rectangular coordinate system

1. Let A be the point (2, 3).
Suppose its x–coordinate is
increased by 4 to (2 + 4, 3)
= (6, 3) – to the point B,
this corresponds to
moving A to the right by 4.
Rectangular Coordinate System
A B
Similarly if the x–coordinate is
decreased by 4 to (2 – 4, 3)
= (–2, 3) – to the point C,
Hence we conclude that changes in the x–coordinates
correspond to moving the point right and left.
If the x–change is +, the point moves to the right.
If the x–change is – , the point moves to the left.
C
x–coord.
increased
by 4
x–coord.
decreased
by 4
(2, 3) (6, 3)(–2, 3)
this corresponds to moving A to the left by 4.

2. Again let A be the point (2, 3).
If the y–coordinate is
increased by 4 to (2, 3 + 4)
= (2, 7) – to the point D,
this corresponds to
moving A up by 4.
Rectangular Coordinate System
A
D
If the y–coordinate is
decreased by 4 to (2, 3 – 4)
= (2, –1) – to the point E,
Hence we conclude that changes in the y–coordinates
correspond to moving the point up and down.
If the y–change is +, the point moves up.
If the y–change is – , the point moves down.
E
y–coord.
increased
by 4
y–coord.
decreased
by 4
(2, 3)
(2, 7)
(2, –1)
this corresponds to moving A down by 4.

3. Let (x1, y1) and (x2, y2) be two points and
D = the distance between them, then D2 = Δx2 + Δy2,
where Δx = difference in the x's = x2 – x1,
Δy = difference in the y's = y2 – y1,
Hence D = Δx2 + Δy2
or
D = (x2 – x1)2+(y2 – y1)2
Example A. Find the distance
between (–1, 3) and (2, –4).
(–1, 3)
– ( 2, –4)
–3, 7
D = (–3)2 + 72
= 58  7.62
DD
7
-3
The Distance Formula
(2, –4)
(–1, 3)
Δx Δy

4. The Mid-Point Formula
The mid-point m between two numbers a and b is the
average of them, that is m = .a + b
2
For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3.
In picture: a b(a+b)/2
mid-pt.
The mid-point formula
extends to higher
dimensions.
In the x&y coordinate
the mid-point of
(x1, y1) and (x2, y2) is
x1 + x2
2 ,(
y1 + y2
2
)
In 2D
(x1, y1)
(x2, y2)
x1
y1
y2
x2
(x1 + x2)/2
(y1 + y2)/2

5. A. Find the coordinates of the following points.
Sketch both points for each problem.
Rectangular Coordinate System
1. Point A that is 3 units to the left and 6 units down
from (–2, 5).
2. Point A that is 1 unit to the right and 5 units up
from (–3, 1).
3. a. Point B is 3 units to the left and 6 units up from
point A(–8, 4). Find the coordinate of point B.
b. Point A(–8, 4) is 3 units to the left and 6 units up from
point C, find the coordinate of point C
4. a. Point A is 37 units to the right and 63 units down from
point B(–38, 49), find the coordinate of point A.
b. Point A(–38, 49) is 37 units to the right and 63 units down
from point C, find the coordinate of point C.

6. Linear Equations and Lines
C. Find the coordinates of the following points assuming
all points are evenly spaced.
1.
1 4
2.
–1 5
1 3 11
3. a. Find x and y.
x zy
The number z is a “weighted average” of {1, 3, 11}
whose average is 5. In this case z is the average of
{1, 3, 3,11} instead because “3” is used both for
calculating x and y.
1 3 11
b. Find z the mid-point of x and y.
x y
Find all the locations of the points in the figures.
(–4, 7)
(2, 3) (0, 0) (8, 0)
(2, 6)4. 5.

7. (Answers to odd problems) Exercise A.
1. B=(-5,-1) 3. B=(-11,10), C=(-5,-2)
Rectangular Coordinate System

8. 1. x – y = 3 3. –y – 7= 0 5. y = –x + 4
Exercise B.
x y
0 -3
3 0
x y
0 4
4 0y=7
Linear Equations and Lines

9. 7. 2 = 6 – 2y 9. 2x + 3y = 0 11. 3x = 4y
x y
0 0
1 -2/3
x y
0 0
1 3/4
y=4
Linear Equations and Lines

10. x y
0 -6
1 0
x y
0 0
1 5/2
13. 3(2 – x) = 3x – y 15. 5(x + 2) – 2y = 10
Linear Equations and Lines

11. Exercise C.
1.
1 4
1 3 11
3. a.
2 4.57
1 3 11
b.
2 7
1.75 2.5 3.25
(0, 0) (8, 0)
(2, 6)5.
(4, 0)
(1, 3)
(6, 6)
(10, 6)
(9, 3)
(5, 3)
(3.5, 4.5) (7.5, 4.5)
(6.5, 1.5)(2.5, 1.5)
Linear Equations and Lines