1. SLOPE a conceptual approach
Standards 2, 5, 25
SLOPE a conceptual approach
SLOPE FORMULA & examples
SLOPE FORMULA problems
SLOPE FORMULA VS. run and rise variation problems
Falling to the right or to the left?
SUMMARIZING SLOPE CASES
PERPERDICULAR OR PARALLEL LINES?
PERPENDICULAR SEGMENTS
END SHOW
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2. ALGEBRA II STANDARDS THIS LESSON AIMS:
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.
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3. Find the slope for the segments at the right:
Standard 5
Slope Formula: m= y 1 2 x - 1 2 3 4 5 6 7 8 9 10 x y
Find the slope for the segments at the right:
Segment AB:
A(3,2), B(7,5) B x 2 y 2 x 1 y 1
m = - AB 2 5 = -3 -4 = 3 4 3 7 A C D
Segment BC:
C(8,2), D(10,1) x 2 y 2 x 1 y 1
m = - BC 2 1 = 1 -2 8 10
The slope is the inclination of a line or segment.
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4. Using the slope formula below, find the slope for the lines in the coordinate plane:
Standard 5 m= y 1 2 x -
Line a:
(1,5) (-1,1) x 2 1 y 2 5 y 1 x 1 -1
m = - a 4 = 4 2 4 2 6 -2 -4 -6 8 10 -8
-10 x y = =2 d
1+1 a
(-8, 7)
Line b:
(3,2) (10,2)
(1, 5) x 2 3 y 2 x 1 10 y 1 2
m = - b = -7 b
(-1, 1) =0
(3, 2)
(10, 2)
Line c:
(6,-8) (2,-5)
(2, -5)
(-4, -4) y 2 -8 x 2 6 y 1 -5 x 1 2 c
m = - c
-8+5 = -3 4
(6, -8)
= - 3 4 = 4
Line d:
(-8,7) (-4,-4) y 2 7 x 2 -8 y 1 -4 x 1 -4
m = - d
7+4 = 11 -4
= - 11 4 =
-8+4
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5. Using the run vs. the rise method, find the slope for the lines in the coordinate plane:
Standard 5 m= y 1 2 x - + -
rise =
run +
Line a: - -4 = 4 2 4 2 6 -2 -4 -6 8 10 -8
-10 x y
m = a =2 d -2 a 4 + 2 -
Line b: - 11 - 4
m = b =0 +7 b 7 +
Line c: +3
= - 3 4
m = c c -4 3 + 4 -
Line d:
-11
= - 11 4
m = d +4
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6. + + + + Standard 5 FALLING TO THE RIGHT y - - - x - c Slope for c: +
m = c
= - -
When we move to the right from one point to the other, we go down the right and the slope is negative!
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7. - - - - FALLING TO THE LEFT Standard 5 x y a - - - - Slope for a: -
m = a =+ -
When we move to the left from one point to the other, we go down and the slope is always positive!
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8. - + - Standard 5 SUMMARIZING FINDINGS y y a b - x x Slope for b:-
m = b =0
m = a =+ + -
Slope of a horizontal line is always 0!
Slope of a line that falls to the left is always POSITIVE! x y x y d c + + -
Slope for c:
Slope for d: +
m = c
= - + -
m = d
= undefined!
The slope of a line that falls to the right is always NEGATIVE!
The slope of a vertical line is UNDEFINED!
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9. + + - + PARALLEL VS. PERPENDICULAR Standard 2 x y PARALLEL4 2 6 -2 -4 -6 8 10 -8
-10 x y
PARALLEL
PERPENDICULAR 4 2 6 -2 -4 -6 8 10 -8
-10 x y a c b 2 + d 2 + 2 + 2 + 2 + 2 + 2 - 2 + +2 +2
Line a:
m = a =1
Line c:
m = c
=-1
m = d m c
(-1)(1)
=-1 +2 -2 +2 +2
Line b:
m = b =1
Line d:
m = d =1 +2 +2
The slope product of perpendicular lines is -1!
Parallel Lines have the same slope!
m = a m b
m = d m c -1
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10. Prove that segments AB and BC are perpendicular:
Standard 2
Prove that segments AB and BC are perpendicular:
Using the slope formula: 1 2 3 4 5 6 7 8 9 10 x y m= y 1 2 x -
Segment AB:
A(2,1), B(6,4) B x 2 y 2 x 1 y 1
m = - AB 1 4 = -3 -4 = 3 4 2 6 A C
Segment BC:
B(6,4), C(9,0) x 2 y 2 x 1 y 1
m = - BC 4 = 4 -3 6 9
Is the product of the slopes -1?
The product of the slopes is -1. So, They are perpendicular. 3 4 = 4 -3 12
-12
= -1
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