Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Notes
The Distance Formula
d
 x2  x1    y2  y1 
2
2
The Midpoint Formula
 x1  x2 y1  y2 
,


2
2


Vertical and Horizontal Lines
A vertical line is of the form x = constant.
A horizontal line is of the form y = constant.
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Example #1
Find the coordinates of the points.
1) A
(-4, 1)
2) B
(-2, -1)
3) C
(0, 0)
4) D
(0, 3)
5) E
(4, 2)
6) G
(4, -2)
7) H
(1, 0)
8) I
(0, -3)
9) J
(-1, 0)
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Example #2
Describe the set of all point P(x, y) in a coordinate plane that
satisfy the given conditions.
2) x = 4
Remember x = constant is a vertical line.
One way of answering
A vertical line through the point (4, 0).
The book’s way of answering is
A line parallel to the y-axis that intersects the x-axis at (4, 0).
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
You Try #1
Describe the set of all point P(x, y) in a coordinate plane that
satisfy the given conditions.
1) y = -5
A)
B)
C)
D)
A vertical line through the point (-5, 0).
A horizontal line through the point (-5, 0).
A vertical line through the point (0, -5).
A horizontal line through the point (0, -5).
Remember y = constant is a horizontal line.
The answer choices will appear after you have had time
to try to work the problem.
A horizontal line through the point (0, -5).
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Example #3
Describe the set of all point P(x, y) in a coordinate plane that
satisfy the given conditions.
3) y > 2
Remember y = constant is a horizontal line.
So y is greater than a horizontal line through the point (0, 2).
Greater than a horizontal line would be above the line,
but not including the line.
So all the points strictly above the horizontal line through the
point (0, 2).
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Example #4
Find the distance d(A, B) between A and B. Leave the answer
in exact terms.
4) A(-5, 6), B(8, -3)
d
 x2  x1    y2  y1 
d
8  5   3  6 
d
13   9 
2
2
2
d  169  81
d  250
d  5 10
2
2
2
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
You Try #2
Find the distance d(A, B) between A and B. Leave the answer
in exact terms.
2) A(-4, -8), B(12, 5)
A)
73
B)
425
C) 5 17
d
 x2  x1    y2  y1 
2
d
12  4    5  8
2
d
16   13
2
2
2
time
D)The
73answer choices will appear after you have
d had256
 169
to try to work the problem. c
d  425
d  5 17
2
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Example #5
Find the midpoint of segment AB.
5) A(-5, 6), B(8, -3)
 x1  x2 y1  y2 
,


2 
 2
 5  8 6  3 
,


2
2


3 3
 , 
2 2
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
You Try #3
Find the midpoint of segment AB.
3) A(-4, -8), B(12, 5)
 3
A)  4, 
 2
 3 
B)  4, 
 2 
C)
 x1  x2 y1  y2 
,


2 
 2
 4  12 8  5 
,


2 
 2
8 3had

 time
The
answer
choices
will
appear
after
you
have
 3 
,


4 work the problem. c
to
 try,to
2 2 
 2 
3 
D)  , 4 
2 
 3 
 4, 
 2 
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
Example #6
Show that the triangle with vertices A, B, and C is a right
triangle, and find its area.
5) A(-3, 2), B(-1, 0), C(2, 3)
To show this is a right triangle,
we find the length of each
segment and see if it satisfies
the Pythagorean Theorem.
AB 
 1  3   0  2 
BC 
 2  1   3  0 
2
AC 
 2  3   3  2 
2
2
2
2
 2 2   3 2  ? 
2
2
26

2
2

 2    2 
2

 3   3

 5  1
2
2
2
2
Area 
2
2 2
3 2
 26
 8  18 ? 26  26  26
bh
2
2 2  3 2 

A
2
62
A
2
A6
Precalculus Functions & Graphs
2.1 Rectangular Coordinate System
You Try #4
Show that the triangle with vertices A, B, and C is a right
triangle, and find its area.
10  10 

A
5) A(-1, 1), B(2, 0), C(1, -3)
2
To show this is a right triangle,
we find the length of each
segment and see if it satisfies
the Pythagorean Theorem.
AB 
2
 2  1   0  1
2
BC 
1  2    3  0 
2
AC 
1  1   3  1
2
2
 10    10  ?  2 5 
2
2
2
2
10
A
2
A5

 3   1

 1   3

2
2
2
 2    4 
2
 10
2
 10
2
 20  2 5
 10  10? 20  20  20