
Distance
Presentation
•
Mathematics
•
University
•
Practice Problem
•
Easy
Azi Saban
Used 2+ times
FREE Resource
13 Slides • 6 Questions
1
Slide - 7
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Objective 1 Use the Distance Formula
2
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 8
Example 1: Finding the Distance
Between Two Points (1 of 2)
Find the distance d between the points (2, 1) and (6, 6).
Solution:
First plot the points (2, 1) and
(6, 6) and connect them with a
straight line.
To find the length d, begin by
drawing a horizontal line from
(2, 1) to (6, 1) and a vertical
line from (6, 1) to (6, 6), forming
a right triangle, as shown.
3
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 9
Example 1: Finding the Distance
Between Two Points (2 of 2)
One leg of the triangle is of length 4
6
2 = 4)
(since
−
and the other is of length 5
6
1 = 5).
(since
−
By the Pythagorean Theorem,
the square of the distance d that
we seek is
2
2
2
= 4 + 5 = 16 + 25 = 41
d
=
41
6.40
d
4
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 10
Theorem 1
Distance Formula
The distance between two points 1
1
1
)
P
x , y
and
2
2
2
1
2
)
( ,
),
,
, denoted by
is
P
x y
d P P
2
2
1 2
2
1
2
= (
) =
(
) + (
)
,
−
−
(1)
d P P
x
x
y
y
Illustration of the
Distance Formula
= (
= (
5
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 11
Example 2: Finding the Length
of a Line Segment
Find the length of the line segment between the points
1
2
( –1, 4)
(5, 3).
P
P
and
Solution:
The length of the line segment is the distance between
the points
1
1
1
2
2
2
( ,
)
( 1, 4)
(
,
)
(5, 3).
P
x y
P
x y
=
= −
=
=
and
Using the distance formula
with
1
1
2
1,
4,
5,
x
y
x
= −
=
=
and
2
3,=y
the length d is
2
2
2
1
2
1
2
2
2
2
=
(
) + (
)
=
(5
( 1)) + (3
4)
=
6 + ( 1) =
36 +1
=
37
6.08.
d
x
x
y
y
−
−
− −
−
−
6
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 12
Example 3: Using Algebra to Solve
Geometry Problems (1 of 5)
Consider the three points A = (−4, 2), B = (1, 2),
and C = (0, 4).
(a) Plot each point and form the triangle ABC.
(b) Find the length of each side of the triangle.
(c) Verify that the triangle is a right triangle.
(d) Find the area of the triangle.
7
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 13
Example 3: Using Algebra to Solve
Geometry Problems (2 of 5)
Solution:
(a) The figure shows points A, B, C, and triangle
ABC.
8
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 14
Example 3: Using Algebra to Solve
Geometry Problems (3 of 5)
(b) To find the length of each side of the triangle,
use the distance formula.
2
2
( , )
[1 ( 4)]
(2
2)
25
0
25
5
d A B =
− −
+
−
=
+
=
=
2
2
( , )
(0 1)
(4
2)
1
4
5
d B C =
−
+
−
=
+
=
2
2
( , )
[0
( 4)]
(4
2)
16
4
20
2 5
d A C =
− −
+
−
=
+
=
=
9
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 15
Example 3: Using Algebra to Solve
Geometry Problems (4 of 5)
(c) If the sum of the squares of the lengths of two of
the sides equals the square of the length of the third
side, the triangle is a right triangle. Looking at the
figure, we conjecture that the angle at vertex C might
be a right angle.
2
2
2
( , )
( , )
( , )
d A C
d B C
d A B
+
=
2
2
2
2 5
5
5
+
=
20
5
25
+
=
Triangle ABC is
a right triangle.
10
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 16
Example 3: Using Algebra to Solve
Geometry Problems (5 of 5)
(d) Because the right angle is at vertex C, the sides
AC and BC form the base and height of the triangle.
Its area is
(
)(
)
1
1
Area
(Base)(Height)
2 5
5
5 square units
2
2
=
=
=
11
Slide - 17
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Objective 2 Use the Midpoint Formula
12
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 18
Theorem 2
Midpoint Formula
The midpoint M = (x, y) of the line segment from
(
)
(
)
1
1
1
2
2
2
=
,
=
,
to
is
P
x y
P
x
y
(
)
1
2
1
2
+
+
=
,
=
,
2
2
(2)
x
x
y
y
M
x y
13
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Slide - 19
Example 4: Finding the Midpoint
of a Line Segment
Find the midpoint of a line segment from
1
2
( 6,2)
(4, 6).
and
−
P
P
Solution:
Apply the midpoint formula using
1
1
6,
2,
= −
=
x
y
2
2
4,
6.
and
=
=
x
y
Then the coordinates of the midpoint M are
1
2
6
4
1
2
2
+
− +
=
=
= −
x
x
x
and
1
2
2
6
4
2
2
y
y
y
+
+
=
=
=
That is,
(
)
= –1, 4 .
M
14
Open Ended
Find the distance between P1 ad P2
15
Open Ended
The midpoint of the line segment from P1 to P2 is (- 1, 4). If P1 = (- 3, 6), what is P2?
16
Open Ended
A major league baseball “diamond” is actually a square 90 feet on a side (see the figure). What is the distance directly from home plate to second base (the diagonal of the square)?
17
Open Ended
Refer to the major league problem, overlay a rectangular coordinate system on a major league baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement.
18
Open Ended
Refer to previous question, If the center fielder is located at (310, 15), how far is it from the center fielder to third base?
19
Open Ended
Refer to previous question, If the center fielder is located at (300, 300), how far is it from the center fielder to third base?
Slide - 7
Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved
Objective 1 Use the Distance Formula
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