Transcript X 2

Distance Formula
Goal: to find the length of a line
segment using the distance formula.
Method 1




Whenever the segments
are horizontal or
vertical, the length can
be obtained by counting.
When we need to find the
length (distance) of a
segment such as we simply
COUNT the distance from
point A to point B.
(AB = 7)
We can use this same
counting approach for .
(CD = 3)
Unfortunately, this counting
approach does NOT work
for
which is a diagonal
segment.
Method 2

In the last slide, we
found the distance
between two
numbers on a
number line. But
what happens if the
points are not on a
straight line like
points A and B
above?
Method 2
• When working with
diagonal segments, the
Pythagorean theorem
can be used to
determine the length.
Method 2
•Draw a right triangle using
points E and F as corners.
•When you have a right
triangle you can use the
Pythagorean theorem to find
the distance.
•c2=a2+b2 (Solve for c)
c  a b
2
2
•Find the distance
between points E and F
using the Pythagorean
theorem.
Work:
a b c
2
•Use this method of
working with the
Pythagorean Theorem
whenever you forget
the Distance Formula!
2
3 4  x
2
9  16  x
2
25  x
2
•The distance between
E and F is 5 units.
2
2
2
25  x
25  x
5x
2
Method 3

When working with diagonal
segments, use the Distance
Formula to determine the length.

Now let’s use the Pythagorean theorem to develop
the distance formula.

The distance between
the points E = (x1, y1)
and F = (x2, y2) is
given by the formula:
d
( x2  x1 )2  ( y2  y1 )2
Length of
Hypotenuse
Length of
Side 1 (a2)
Length of
Side 2(b2)
Distance Formula:
E
b: Length of
Side 2
hypotenuse/
distance
F
a: Length of Side 1
d  ( x2  x1 )  ( y2  y1 )
2
Length of
Hypotenuse
Length of
Side 1 (a2)
Length of
Side 2(b2)
2
IMPORTANT!
Note: x2 is not the same as x2!!!!!!
 X2 means x - point number 2 and x2
means the square of x.
Now use the distance formula to find the distance for
the first example:



Work:
Coordinates of E: (1, -1)
Coordinates of F: (4, -5)
d  ( x2  x1 )2  ( y2  y1 )2
d  (1  4)  (1  (5))
2
d  (3)  (4)
d  9  16
2
d  25
d 5
2
2
Distance Formula:



It doesn't matter which point you start
with. Just start with the same point for
reading both the x and y coordinates
The Distance Formula can be used to find
the lengths of all forms of line
segments: horizontal, vertical and
diagonal.
The advantage of the Distance Formula is
that you do not need to draw a picture to
find the answer. All you need to know
are the coordinates of the endpoints of
the segment.
Use the distance formula to find the
distance between the following points.

1. (1,6) and (5,1)
d  (1  5) 2  (6  1) 2
d  (4)  (5)
2
d  16  25
d  41
d  6.4
2
Use the distance formula to find the
distance between the following points.

2. (-5,2) and (3, -1)
d  (5  3) 2  (2  (1)) 2
d  (8)  (3)
2
d  64  9
d  73
d  8.5
2
The expression represents the distance
formula. What are the original points?
d  (0  11)2  ( 3  2)2
x2
x1
y2
y1
Coordinates (x2, y2) and (x1, y1)
(0, -3) and (11, 2)
Midpoint
Formula
The point halfway between the endpoints of a line
segment is called the midpoint.
A midpoint divides a line segment into two equal
parts.
In Coordinate Geometry, there are several ways to
determine the midpoint of a line segment.
Method 1:If the line segments are vertical or
horizontal, you may find the midpoint by simply
dividing the length of the segment by 2 and counting
that value from either of the endpoints.
You are asking yourself “What point is halfway between the two?”
Example:

Find the midpoints
of line segments
AB and CD.
•The length of line
segment AB is 8
(by counting). The
midpoint is 4 units
from either
endpoint. On the
graph, this point is
(1,4).
The
length of line
segment CD is 3 (by
counting). The
midpoint is 1.5 units
from either
endpoint. On the
graph, this point is
(2,1.5)
However, …


If the line segments are diagonal, more
thought must be paid to the
solution. When you are finding the
coordinates of the midpoint of a segment,
you are actually finding the average
(mean) of the x-coordinates and the
average (mean) of the y-coordinates.
This concept of finding the average of the
coordinates can be written as a formula:
The Midpoint Formula:

The midpoint of a segment with endpoints
(x1 , y1) and (x2 , y2) has coordinates:
 x1  x2 y1  y2 
,


2 
 2
"The Midpoint Formula"
sung to the tune of "The Itsy Bitsy Spider"
by Halyna Reynolds
Surf City, NJ
When finding the midpoint of two points on a graph,
Add the two x's and cut their sum in half.
Add up the y's and divide 'em by a two,
Now write 'em as an ordered pair
You've got the middle of the two.
Example:
Find the midpoint of line segment AB.
A(-3,4)
B(2,1)
Find the midpoint of the two points using the
Midpoint formula.
1. (-2, 4) and (3, 4)
2. (8, -3) and (5, -4)
 x1  x2 y1  y2 
,


2 
 2
 x1  x2 y1  y2 
,


2 
 2
 1 8 
 , 
 2 2
 0.5,4
 8  5  3  4 
,


2 
 2
 13  7 
 ,

2 2 
6.5,3.5

Example 3:
Southwestern Telephone Company uses GPS to map the
locations of its telephone poles. IT is determined that an
additional pole is needed exactly halfway between 2 poles
located at coordinates (20, 35) and (40, 15). What are the
coordinates of the location of the new pole?
 x1  x2 y1  y2   20  40 35  15 
,
,


 
2 
2   2
 2
 60 50 
 , 
 2 2
30,25
4. The expression represents the midpoint
formula. What are the original points?
 5  1 0  2 
,


2 
 2
x2
x1
y2
opposite
sin A 
hypotenuse
y1
Coordinates (x2, y2) and (x1, y1)
(-1, 2) and (5, 0)
Trigonometry: For Right Triangles Only!
leg
hypotenuse - always opposite
the right angle
leg
Basic Trigonometry Rules:





With right triangles you can use three
special ratios to solve problems.
These ratios ONLY work in a right
triangle.
The hypotenuse is across from the right
angle.
Questions usually ask for an answer to
the nearest units.
You need a scientific or graphing
calculator.
Definitions:
Leg a is opposite <A
and adjacent to <B.
A
c
b
C
Leg b is opposite <B
and adjacent to <A.
Hypotenuse
a
B
Sine

The sine (sin) of
an acute angle of
a right triangle is
the ratio that
compares the
length of the leg
opposite the
acute angle to the
length of the
hypotenuse.
A
Hypotenuse
c
b
C
a
a
sin A 
c
b
sin B 
c
B
Cosine

The cosine (cos)
of an acute angle
of a right triangle
is the ratio that
compares the
length of the leg
adjacent the
acute angle to the
length of the
hypotenuse.
A
Hypotenuse
c
b
C
a
a
cos B 
c
B
Tangent

The tangent (tan)
of an acute angle of
a right triangle is
the ratio that
compares the
length of the leg
opposite the acute
angle to the length
of the leg adjacent
the acute angle .
A
Hypotenuse
c
b
C
a
a
tan A 
b
b
tan B 
a
B
Formulas: Soh Cah Toa

A represents the
angle of reference
A
Hypotenuse
Leg
Adjacent to A
Leg
Opposite A
Remember: 



The formulas can be remembered
by:
soh cah toa
The formulas can be remembered
by:
oscar had a heap of apples
The formulas can be remembered
by:
oh heck, another hour of algebra!
Examples:
Find the sine, cosine and tangent ratios for both acute
angles of the following right triangle.
o
8
4
sin A  

h 10 5
a
3
6
cos A  

h 10
5
4
8
o


tan A 
3
a
6
B
10
8
A
6
C
Examples:
Find the sine, cosine and tangent ratios
for both acute angles of the following
right triangle.
B
o
6
3
sin B  

h 10
5
8
4
a

cos B  
h 10 5
o 6
3
tan B  

a 8
4
10
8
A
6
C
Using Trig Ratio Tables/Calculator
Find the following values to the nearest
ten-thousandth:
0.6691
•sin 42°
cos
30°
•tan 27°
•sin 73°
cos 36°
 tan 81°
0.8660
0.5095
0.9563
0.8090
6.3138
Applications:



In order to determine how an object is
grown, you will need to determine the
height of the object. You can use
trigonometry, or the study of
triangles, to find the height of an
object.
The tangent function can help find the
height of objects.
To determine the height of the
flagpole, set up a triangle with one
side being the height of the flagpole
(a), another side being a distance
from the flagpole to a point on the
ground (b), and the third side being
the distance from that point to the top
of the flagpole (c). Assume the
flagpole meets the ground at a right
angle.
Solution:


You have two options to solve this
problem:
1. Use inverse operations:
measure _ opposite
tan A 
measure _ adjacent
a
tan 45 
8
a
8  tan 45   8
8
8 1  a
The height of the flagpole is
8 meters high.
8a
Method 2:

Use a proportion.
tan A 
measure _ opposite
measure _ adjacent
tan 45 a

1
8
The height of the flagpole is
8 meters high.
8  tan 45  a
a  8  tan 45
a  8 1
a 8
Example 2:
In right triangle ABC, hypotenuse AB=15 and angle A=35º.
Find leg BC to the nearest tenth.
measure _ opposite
sinA 
measure _ hypotenuse
sin 35 x

1
15
x  15  sin35
x  15  0.5736
x  8.6
Example 3:
In right triangle ABC, leg BC=20 and angle B = 41º.
Find the hypotenuse BA to the nearest hundredth
measure _ adjacent
cos A 
measure _ hypotenuse
cos 41 20

1
x
x  cos 41  20
x  cos 41
20

cos 41
cos 41
20
x
cos 41
20
x
0.7547
x  26.50
Example 4:

A ladder 6 feet long leans against a
wall and makes an angle of 71º with
the ground. Find to the nearest
tenth of a foot how high up the wall
the ladder will reach.
B
measure _ opposite
sinA 
measure _ hypotenuse
sin 71 x

1
6
x  6  sin 71
x  15  0.5736
71
6
A
x  8.6
x
C
The person in the drawing is
using a hypsometer.





A hypsometer is an instrument you can use
to find the height of very tall objects.
How is this example different from the
one in our example 1?
How will that affect the answer?
What will you have to do to compensate
for this difference?
Will your answer be the same?
Using a Hypsometer:

Hold the hypsometer up
and look through the
straw at the top of a tall
object. The imaginary
line that goes from your
eye to the top of the
object is called the line of
sight. Once you have
found your line of sight,
look at the point at which
the string crosses the
semicircle. That is the
measure of the angle
from your line of site and
an imaginary horizontal
line.
Work
x
measure _ opposite
tan A 
measure _ adjacent
tan 45 x

1
8
x  8  tan 45
x  8 1
x 8
Now a = x + 2
A = 10 m
Height of
person
from toes
to eye
level.
Another Example
h
measure _ opposite
tan A 
measure _ adjacent
tan 60 h

1
10
h  10  tan 60
h  10 1.732
h  17.32
Now h = x + 2
A = 19.32 m
Height of
person
from toes
to eye
level.
Activity: You are now going to use your hypsometer to
measure tall objects. Complete the data table on the back





To make hypsometer, you
will need a straw, some
string, a paper clip, and
index cared and a
protractor (paper).
Glue the semi-circle paper
protractor) to the index
card and cut it out.
Attach the straw to the
semicircle, with the
middle of the straw at the
dot labeled A.
Attach a piece of string 6
inches long to the center
of the straw.
Attach a paper clip to the
other end of the string.
Make sure that the paper
clip hangs freely.
Date Table:
Length of your foot (in feet):
Your height from foot to eye (in feet):
Object
Angle
Ground Distance from
Object
Tree
Building
Light Pole
Your Choice
Now use your data to calculate the height of each object. Show ALL work!