 2: Los estudiantes resuelven sistemas de ecuaciones lineares y desigualdades (en 2 o tres variables) por substitución, con gráficas o con matrices. Standard 5: Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane. Estándar 5: Los estudiantes demuestran conocimiento de como los números reales y complejos se relacionan tanto aritméticamente como gráficamente. En particular, ellos grafican números como puntos en el plano. Standard 25: Students use properties from number systems to justify steps in combining and simplifying functions. Estándar 25: Los estudiantes usan propiedades de sistemas numéricos para justificar pasos en combinar y simplificar funciones. ALGEBRA II STANDARDS THIS LESSON AIMS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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3 Standard 5 m= y 1 y 2 x 1 x Segment AB:A(3,2), B(7,5) y 1 y 2 x 1 x = = 3 4 Segment BC:C(8,2), D(10,1) y 1 y 2 x 1 x = 1 -2 m = - - AB m = - - BC Slope Formula: Find the slope for the segments at the right: x y A B C D The slope is the inclination of a line or segment. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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4 Standard 5 m= y 1 y 2 x 1 x x y Using the slope formula below, find the slope for the lines in the coordinate plane: a b c d (-8, 7) (-4, -4) (-1, 1) (1, 5) (3, 2) (10, 2) (2, -5) (6, -8) (1,5) (-1,1) x 2 1 x 1 y 2 5 y 1 1 = = 4 2 m = - - a Line a:a: = =2 (3,2) (10,2) x 2 3 x 1 10 y 2 2 y 1 2 = 0 -7 m = - - b Line b:b: =0 (6,-8) (2,-5) x 2 6 x 1 2 y 2 -8 y 1 -5 = -3 4 m = - - c Line c:c: = = (-8,7) (-4,-4) x 2 -8 x 1 -4 y 2 7 y 1 = m = - - d Line d:d: = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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5 Standard 5 m= y 1 y 2 x 1 x x y Using the run vs. the rise method, find the slope for the lines in the coordinate plane: a b c d = = m = a = m = b = m = c = m = d Line a:a: b:b: c:c: d:d: = rise run PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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6 Standard 5 x y c + - FALLING TO THE RIGHT When we move to the right from one point to the other, we go down the right and the slope is negative! + - m = c Slope for c:c: = - PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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7 Standard 5 x y a m = a =+=+ Slope for a:a: FALLING TO THE LEFT When we move to the left from one point to the other, we go down and the slope is always positive! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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8 Standard 5 x y b m = b =0=0 Slope for b:b: Slope of a horizontal line is always 0! x y a m = a =+=+ Slope for a:a: Slope of a line that falls to the left is always POSITIVE! x y c m = c Slope for c:c: = - The slope of a line that falls to the right is always NEGATIVE! x y + 0 m = d Slope for d:d: = undefined! The slope of a vertical line is UNDEFINED! SUMMARIZING FINDINGS d + PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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9 Standard x y Parallel Lines have the same slope! +2 m = a =1 +2 m = b =1 a b PARALLEL VS. PERPENDICULAR x y c d m = c =-1 +2 m = d =1 The slope product of perpendicular lines is -1! m = d m c (-1)(1) m = a m b d m c PARALLEL PERPENDICULAR =-1 Line a:a: b:b: c:c: d:d: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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10 Standard 2 m= y 1 y 2 x 1 x Segment AB:A(2,1), B(6,4) y 1 y 2 x 1 x = = 3 4 Segment BC:B(6,4), C(9,0) y 1 y 2 x 1 x = = = -1 m = - - AB m = - - BC Using the slope formula: Prove that segments AB and BC are perpendicular: x y A B C Is the product of the slopes -1? The product of the slopes is -1. So, They are perpendicular. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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