Geometry Section 3-3 1112

Section 3-3
Slopes of Lines

Monday, December 19, 2011

Essential Questions

How do you ﬁnd slopes of lines?

How do you use slope to identify parallel
and perpendicular lines?


Vocabulary
1. Slope:

2. Rate of Change:


Vocabulary
1. Slope: The ratio of the vertical change
to the horizontal change between two
points

2. Rate of Change:


Vocabulary
points; Change in y over change in x

2. Rate of Change:


Vocabulary
rise over run

2. Rate of Change:


Vocabulary
rise over run

2. Rate of Change: A way to describe slope


Explore
Graph the points A(−2, −4) and B(3, 3). Then
draw both a vertical and horizontal line
through both and determine the distance
between the vertical lines, then between
the horizontal lines.
Distance between vertical

Distance between horizontal


Explore
y the horizontal lines.


x


Explore


x

A


Explore


B
x

A


Explore

5 Units
B
x

A


Explore

5 Units
B
x
7 Units
A


Explore

5 Units
B
x
7 Units
A
Up 7, Right 5

Slope Formula


Slope Formula
y 2 − y1
m=
x 2 − x1
for points
(x 1,y 1),(x 2 ,y 2 )


Example 1
Find the slope of the line that goes
through the following pairs of points.
a. C(−3, 4) and D(8, 1)


Example 1
a. C(−3, 4) and D(8, 1)
(x 1,y 1)


Example 1
a. C(−3, 4) and D(8, 1)
(x 1,y 1) (x 2 ,y 2 )


Example 1
a. C(−3, 4) and D(8, 1)
(x 1,y 1) (x 2 ,y 2 )

y 2 − y1
m=
x 2 − x1


Example 1
a. C(−3, 4) and D(8, 1)
(x 1,y 1) (x 2 ,y 2 )

y 2 − y1 1− 4
m= =
x 2 − x 1 8 − (−3)


Example 1
a. C(−3, 4) and D(8, 1)
(x 1,y 1) (x 2 ,y 2 )

y 2 − y1 1− 4 −3
m= = =
x 2 − x 1 8 − (−3) 11


Example 1
a. C(−3, 4) and D(8, 1)
(x 1,y 1) (x 2 ,y 2 )

y 2 − y1 1− 4 −3
m= = =
x 2 − x 1 8 − (−3) 11

Down 3, Right 11


Example 1
b. E(5, −1) and F(−3, 7)


Example 1
b. E(5, −1) and F(−3, 7)

y 2 − y1
m=
x 2 − x1


Example 1
b. E(5, −1) and F(−3, 7)

y 2 − y 1 7 − (−1)
m= =
x 2 − x 1 −3 − 5


Example 1
b. E(5, −1) and F(−3, 7)

y 2 − y 1 7 − (−1) 8
m= = =
x 2 − x 1 −3 − 5 −8


Example 1
b. E(5, −1) and F(−3, 7)

y 2 − y 1 7 − (−1) 8
m= = = = −1
x 2 − x 1 −3 − 5 −8


Example 1
b. E(5, −1) and F(−3, 7)

y 2 − y 1 7 − (−1) 8
m= = = = −1
x 2 − x 1 −3 − 5 −8

Down 1, Right 1


Example 1
c. G(−1, 2) and H(−1, 7)


Example 1
c. G(−1, 2) and H(−1, 7)

y 2 − y1
m=
x 2 − x1


Example 1
c. G(−1, 2) and H(−1, 7)

y 2 − y1 7−2
m= =
x 2 − x 1 −1− (−1)


Example 1
c. G(−1, 2) and H(−1, 7)

y 2 − y1 7−2 5
m= = =
x 2 − x 1 −1− (−1) 0


Example 1
c. G(−1, 2) and H(−1, 7)

y 2 − y1 7−2 5
m= = =
x 2 − x 1 −1− (−1) 0

Undeﬁned


Example 1
c. G(−1, 2) and H(−1, 7)

y 2 − y1 7−2 5
m= = =
x 2 − x 1 −1− (−1) 0

Undeﬁned

Up 5, Right 0

Example 1
c. G(−1, 2) and H(−1, 7)

y 2 − y1 7−2 5
m= = =
x 2 − x 1 −1− (−1) 0

Undeﬁned

Up 5, Right 0 Vertical Line

Example 1
d. J(3, 4) and K(−2, 4)


Example 1
d. J(3, 4) and K(−2, 4)

y 2 − y1
m=
x 2 − x1


Example 1
d. J(3, 4) and K(−2, 4)

y 2 − y1 4−4
m= =
x 2 − x 1 −2 − 3


Example 1
d. J(3, 4) and K(−2, 4)

y 2 − y1 4−4 0
m= = =
x 2 − x 1 −2 − 3 −5


Example 1
d. J(3, 4) and K(−2, 4)

y 2 − y1 4−4 0
m= = =
x 2 − x 1 −2 − 3 −5

Up 0, Left 5


Example 1
d. J(3, 4) and K(−2, 4)

y 2 − y1 4−4 0
m= = =
x 2 − x 1 −2 − 3 −5

Up 0, Left 5 Horizontal Line


Example 1
d. J(3, 4) and K(−2, 4)

y 2 − y1 4−4 0
m= = = =0
x 2 − x 1 −2 − 3 −5

Up 0, Left 5 Horizontal Line


Example 2
In 2000, the annual sales for one
manufacturer of camping equipment was
$48.9 million. In 2005, the total sales
were $85.9 million. If sales increase at
the same rate, what will the total sales
be in 2015?


Example 2
be in 2015?

85.9 − 48.9 =


Example 2
be in 2015?

85.9 − 48.9 = 37


Example 2
be in 2015?

85.9 − 48.9 = 37
Every 5 years, sales increase by $37 million


Example 2
be in 2015?

85.9 − 48.9 = 37
85.9 + 2(37) =


Example 2
be in 2015?

85.9 − 48.9 = 37
85.9 + 2(37) = 159.9


Example 2
be in 2015?

85.9 − 48.9 = 37
85.9 + 2(37) = 159.9
In 2015, sales should be about $159.9
million

Postulates
Slopes of parallel lines:

Slopes of perpendicular lines:


Postulates
Two lines will be parallel IFF they
have the same slope. All vertical lines
are parallel.


Postulates
Two lines will be parallel IFF they
have the same slope. All vertical lines
are parallel.
Two lines will be perpendicular IFF the
product of their slopes is −1. Vertical
and horizontal lines are perpendicular.


Example 3
Determine whether FG and HJ are
parallel, perpendicular, or neither for
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).
Graph each line to verify your answer.


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y
m(FG ) = 2 1

x 2 − x1


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3)
m(FG ) = 2 1
=
x 2 − x1 −2 − 1


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3) 2
m(FG ) = 2 1
= =
x 2 − x1 −2 − 1 −3


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3) 2
m(FG ) = 2 1
= =
x 2 − x1 −2 − 1 −3

 y −y
m(HJ ) = 2 1

x 2 − x1


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3) 2
m(FG ) = 2 1
= =
x 2 − x1 −2 − 1 −3

 y −y 3−0
m(HJ ) = 2 1
=
x 2 − x1 6 − 5


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3) 2
m(FG ) = 2 1
= =
x 2 − x1 −2 − 1 −3

 y −y 3−0 3
m(HJ ) = 2 1
= =
x 2 − x1 6 − 5 1


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3) 2
m(FG ) = 2 1
= =
x 2 − x1 −2 − 1 −3

 y −y 3−0 3
m(HJ ) = 2 1
= = =3
x 2 − x1 6 − 5 1


Example 3
F(1, −3), G(−2, −1), H(5, 0), and J(6, 3).

 y −y −1− (−3) 2
m(FG ) = 2 1
= =
x 2 − x1 −2 − 1 −3

 y −y 3−0 3
m(HJ ) = 2 1
= = =3
x 2 − x1 6 − 5 1
Neither

Example 4
Graph the line that contains Q(5, 1) and
is parallel to the line through M(−2, 4)
and N(2, 1).
y

x


Example 4
and N(2, 1).
y
y 2 − y1
m=
x 2 − x1

x


Example 4
and N(2, 1).
y
y 2 − y1 4 −1
m= =
x 2 − x 1 −2 − 2

x


Example 4
and N(2, 1).
y
y 2 − y1 4 −1 3
m= = =
x 2 − x 1 −2 − 2 −4

x


Example 4
and N(2, 1).
y
y 2 − y1 4 −1 3
m= = =
x 2 − x 1 −2 − 2 −4
3
m=−
4 x


Example 4
and N(2, 1).
y
y 2 − y1 4 −1 3
m= = =
x 2 − x 1 −2 − 2 −4
3 Q
m=−
4 x


Example 4
and N(2, 1).
y
y 2 − y1 4 −1 3
m= = =
x 2 − x 1 −2 − 2 −4
M
3 Q
m=− N
4 x


Check Your
Understanding

Use problems 1-11 to check the ideas from
this lesson


Problem Set


Problem Set

p. 191 #13-39 odd

“I have found power in the mysteries of
thought.” - Euripides

Geometry Section 3-3 1112

Recommandé

Recommandé

Contenu connexe

En vedette

En vedette (20)

Similaire à Geometry Section 3-3 1112

Similaire à Geometry Section 3-3 1112 (7)

Plus de Jimbo Lamb

Plus de Jimbo Lamb (20)

Dernier

Dernier (20)

Geometry Section 3-3 1112