Intermediate Algebra – 1
HFCC Math Lab
SLOPE – INTERCEPT AND POINT-SLOPE FORMS OF THE LINE
THE EQUATION OF A LINE
Goal I.
Use the slope-intercept form of the line to write the equation of a non-vertical line
A. given its y-intercept and slope
B. given its y-intercept and another point on the line
Goal II.
Write the equation of a horizontal line through a given point.
Goal III.
Use the point-slope form of the line to write the equation of a non vertical line
A. given its slope and a point on the line
B. given two points on the line
Goal IV.
Write the equation of a vertical line through a given point.
Goal V.
Given the equation of a line, find the slope and the y-intercept.
TERMINOLOGY
m = slope of the line
b = the y-intercept (where the line crosses the y-axis)
( x1 , y1 ), (x2 , y2 ) two given points on the line ( may be the intercept points)
I.
SLOPE-INTERCEPT FORM OF A LINE: The equation of the line with slope m
and y-intercept b is given by y = mx + b
Examples:
A.
1. Find the equation of the line with slope 5 and y-intercept 7.
m = 5, b = 7; y = mx + b
Therefore, y = 5x + 7 is the equation of the line.
2.
Find the equation of the line with slope
m = 32 , b = 0; y = mx + b
Therefore,
y
2
3
x 0
y
2
3
x is the equation of the line.
Revised 11/09
1
2
and y-intercept 0.
3
B.
1. Find the equation of the line with y-intercept 6 and passing through the point
(1, 4).
First, we need to find the slope of the line, m:
The line passes through the points (0, 6) and (1, 4). We can use the slope formula
y2 y1
m
where ( x1 , y1 ) (0, 6) and (x 2 , y2 ) (1, 4)
x2 x1
6 4
m
2
0 1
Therefore, m = -2 and b = 6
y = mx + b
y = -2x + 6.
2.
Find the equation of the line with y-intercept -3 and passing through the point
(4. -3).
First, we need to find the slope of the line, m:
y2 y1
3 ( 3) 0
m
= 0 where (x1 , y1 ) (4, 3) & (x2 , y2 ) (0, 3)
x2 x1
0 4
4
Therefore, m = 0 and b = -3
y = mx + b
y = 0x + -3 or y = -3.
II.
EQUATION OF HORIZONTAL LINE: The slope of a horizontal line is zero.
y1 .
The equation of a horizontal line through a point ( ( x1 , y1 ) is y
Examples:
1.
Find the equation of the horizontal line passing through the point ( -2, -5).
y = -5
2.
Find the equation of the line having slope of 0 and y-intercept of 9.
Slope of 0 implies the line is horizontal.
( x1 , y1 ) (0,9) so the equation is y = 9.
3.
Find the equation of the line passing through the points (2, 5) and (-2, 5).
The slope of the line is
m
y2
x2
y1
x1
5 5
1 2
0
= 0 where (x1 , y1 ) (2,5) & (x2 , y2 ) ( 1,5)
3
Therefore, the line is horizontal and has an equation y = 5.
Revised 11/09
2
III.
POINT-SLOPE FORM OF A LINE: The equation of the line with slope m and
passing through the point ( x1 , y1 ) is given by the equation y y1 m( x x1 ) .
Examples:
A. 1. Find the equation of the line with slope -6 and passing through the point (2, 5).
m
y
6; ( x1 , y1 ) (2,5)
y1
m( x x1 )
y 5
6( x 2)
[You can rewrite the equation in slope-intercept form by solving for y]
y - 5 -6 x 12
y 5 5
6 x 12 5
y
6 x 17.
2. Find the equation of the line with slope 2 and passing through the point (-3, -1).
m 2; ( x1 , y1 ) ( 3, 1)
y
y1
m( x x1 )
y ( 1) 2( x ( 3)) or y 1 2( x 3)
[You can rewrite the equation in slope-intercept form by solving for y]
y 1 2x 6
y 1 31 2 x 6 1
y 2 x 5.
B. 1. Find the equation of the line passing through the point (6, 3) and (-2, 7)
First, we need to find the slope m:
y2 y1
7 3
4
1
m
= - where (x1 , y1 ) (6,3) & (x2 , y2 ) ( 2, 7)
x2 x1
2 6
8
2
m
y
1
; ( x1 , y1 ) (6,3)
2
y1 m( x x1 )
1
2
y 3
or y
Revised 11/09
-1
2
( x 6)
x 6 in slope-intercept form.
3
2.
Find the equation of the line passing through the point (3, 5) and (3, -4)
First, we need to find the slope m:
y2 y1
4 5
9
m
undefined where (x1 , y1 ) (3,5) & (x2 , y2 ) (3, 4)
x2 x1
3 3
0
m is undefined, the line is vertical. Therefore, the equation of the line is x = 3.
IV.
EQUATION OF A VERTICAL LINE: The slope of a vertical line is undefined.
x1
The equation of a vertical line passing through the point ( x1 , y1 ) is x
Examples:
1. Find the equation of the vertical line passing through the point (-7, 6).
The equation of the line is x = -7.
2. Find the equation of the line with undefined slop and x-intercept 8.
( x1 , y1 ) = (8, 0). Therefore, the equation of the line is x = 8.
3. Find the equation of the line passing through the points (4, 3) and (4, -2).
First, we need to find the slope m:
y2 y1
2 3
5
m
undefined where (x1 , y1 ) (4,3) & (x2 , y2 ) (4, 2)
x2 x1
4 4
0
Therefore, the equation of the vertical line is x = 4.
V.
Given the equation of a non-vertical line, you can find the slope and y-intercept by
rewriting the equation in slope-intercept form (y = mx+b) and then reading the values
of m and b.
Examples:
1. Find the slope and y-intercept of the line 2x + y = 5.
First, we need to solve for y:
2x y 5
2x 2x y
2 x 5 [subtract 2x from both sides]
y
2x 5
y mx b, m = -2 and b = 5
Therefore, the slope is -2 and the y-intercept (0, 5).
2. Find the slope and y-intercept of the line 3x +6y = 2.
First, we need to solve for y:
Revised 11/09
4
3x 6 y 2
3x 3x 6 y
3 x 2 [subtract 3x from both sides]
6y
3x 2
6y
3x 2
[divide both sides by 6]
6
6
6
-1
1
y
x
2
3
y mx b, m = - 12 and b = 13
Therefore, the slope is
1
1
and the y-intercept is (0, ).
2
3
EXERCISES
A. Write the equation of the line described in each of the exercises below. Write your
answer in slope-intercept form if possible.
2
and y-intercept of -3.
5
1.
slope
2.
slope 4 and y-intercept
3.
y- intercept of 7 and passing through the point (4, 3).
4.
y- intercept of 0 and passing through the point (1, 5).
5.
y-intercept of -1 and x-intercept 4.
6.
a horizontal line with y-intercept 9.
7.
slope 3 and passing through the point (-5, 2).
8.
slope of -1 and passing through the point (3, -1).
9.
slope of
1
.
2
1
and x-intercept of 8.
4
10. a vertical line with x-intercept 15.
Revised 11/09
5
11. passing through the points (-3, 5) and (6, -1).
12. passing through (-4, 3) and (-6, 7).
13. passing through the points (-5, 0) and (6, 0).
14. x-intercept 5 and passing through the point (3, 12).
15. passing through the point (9, -1) and parallel to the line with slope
1
.
3
16. parallel to the y-axis and passing through the point (-2, 4).
17. Passing through the point (2, 1) and perpendicular to the line whose slope is
B. Find the slope and the y-intercept of the given line.
18. 8x 2 y 3
19.
3x 6 y 10
20. 4 y 8
21. y
2
x
3
22. y
2
x
23. x 5
ANSWERS AND SOLUTIONS
2
x 3.
5
1.
y
2.
y
3.
y
4.
y 5x 0 or y 5 x since the slope m
4x
Revised 11/09
1
.
2
x 7 since the slope m
6
7 3
0 4
4
4
5 0
1 0
1.
5
1
5.
1
2
5.
y
1
4
6.
y
9
7.
y 2 3( x 5) or y
8.
y 1
1( x 3)
9.
y 0
1
( x 8)
4
x 1 since the slope m
since y
10. x 15
y1
0 ( 1)
4 0
1
.
4
9.
3x 17 (in slope-intercept form) .
or
- x 2.
y
1
x 2 we used (x1 , y1 ) (8, 0) .
4
or y
since x = x1 15 .
11. y 5
2
( x 3) or y
3
2
x+3 since m
3
12. y 3
2( x 4)
y
2x 5 .
13. y
since the slope m
0 0
6 ( 5)
0
or
1 5
6 ( 3)
6
9
2
.
3
0
0 , and passes through the point (-5, 0).
11
14. First, we find the slope m:
12 0 12
m
= = -6 where ( x1 , y1 ) (5, 0) and ( x2 , y2 ) (3, 12)
3 5 -2
y - y1 m( x x1 )
y 0
y
6( x 5)
6 x 30
or y 13 x 4 ( m = 13 )
15. y 1 13 ( x 9)
N.B. If two lines have the same slope, you can conclude that the two lines are
parallel and vice versa.
16. Since the line is parallel to the y-axis, then it is a vertical line.
The equation is x = -2.
17. m 4, b =
-3
since
2
2y
8 x 3 and by solving for y, we obtain y
4x
3
2
.
18. Note here that two lines are perpendicular if the product of their slopes is -1.
Therefore if the slope of the given line is ½ then the slope of the line in question must
1
be -2 since 2
1
2
Revised 11/09
7
The line passes through the point (2, 1) with slope of -2 has equation
y 1 2( x 2)
y 1 2x 4
y
2x 5
19. m
1
, b
2
5
.
3
20. m
1
, b =-2 since 4 y
4
21. m
2
, b =0 since y
3
22. m 0, b =-2 since y
x 8 and by solving for y, we obtain y
1
4
x 2.
2
x 0 which is y = mx + b.
3
2 0 x 2 (this is a horizontal line) .
23. m is undefined (this is a vertical line) and no y-intercept since the line is parallel to
the y-axis.
***************************************************************
NOTE: You can get additional instruction and practice by going to the
following websites:
http://www.purplemath.com/modules/strtlneq.htm
Many examples explaining in details how to write an equation of a line in
slope-intercept form.
http://www.analyzemath.com/line/Tutorials.html
This is a tutorial on how to find the slopes and equations of lines. A review of
the main results concerning lines and slopes and then examples with detailed
solutions are presented.
Revised 11/09
8