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Chapter 1
Linear Equations
and Graphs
Section 2
Graphs and Lines
Table of Content





Cartesian Coordinate System
Graph of Ax + By = C
Slope of a Line
Equations of Lines: Special Forms
Applications
It will be considered one of the most basic geometric figures – a line.
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Learning Objectives for Section 1.2
Graphs and Lines
 The student will be able to identify and work with the
Cartesian coordinate system.
 The student will be able to draw graphs for equations of the
form Ax + By = C.
 The student will be able to calculate the slope of a line.
 The student will be able to graph special forms of equations
of lines.
 The student will be able to solve applications of linear
equations.
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Terms








cartesian coordinate system
horizontal axis
vertical axis
coordinate axes
x axis
y axis
quadrants
ordered pair
Barnett/Ziegler/Byleen College Mathematics 12e








abscisa
ordinate
x coodinate
y coodinate
origin
slope of a line
slope-intercept form
pont-slope form
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The Cartesian Coordinate System
 The Cartesian coordinate system was named after René
Descartes. It consists of two real number lines, the
horizontal axis (x-axis) and the vertical axis (y-axis)
which meet in a right angle at a point called the origin.
The two number lines divide the plane into four areas
called quadrants.
 The quadrants are numbered using Roman numerals as
shown on the next slide. Each point in the plane
corresponds to one and only one ordered pair of numbers
(x,y). Two ordered pairs are shown.
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The Cartesian Coordinate System
(continued)
II
I
(3,1)
x
III
(–1,–1)
Two points, (–1,–1)
and (3,1), are plotted.
Four quadrants are as
labeled.
IV
y
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The Cartesian Coordinate System
(continued)
y
Abscisa
10
II
I
b
Q = (-5,5)
-10
5
P = (a,b)
Coordenadas
Origen
5
-5
Ordenada
a
x
10
Ejes
-5
IV
III
-10
R = (10,-10)
Sistema de Coordenadas Cartesianas (Rectangulares)
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The Cartesian Coordinate System
(continued)
 If we define an arbirtrary point P, the coordinates are the
horizontal and vertical lines through this point.
 The intersections are at a point with coordinate a and b.
 Those two numbers, written as the ordered pair (a, b)
form the coordinates of the point P.
 They are called the abscisa and the ordinate of P.
 The coordinates of a point are also referenced in terms of
the axis labels: x coordinate and y coordinates.
 The point with coordinates (0,0) is called the origin.
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The Cartesian Coordinate System
(continued)
 The procedure just described assigns to each point P a
unique pair of real numbers (a, b).
 Conversely, if we are given an ordered pair of real numbers
(a, b), then, reversing this procedure, we can determine a
unique point P in the plane.
 Thus
There is a one-to-one correspondance between the points
in a plane and the elements in the set of all ordered pairs of
real numbers.
 This is often referred as the
fundamental theorem of analytic geometric.
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Linear Equations in Two Variables
Graphs of Ax + By = C
 A linear equation in two variables is an equation that can
be written in the standard form Ax + By = C, where A, B,
and C are constants (A and B not both 0), and x and y are
variables.
 A solution of an equation in two variables is an ordered
pair of real numbers that satisfy the equation. For example,
(4,3) is a solution of 3x - 2y = 6.
 The solution set of an equation in two variables is the set
of all solutions of the equation.
 The graph of an equation is the graph of its solution set.
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Linear Equations in Two Variables
Graphs of Ax + By = C
 Theorem 1 – Graph of a Linear Equation in Two
Variables.
The graph of any equation of the form
Ax + By = C
(A and B not both 0)
is a line, and any line in a Cartesian coordinate system is
the graph of an equation of this form.
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Linear Equations in Two Variables
(continued)
 If A is not equal to zero and B is not equal to zero, then
Ax + By = C can be written as
This is known as
slope-intercept form.
A
C
y   x   mx  b
B
B
 If A = 0 and B is not equal to zero,
then the graph is a horizontal line
 If A is not equal to zero and B = 0,
then the graph is a vertical line
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C
y
B
C
x
A
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Using Intercepts to Graph a Line
Graph 2x – 6y = 12.
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Using Intercepts to Graph a Line
Graph 2x – 6y = 12.
x
y
0
–2
y-intercept
6
0
x-intercept
3
–1 check point
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Special Cases
 The graph of x = k is the graph of a vertical line k units from
the y-axis.
 The graph of y = k is the graph of the horizontal line k units
from the x-axis.
 Examples:
1. Graph x = –7
2. Graph y = 3
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Solutions
x = –7
y=4
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Slope of a Line
 Slope of a line:
y2  y1 rise
m

x2  x1 run
 x1 , y1 
rise
 x2 , y2 
run
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Slope of a Line
 For a horizontal line, y does not change; its slope is 0. For a
vertical line, x does not change; x1 = x2 so its slope is not
defined. In general, the slope of a line may be positive,
negative, 0, or not defined. See next table.
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Slope of a Line
TABLA 2 – Interpretación Geométrica de la
Pendiente
Línea
Ascendente a medida
que x se mueve de
izquierda a derecha
Descendente a medida
que x se mueve de
izquierda a derecha
Pendiente
Ejemplo
Positiva
Negativa
Horizontal
0
Vertical
No definida
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Slope of a Line - Example
 Finding Slopes. Sketch a line through each pair of points, and
find the slope of each line.
A) (-3, -2), (3, 4)
B) (-1, 3), (2, -3)
C) (-2, -3), (3, -3)
C) (-2, 4), (-2, -2)
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Slope of a Line - Example
A)
B)
y
x
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y
x
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Slope of a Line - Example
C)
D)
y
y
x
x
Slope is not defined
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Slope-Intercept Form
The equation
y = mx+b
is called the slope-intercept form of an equation of a line.
The letter m represents the slope and b represents the
y-intercept.
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Slope-Intercept Form
The constants a and m have the following interpretations:
 If we let x = 0, then y = b. So the graph of y = mx + b crosses
the y axis at (0,b). The constant b is the y intercept.
 For m, if y = mx + b, then by setting x = 0 and x = 1, we
conclude that (0,b) and (1, m + b) are points in the line. The
slope of such line is:
So m is the slope of the line given by y = mx + b.
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Find the Slope and Intercept
from the Equation of a Line
Example: Find the slope and y intercept of the line
whose equation is 5x – 2y = 10.
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Find the Slope and Intercept
from the Equation of a Line
Example: Find the slope and y intercept of the line
whose equation is 5x – 2y = 10.
Solution: Solve the equation
for y in terms of x. Identify the
coefficient of x as the slope and
the y intercept as the constant
term.
5 x  2 y  10
2 y  5 x  10
5 x 10 5
y

 x5
2 2 2
Therefore: the slope is 5/2 and
the y intercept is –5.
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Write the Equation of the Line
knowing the Slope and the Intercept
Example: Find the equation of the line with slope 2/3
and intercept -2.
Solution:
m = 2/3 and b = -2 so,
Barnett/Ziegler/Byleen College Mathematics 12e
y = 2/3 x - 2
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Point-Slope Form
The point-slope form of the equation of a line is
y  y1  m( x  x1 )
where m is the slope and (x1, y1) is a given point.
It is derived from the definition of the slope of a line:
y2  y1
m
x2  x1
Barnett/Ziegler/Byleen College Mathematics 12e
Cross-multiply and
substitute the more
general x for x2
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Example
Find the equation of the line through the points (–5, 7) and (4, 16).
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Example
Find the equation of the line through the points (–5, 7) and (4, 16).
Solution:
16  7 9
m
 1
4  (5) 9
Now use the point-slope form with m = 1 and (x1, x2) = (4, 16).
(We could just as well have used (–5, 7)).
y  16  1( x  4)
y  x  4  16  x  12
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Summary
Equations of a Line
Standard Form
Ax + By = C
A and B not both 0
Slope-intercept form
y = mx + b
Slope: m; y intercept: b
Point-slope form
y – y1 = m(x – x1)
Slope: m; point: (x1, y1)
Horizontal line
y=b
Slope: 0
Vertical line
x=a
Slope: Undefined
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Application
Office equipment was purchased for $20,000 and will have a
scrap value of $2,000 after 10 years. If its value is depreciated
linearly, find the linear equation that relates value (V) in dollars
to time (t) in years:
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Application
Office equipment was purchased for $20,000 and will have a
scrap value of $2,000 after 10 years. If its value is depreciated
linearly, find the linear equation that relates value (V) in dollars
to time (t) in years:
Solution: When t = 0, V = 20,000 and when t = 10, V = 2,000.
Thus, we have two ordered pairs (0, 20,000) and (10, 2000).
We find the slope of the line using the slope formula.
The y intercept is already known (when t = 0, V = 20,000, so
the y intercept is 20,000).
The slope is (2000 – 20,000)/(10 – 0) = –1,800.
Therefore, our equation is V(t) = –1,800t + 20,000.
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Supply and Demand
 In a free competitive market, the price of a product is
determined by the relationship between supply and
demand. The price tends to stabilize at the point of
intersection of the demand and supply equations.
 This point of intersection is called the equilibrium point.
 The corresponding price is called the equilibrium price.
 The common value of supply and demand is called the
equilibrium quantity.
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Supply and Demand
Example
Use the barley market data in the following table to find:
(a) A linear supply equation of the form p = mx + b
(b) A linear demand equation of the form p = mx + b
(c) The equilibrium point.
Year
Supply
Mil bu
Demand
Mil bu
Price
$/bu
2002
340
270
2.22
2003
370
250
2.72
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Supply and Demand
Example (continued)
(a) To find a supply equation in the form p = mx + b, we must
first find two points of the form (x, p) on the supply line.
From the table, (340, 2.22) and (370, 2.72) are two such
points. The slope of the line is
2.72  2.22 0.5
m

 0.0167
370  340 30
Now use the point-slope form to find the equation of the line:
p – p1 = m(x – x1)
p – 2.22 = 0.0167(x – 340)
p – 2.22 = 0.0167x – 5.678
p = 0.0167x – 3.458
Barnett/Ziegler/Byleen College Mathematics 12e
Price-supply equation.
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Supply and Demand
Example (continued)
(b) From the table, (270, 2.22) and (250, 2.72) are two points
on the demand equation. The slope is
2.72  2.22
.5
m

 0.025
250  270 20
p – p1 = m(x – x1)
p – 2.22 = –0.025(x – 270)
p – 2.22 = –0.025x + 6.75
p = –0.025x + 8.97
Barnett/Ziegler/Byleen College Mathematics 12e
Price-demand equation
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Supply and Demand
Example (continued)
(c) The equilibrium point occurred when the price p is the
same, that is, when p = 0.0167x – 3.458 and p = –0.025x + 8.97
are equal. Then,
0.0167x – 3.458 = –0.0250x + 8.97
0.0167x = –0.0250x + 12.428
adding 3.458 in both sides
0.0417x = 12.428
adding 0.0250x in both sides
x = 12.428/0.0417 = 298 rounded
and p = 1.52
if we substitute x = 298 in any of the two equations
[p = 0.0167(298) – 3.458 = 4.9766 – 3.4580]
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Supply and Demand
Example (continued)
(c) If we graph the two equations on a graphing
calculator, set the window as shown, then use the
intersect operation, we obtain:
The equilibrium point is approximately (298, 1.52). This
means that the common value of supply and demand is
298 million bushels when the price is $1.52.
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Chapter 1
Linear Equations
and Graphs
Section 2
Graphs and Lines
END
Last Update: Juanuary 2013