Section 2.1 – Linear Equations in Two Variables
Match the graph with the appropriate slope:
A) No Slope (undefined)
B) Zero Slope
C) Positive Slope
D) Negative Slope
1
2
3
4
C
D
B
A
Based upon each graph below, identify the slope as positive,
negative, zero, or no slope/undefined.
3
4
2
1
positive
zero
undefined
negative
5
6
7
8
zero
undefined
negative
positive
Comment on the equation of the line given the graph:
1
2
3
4
y = mx + b
m is positive
y = mx + b
m is negative
y=k
m is zero
x=k
m is undefined
Comment on the equation of the line given the graph below:
3
4
2
1
y = mx +b
m is positive
y=k
m is zero
5
6
y=k
m is zero
y = mx + b
x=k
m is undefined m is negative
7
8
y = mx + b
y = mx +b
x=k
m is undefined m is negative m is positive
y1  y 2
m
x1  x 2
a) Find the slope of the line given the two points
b) Comment briefly on the graph of the line connecting the
two points. (up to the right, down to the right, vertical, or
horizontal)
 3, 6 ,  2, 7 
(2, 0), (8, 12)
76
1
m

2   3  5
12  0
m
2
82
graph is up to right
graph is up to right
 5, 9 , 5, 12
m
12  9
 und.
55
Vertical; x = 5
y1  y 2
m
x1  x 2
a) Find the slope of the line given the two points
b) Comment briefly on the graph of the line connecting the
two points. (up to the right, down to the right, vertical, or
horizontal)
 2,  3 ,  4,  3 
3   3 
m
0
42
Horizontal
y = -3
(-1, 5), (2, 4)
45
1
m

2   1 3
graph is down
to right
 5, 9 , 5, 12
m
12  9
 und.
55
Vertical
x=5
ALL Equations – Point Slope Form y  y1  m  x  x1 
Find the equation of the line which passes through (2, 3) and (3, 5)
53
m
2
32
y  3  2  x  2
or
y  5  2  x  3
Find the equation of the line which passes through (3, 0) and (3, 3)
30
x3
m
 und
33
Find the equation of the line which passes through (6, 7) and (2, 7)
77
m
0
26
y3
Find the equation of the line with slope 5 passing through (3, -1)
y  1  5  x  3
Find the equation of the line passing through (2, 3) and (7, 5)
2
y  3   x  2
5
53 2
or
m

72 5
2
y  5   x  7
5
Find the equation of the line passing through (4, 6) and (4, -1)
m
6   1
44
 und
x4
Parallel Lines – Same Slope m1  m2
1
Normal Lines – Negative/Reciprocal Slopes m1 
m2
Determine if the lines connecting the two points below are parallel,
perpendicular, or neither.
 2, 3 ,  2, 4 
1, 5 , 7, 3 
43
1
m

2  2 4
m
5  3 2 1


1 7
6
3
neither
Determine if the lines connecting the two points below are parallel,
perpendicular, or neither.
 6, 2,  9, 3 
1, 11,  3, 5 
32 1
5  11 6
m

m

 3
96 3
3 1
2
perpendicular
Find the equation of the line parallel to 2x – 5y = -3 which
passes through (3, 1).
2x  5y  3
5y  2x  3
2
y x
5
2
y  1   x  3
5
Rewrite your equation in Ax + By = C form.
2
y  1   x  3
5
5  y  1  2  x  3 
5y  5  2x  6
1  2x  5y
2x  5y  1
Find the equation of the line perpendicular to 7x – y = 4 which
passes through (2, -5).
7x  y  4
 y  7x  4
y  7x  4
1
y  5   x  2
7
Rewrite your equation in Ax + By = C form.
1
y  5   x  2
7
7  y  5   1 x  2
7y  35  x  2
x  7y  33
Your salary was $28500 in 1998 and $32900 in 2000. If your
salary follows a linear growth pattern, what will your salary
be in 2003?
32900  28500
m
 2200
2000  1998
y  28500  2200  x  1998 
y  28500  2200  2003  1998 
y  39500
A business purchases a piece of equipment for $875. After 5
years the equipment will be outdated and have no value. Write
a linear equation giving the value V of the equipment during the
5 years it will be used.
(0, 875)
and
(5, 0)
875  0
m
 175
05
y  875  175  x  0 
A contractor purchases a piece of equipment for $36500. The
equipment requires an average expenditure of $5.25 per hour
for fuel and maintenance, and the operator is paid $11.50
per hour.
a) Write a linear equation giving the total cost C of operating
this equipment for t hours.
C  36500  5.25t  11.50t
C  16.75t  36500
b) Assuming that customers are charged $27 per hour of
machine use, an equation which represents the profit.
P  27t  16.75t  36500 
P RC
c) Find the ‘break-even’ point.
0  27t  16.75t  36500 
t  3561