Download Linear Equations: Slope and Equations of Lines - UH

Section 1.1 – Linear Equations: Slope and Equations of Lines
Slope
The measure of the steepness of a line is called the slope of the line. It is the amount of change in
y, the rise, divided by the amount of change in x, the run.
Let ( x1 , y1 ) and ( x2 , y2 ) be two arbitrary points on the coordinate plane. The
slope of the line that passes through these two points, denoted by m, is given by
m=
y2 − y1 rise
vertical change
=
=
,
x2 − x1 run horizontal change
provided that x2 − x1 ≠ 0 .
• If the slope is positive, the line rises to the right.
• If the slope is negative, the line falls to the right.
• If the slope is zero, the line is a horizontal line.
• If the slope is undefined, the line is a vertical line. (This occurs when x2 − x1 = 0 .)
Example 1: Find the slope of each line shown below.
A.
B.
C.
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Page 1 of 19
Section 1.1
Solution:
A.
To find the slope of the given line, choose two
points on the graph with integer coordinates, like
( 0, − 2 ) and ( 2, 0 ) , and keep in mind that the
slope is “rise over run”. Let us start at ( 0, − 2 ) .
To get to ( 2, 0 ) , we “rise” 2 units, then “run” two
units. Hence, we get m =
B.
2
= 1.
2
To find the slope of the given line, choose two
points on the graph with integer coordinates, like
( 0, 4 ) and ( 2, 0 ) , and keep in mind that the slope
is “rise over run”. Let us start at ( 0, 4 ) . To get to
( 2, 0 ) , we move down four units, then to the right
two units. Hence, we get m =
C.
−4
= −2 .
2
To find the slope of the given line, choose two
points on the graph with integer coordinates, like
( −2, − 1) and (1, − 2 ) , and keep in mind that the
slope is “rise over run”. Let us start at (1, − 2 ) . To
get to ( −2, − 1) , we move up 1 unit, then to the
left 3 units. Hence, we get m =
1
1
=− .
−3
3
***
Example 2: Find the slope of the line that passes through the following points.
A.
( −3, 1) and ( 4, − 5)
B.
( 5, − 10 ) and ( −6, − 7 )
C.
 3

 1 1
 − , − 8  and  − , 
 4

 6 2
D.
( −2, 9 )
E.
5

5 
 , − 3  and  , 1
6

6 
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 4 
and  − , 9 
 5 
Section 1.1
Solution:
A.
Let ( x1 , y1 ) represent ( −3, 1) , and ( x2 , y2 ) represent ( 4, − 5) .
m=
y2 − y1
−5 − 1
−6
6
=
=
=−
x2 − x1 4 − (−3) 7
7
Notice that if we had switched the selection of ( x1 , y1 ) and ( x2 , y2 ) , we would obtain the
same result: Let ( x1 , y1 ) represent ( 4, − 5) , and ( x2 , y2 ) represent ( −3, 1) . Then
m=
B.
Let ( x1 , y1 ) represent ( 5, − 10 ) , and ( x2 , y2 ) represent ( −6, − 7 ) .
m=
C.
y2 − y1 1 − ( −5 ) 6
6
=
=
=−
x2 − x1
−3 − 4 −7
7
y2 − y1 −7 − ( −10 )
3
3
=
=
=−
x2 − x1
−6 − 5
−11
11
 3

 1 1
Let ( x1 , y1 ) represent  − , − 8  , and ( x2 , y2 ) represent  − ,  .
 4

 6 2
1
1
− ( −8 )
+8
2
= 2
1 3
1  3
− +
− −− 
6 4
6  4
y −y
m= 2 1 =
x2 − x1
To simplify the complex fraction, we first simplify the numerator and denominator
separately (by finding a common denominator for each).
1 16
17
+
m= 2 2 = 2
2 9
7
− +
12 12
12
To divide the two fractions, we multiply by the reciprocal of the denominator and then
simplify.
6
17 12 17 12 17 6 102
m= ⋅ = ⋅
= ⋅ =
2 7
1 7
7
2 7
1
Note: Another way to simplify the complex fraction is to multiply both the numerator and
denominator by the least common multiple of 2, 6, and 4 (which is 12).
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Section 1.1
1

12  + 8 
2
 = 6 + 96 = 102
m=
7
 1 3  −2 + 9
12  − + 
 6 4
D.
 4 
Let ( x1 , y1 ) represent ( −2, 9 ) , and ( x2 , y2 ) represent  − , 9  .
 5 
m=
y2 − y1
9−9
=
=0
x2 − x1 − 4 − (−2)
5
We do not need to compute the denominator, since zero divided by a nonzero number is
zero. Recall that if the slope of a line is zero, the line is a horizontal line.
E.
5

5 
Let ( x1 , y1 ) represent  , − 3  , and ( x2 , y2 ) represent  , 1 .
6

6 
m=
y2 − y1 1 − (−3) 4
=
= , which is undefined
5 5
x2 − x1
0
−
6 6
Division by zero is undefined. Recall that if the slope of a line is undefined, the line is a
vertical line.
***
Intercepts
The y-intercept is the y-coordinate of the point where the graph crosses the y-axis, and is found
by setting x = 0 in the equation and solving for y. Similarly, the x-intercept is the x-coordinate
of the point where the graph crosses the x-axis, and is found by setting y = 0 and solving for x.
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Section 1.1
Equations of Lines
There are four forms of an equation of a line:
Point-Slope Form: y − y1 = m ( x − x1 ) ,
where m is the slope and ( x1 , y1 ) is a point on the line.
Slope-Intercept Form: y = mx + b ,
where m is the slope and b is the y-intercept of the line.
Standard Form: Ax + By = C ,
where A, B, and C real numbers and are written as integers whenever possible*,
and A and B cannot both be equal to zero.
General Form: Ax + By + C = 0 ,
where A, B, and C are real numbers and are written as integers whenever
possible*, and A and B cannot both be equal to zero.
*Notes about standard and general form: Standard form and general form are to be written such
that A, B, and C are integers whenever possible. In this course, you will be given problems where
it is always possible to change the equation so that A, B, and C are integers. There are cases
where it is not possible to change the coefficients to integers, such as the equation
2 x + π y = 7 , but such examples will not be used in this course.
The equations for standard and general form are not unique, as seen in the example below.
However, textbooks often display standard and general form so that A > 0 , and so that A, B, and
C are relatively prime.
Example 3: Change each of the following equations to slope-intercept form, standard form, and
general form.
5
4
5
7
A. y − 4 = − ( x − 6 )
B. −2 x − 6 = 8 y
C.
y− x=
7
9
12
6
Solution:
A.
To change to slope-intercept form, y = mx + b , we want to distribute the −
5
and then
7
solve for y.
y−4 = −
Math 1313
5
5
30
( x − 6 ) , so y − 4 = − x +
7
7
7
Page 5 of 19
Section 1.1
5
30
5
30
5
30 28
5
58
y−4 = − x+
, so y = − x + + 4 = − x + +
=− x+ .
7
7
7
7
7
7
7
7
7
5
58
Slope-intercept form is y = − x + .
7
7
We want to clear all fractions in order to put the equation in standard or general form. We
can do this by multiplying by 7 (the common denominator of the fractions):
58 
 5
7 ( y ) = 7  − x +  , so 7 y = −5 x + 58 .
7 
 7
We can then rearrange the terms to obtain standard and general form. The equation
5 x + 7 y = 58 is in standard form. The equation 5 x + 7 y − 58 = 0 is in general form. Note
that standard form and general form are not unique. Other acceptable answers can be
obtained by choosing any nonzero integer and then multiplying each term in the equation
by that integer.
B.
To change to slope-intercept form, y = mx + b , we can divide both sides of the equation
by 8 to solve for y.
−2 x − 6 = 8 y , so y =
−2 x − 6
2
6
1
3
= − x− = − x− .
8
8
8
4
4
1
3
Slope-intercept form is y = − x − .
4
4
Since the original equation already has terms with integer coefficients, we can simply
rearrange the terms of −2 x − 6 = 8 y to obtain standard and general form. The equation
−2 x − 8 y = 6 is in standard form, and the equation −2 x − 8 y − 6 = 0 is in general form.
Although both of the previous answers are acceptable, notice that the first term is
negative, and that all three terms have a common factor of 2. We can divide each term by
−2 to obtain x + 4 y = −3 for standard form and x + 4 y + 3 = 0 for general form. These
are the answers that would most commonly be given in a textbook answer key. Note that
standard form and general form are not unique. Other acceptable answers can be obtained
by choosing any nonzero integer and then multiplying each term in the equation by that
integer.
C.
4
5
7
y − x = contains multiple fractions, we will choose to clear the
9
12
6
equation of fractions by multiplying each term by the common denominator of 9, 12, and
6 – which is 36.
Since the equation
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Section 1.1
36 ⋅ 4 y 36 ⋅ 5 x 36 ⋅ 7
5 
4
7
−
=
. Simplifying each term, we obtain
36  y − x  = 36   , so
9
12
6
12 
9
6
4
3
6
36 ⋅ 4 y 36 ⋅ 5 x 36 ⋅ 7
−
=
, which gives us the equation 16 y − 15 x = 42 .
9
12
6
1
1
1
To change to slope-intercept form, y = mx + b , we need to solve for y:
16 y − 15 x = 42 , so 16 y = 15 x + 42 . Dividing by 16, y =
15 x + 42 15
21
= x+ .
16
16
8
To change 16 y − 15 x = 42 to standard form and general form, we can simply rearrange
the terms, since the coefficients and the constant term are already integers. The equation
−15 x + 16 y = 42 is in standard form, and the equation −15 x + 16 y − 42 = 0 is in general
form. Although both of the previous equations are acceptable, it is common to make the
first term positive by multiplying each term by −1 , to obtain a standard form of
15 x − 16 y = −42 , and a general form of 15 x − 16 y + 42 = 0 . Note that standard form and
general form are not unique. Other acceptable answers can be obtained by choosing any
nonzero integer and then multiplying each term in the equation by that integer.
***
Example 4: Write an equation in slope-intercept form for the line shown below.
Solution: Slope-intercept form is y = mx + b , where m is the slope and b is the y-intercept. We
can see from the graph that the y-intercept is 4, so b = 4 . We now need to find the slope, m.
Choose any two points on the graph with integer coordinates; we will use ( 0, 4 ) and (1, 0 ) .
From the point ( 0, 4 ) , we move down 4 units, then to the right 1 unit to get to (1, 0 ) .
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Section 1.1
rise −4
=
= −4 . We can now substitute m = −4 and b = 4 into y = mx + b and obtain the
run
1
equation of the line: y = −4 x + 4
So m =
***
Example 5: Write an equation in slope-intercept form for the line shown below.
Solution: Slope-intercept form is y = mx + b , where m is the slope and b is the y-intercept. We
can see from the graph that the y-intercept is 0, so b = 0 . We now need to find the slope, m.
Choose any two points on the graph with integer coordinates; we will use ( 0, 0 ) and ( 3, 2 ) .
From the point ( 0, 0 ) , we move up 2 units, then to the right 3 units to get to ( 3, 2 ) .
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Section 1.1
rise 2
2
= . We can now substitute m = and b = 0 into y = mx + b and obtain the
run 3
3
2
equation of the line: y = x
3
So m =
***
Example 6: Write an equation in slope-intercept form for the line shown below.
Solution: Slope-intercept form is y = mx + b , where m is the slope and b is the y-intercept.
Notice on the graph that the y-intercept is not an integer; we will use other means to find that
later. Let us first find the slope, m. Choose any two points on the graph with integer coordinates;
we will use ( −4, 2 ) and ( −1, − 2 ) . From the point ( −4, 2 ) , we move down 4 units, then to the
right 3 units to get to ( −1, − 2 ) .
4
So m = − . We can now use the slope along with one of the two chosen points and substitute
3
into either y = mx + b or y − y1 = m ( x − x1 ) to find the equation of the line. (There are four
means of solving this problem, using either of the two points with either of the two equations.
All methods yield the same result. We will show two of the four methods below.)
4
First, let us plug the point ( −4, 2 ) and the known slope, − , into the equation y = mx + b .
3
Math 1313
Page 9 of 19
Section 1.1
y = mx + b
4
2 = − ( −4 ) + b
3
16
2 = +b
3
b = 2−
16 6 16
10
= − =−
3 3 3
3
Now that we have found the slope and the y-intercept, we can substitute m = −
4
10
and b =
into
3
3
y = mx + b to obtain the equation of the line.
4
10
y =− x−
3
3
If we instead use the point ( −1, − 2 ) with the point-slope formula, we obtain the same result:
y − y1 = m ( x − x1 )
y − ( −2 ) = −
4
( x − ( −1) )
3
4
4
( x + 1) Next, distribute the − and solve for y.
3
3
4
4
y+2=− x−
3
3
4
4
4
4 6
y = − x− −2= − x− −
3
3
3
3 3
4
10
y =− x−
3
3
y+2=−
4
10
This equation y = − x −
is in slope-intercept form, and matches the answer we found using
3
3
the first method. Notice that when using y − y1 = m ( x − x1 ) , we did not need to separately find
the value of b. After solving for y, the equation is in slope-intercept form.
***
Math 1313
Page 10 of 19
Section 1.1
Example 7: Write an equation in slope-intercept form for the line shown below.
Solution: Slope-intercept form is y = mx + b , where m is the slope and b is the y-intercept. Once
again, notice that the y-intercept is not an integer. Let us first find the slope, m. Choose any two
points on the graph with integer coordinates; we will choose ( −3, − 2 ) and (1, 1) . From the point
( −3, − 2 ) , we move up 3 units, then to the right 4 units to get to (1, 1) .
3
. Next, we can use the slope and either of the two chosen points and substitute into
4
either y = mx + b or y − y1 = m ( x − x1 ) , to find the equation of the line. Let us use the point
So m =
( −3, − 2 )
and the known slope,
3
. Two methods are shown, but any one method is sufficient.
4
y = mx + b
y − y1 = m ( x − x1 )
3
( −3 ) + b
4
9
−2 = − + b
4
9
8 9 1
b = −2 + = − + =
4
4 4 4
3
1
y = mx + b = x +
4
4
3
( x − ( −3 ) )
4
3
9
y+2= x+
4
4
3
9
3
9 8
y = x+ −2= x+ −
4
4
4
4 4
3
1
y = x+
4
4
−2 =
Math 1313
y − ( −2 ) =
Page 11 of 19
Section 1.1
The equation of the line is y =
3
1
x+ .
4
4
***
Horizontal and Vertical Lines
The equation of any horizontal line is of the form y = b , where b is the y-intercept of the line.
The line therefore passes through the point ( 0, b ) . The slope of any horizontal line is zero.
The equation of any vertical line is of the form x = a , where a is the x-intercept of the line. The
line therefore passes through the point ( a, 0 ) . The slope of any vertical line is undefined.
Example 8: Write an equation for the line shown below.
Solution: The slope of any horizontal line is zero. The form of any horizontal line is y = b .
Hence, the equation of the line is y = −3 .
***
Example 9: Write an equation for the line shown below.
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Page 12 of 19
Section 1.1
Solution: The slope of any vertical line is undefined. The form of any vertical line is x = a .
Hence, the equation of the line is x = 2 .
***
Example 10: Find an equation of the line in slope-intercept form that passes through ( 0, 7 ) and
has slope 5.
Solution: We are given the y-intercept and the slope, so we can substitute into the slope-intercept
form and obtain the equation of the line: y = 5 x + 7
***
1
Example 11: A line passes through the point (10, − 8) and has slope − . Find an equation of
2
the line in point-slope form. Then write the equation in slope-intercept form.
1
Solution: Let ( x1 , y1 ) represent the point (10, − 8) . It is given that m = − . Substitute this
2
information into the point-slope equation.
y − y1 = m ( x − x1 )
y − ( −8 ) = −
y +8 = −
1
( x − 10 )
2
1
( x − 10 )
2
The above equation is in point-slope form. We will now solve for y to write the equation in
slope-intercept form.
1
( x − 10 )
2
1
y +8 = − x +5
2
1
y = − x +5−8
2
1
y = − x−3
2
y +8 = −
Point-slope form is y + 8 = −
1
1
( x − 10 ) , and slope-intercept form is y = − x − 3 .
2
2
***
Math 1313
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Section 1.1
Example 12: Find an equation of the line in slope-intercept form that passes through the points
( −4, 9 )
and ( −1, 1) .
Solution: We can find the slope of the line by applying the slope formula.
m=
y2 − y1
1− 9
8
=
=−
x2 − x1 −1 − ( −4 )
3
We can now use the slope and either one of the two given points to write the equation of the line.
Two methods are shown below, but one method is sufficient. We will use the point ( −4, 9 ) .
y = mx + b
y − y1 = m ( x − x1 )
8
( −4 ) + b
3
32
+b
9=
3
32 27 32
5
=
−
=−
b =9−
3
3
3
3
8
5
y = mx + b = − x −
3
3
8
( x − ( −4 ) )
3
8
32
y −9 = − x −
3
3
8
32
8
32 27
y = − x− +9 = − x− +
3
3
3
3
3
8
5
y = − x−
3
3
9=−
y −9 = −
8
5
The equation of the line is y = − x − .
3
3
***
Example 13: Find an equation of the line in standard form that has x-intercept 5 and y-intercept
− 0.75 .
Solution: We are given the points ( 5, 0 ) and ( 0, − 0.75) , respectively. First, find the slope.
m=
−0.75 − 0
= 0.15
0−5
Now use the slope and the given y-intercept to write the equation in slope-intercept form.
y = 0.15 x − 0.75
The above equation can alternatively be written with fractions:
y=
15
75
3
3
x−
, so y =
x−
100
100
20
4
We can clear the fractions by multiplying by 20 (the common denominator of 4 and 20).
Math 1313
Page 14 of 19
Section 1.1
3
3
x−
20
4
3
 3
20 ( y ) = 20  x − 
4
 20
20 y = 3 x − 15
y=
We then rearrange the terms to obtain −3 x + 20 y = −15 , which is in standard form, but is not
unique. If we want to make the first term positive (which is common), we can multiply each term
by −1 to obtain 3 x − 20 y = 15 . Other acceptable answers can be obtained by choosing any
nonzero integer and then multiplying each term in the equation by that integer.
***
Example 14: Find an equation of the line that passes through ( −9, 3.5 ) and ( 0.12, 3.5 ) .
Solution: First, we use the slope formula to find the slope of the line.
m=
y2 − y1
3.5 − 3.5
0
=
=
=0
x2 − x1 0.12 − ( −9 ) 9.12
Recall that if the slope of a line is zero, the line is a horizontal line. Hence, we have a horizontal
line, and its equation is y = 3.5 .
***
 2

 2

Example 15: Find an equation of the line that passes through  − , − 11 and  − , 11 .
 7

 7

Solution: First, we need to find the slope of the line.
m=
y2 − y1 11 − ( −11) 22
=
=
, which is undefined.
2  2 0
x2 − x1
− −− 
7  7
Recall that if the slope of a line is undefined, the line is a vertical line. Hence, we have a vertical
2
line and its equation is x = − .
7
***
Parallel and Perpendicular Lines
Two nonvertical lines are parallel if and only if their slopes are the same. (Any two vertical lines
are parallel to each other, but have undefined slope.)
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Section 1.1
Two lines are perpendicular if and only if their slopes are negative reciprocals of each other. The
exception to this rule is when one line is vertical (with undefined slope) and the other line is
horizontal (with slope zero). These lines are clearly perpendicular to each other, but are not
negative reciprocals since we cannot take the reciprocal of ‘undefined’. Note that two numbers c
1
and d are negative reciprocals of each other if d = − , which means that c ⋅ d = −1 .
c
Example 16: Find the negative reciprocal of each of the following numbers.
A. 5
B.
−
2
5
C.
−0.35
D. 10.75
Solution:
A.
B.
C.
D.
1
The negative reciprocal of 5 is − .
5
2
5
The negative reciprocal of − is .
5
2
35
7
20
−0.35 can be written as −
, which is − . Its negative reciprocal is
.
100
20
7
43
4
10.75 can be written as 10 3 4 , which is
. Its negative reciprocal is − .
4
43
***
Example 17: Find an equation of the line in slope-intercept form that passes through the point
1

 , 13  and is parallel to the line y = −12 x − 1 .
3

Solution: Notice that the line y = −12 x − 1 is in slope-intercept form y = mx + b , and has a slope
of −12 . Our desired line is parallel to y = −12 x − 1 , and therefore also has a slope of −12 .
1

We can use the slope, −12 , and the given point,  , 13  , to find the equation of the line. Two
3

different methods are shown below which yield the same result.
Math 1313
Page 16 of 19
Section 1.1
y = mx + b
y − y1 = m ( x − x1 )
1
13 = −12   + b
 3
13 = −4 + b
1

y − 13 = −12  x − 
3

y − 13 = −12 x + 4
b = 17
y = −12 x + 17
y = mx + b = −12 x + 17
The equation of the line in slope-intercept form is y = −12 x + 17 .
***
Example 18: Find an equation of the line in general form that passes through the point
( −4, − 11)
and is perpendicular to the line − x + 2 y = 10 .
Solution: To find the slope of the line − x + 2 y = 10 , we first need to write it in slope-intercept
form.
− x + 2 y = 10
2 y = x + 10
y=
1
x+5
2
1
1
x + 5 , has a slope of . Since we want to write the equation of a line
2
2
1
perpendicular to the line y = x + 5 , the desired line has a slope equal to the negative reciprocal
2
1
of , which is −2 . We can now use the slope, −2 , along with the given point, ( −4, − 11) , to find
2
the equation of the desired line. Two different methods are shown below, which yield the same
result.
The given line, y =
y = mx + b
y − y1 = m ( x − x1 )
−11 = −2 ( −4 ) + b
y − ( −11) = −2 ( x − ( −4 ) )
−11 = 8 + b
y + 11 = −2 ( x + 4 )
b = −11 − 8
b = −19
y = mx + b = −2 x − 19
y + 11 = −2 x − 8
y = −2 x − 19
Using either method above, we obtain the equation y = −2 x − 19 .
Math 1313
Page 17 of 19
Section 1.1
We want to write the equation in general form, Ax + By + C = 0 , where A, B, and C are integers.
We can rearrange the terms so that they are all on one side of the equation.
2 x + y + 19 = 0
Remember that when writing in general form, the answers are not unique. Other acceptable
answers can be obtained by choosing any nonzero integer and then multiplying each term in the
equation by that integer.
***
Example 19: Find an equation of the line in standard form that has x-intercept −8 and is
perpendicular to the line that passes through ( 4, − 7 ) and ( 6, − 10 ) .
Solution: We first use the slope formula to find the slope of the line that passes through ( 4, − 7 )
and ( 6, −10 ) .
m=
y2 − y1 −10 − ( −7 )
3
=
=−
x2 − x1
6−4
2
3
To find the slope of the desired perpendicular line, we find the negative reciprocal of − , which
2
2
is . We know that the line has x-intercept −8 , which means that the graph passes through the
3
2
point ( −8, 0 ) . We can now use the slope, , along with the given point, ( −8, 0 ) , to find the
3
equation of the desired line. Two different methods are shown below, which yield the same
result.
y = mx + b
y − y1 = m ( x − x1 )
2
( −8 ) + b
3
16
0 = − +b
3
16
b=
3
2
16
y = mx + b = x +
3
3
2
( x − ( −8 ) )
3
2
16
y = x+
3
3
0=
y−0 =
Using either method above, we obtain the equation y =
Math 1313
Page 18 of 19
2
16
x+ .
3
3
Section 1.1
We want to write the equation in standard form, Ax + By = C , where A, B, and C are integers.
We can clear the fractions by multiplying each term by 3.
16 
2
3( y ) = 3 x + 
3
3
3 y = 2 x + 16
We then rearrange the terms to obtain −2 x + 3 y = 16 , which is in standard form, but is not
unique. If we want to make the first term positive (which is common), we can multiply each term
by −1 to obtain 2 x − 3 y = −16 . Other acceptable answers can be obtained by choosing any
nonzero integer and then multiplying each term in the equation by that integer.
***
Example 20: Find an equation of the line in slope-intercept form that passes through ( −4, 13)
and is parallel to the line that passes through ( −2, 0 ) and (10, 15 ) .
Solution: We first use the slope formula to find the slope of the line that passes through ( −2, 0 )
and (10, 15 ) .
m=
y2 − y1
15 − 0
15 5
=
=
=
x2 − x1 10 − ( −2 ) 12 4
Since the equation of the line we wish to write is parallel to the line that passes through
5
( −2, 0 ) and (10, 15 ) , we need to use this same slope, , along with the point ( −4, 13) to find
4
the equation of the line. Two different methods are shown below, which yield the same result.
y − y1 = m ( x − x1 )
y = mx + b
13 =
5
( −4 ) + b
4
5
( x − ( −4 ) )
4
5
y − 13 = x + 5
4
5
y = x + 5 + 13
4
5
y = x + 18
4
y − 13 =
13 = −5 + b
b = 13 + 5 = 18
y = mx + b =
5
x + 18
4
The equation of the line is y =
Math 1313
5
x + 18 .
4
Page 19 of 19
***
Section 1.1