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The Slope-Intercept Form of a Line

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter the student is shown how graphs provide information that is not always evident from the equation alone. The chapter begins by establishing the relationship between the variables in an equation, the number of coordinate axes necessary to construct its graph, and the spatial dimension of both the coordinate system and the graph. Interpretation of graphs is also emphasized throughout the chapter, beginning with the plotting of points. The slope formula is fully developed, progressing from verbal phrases to mathematical expressions. The expressions are then formed into an equation by explicitly stating that a ratio is a comparison of two quantities of the same type (e.g., distance, weight, or money). This approach benefits students who take future courses that use graphs to display information. The student is shown how to graph lines using the intercept method, the table method, and the slope-intercept method, as well as how to distinguish, by inspection, oblique and horizontal/vertical lines. Objectives of this module: be more familiar with the general form of a line, be able to recognize the slope-intercept form of a line, be able to interpret the slope and intercept of a line, be able to use the slope formula to find the slope of a line.

Overview

  • The General Form of a Line
  • The Slope-Intercept Form of a Line
  • Slope and Intercept
  • The Formula for the Slope of a Line

The General Form of a Line

We have seen that the general form of a linear equation in two variables is ax+by=c ax+by=c (Section (Reference)). When this equation is solved for y y , the resulting form is called the slope-intercept form. Let's generate this new form.

ax+by = c Subtract ax  from both sides. by = −ax+c Divide both  sides by  b by b = −ax b + c b b y b = −ax b + c b y = −ax b + c b y = −ax b + c b ax+by = c Subtract ax  from both sides. by = −ax+c Divide both  sides by  b by b = −ax b + c b b y b = −ax b + c b y = −ax b + c b y = −ax b + c b

This equation is of the form y=mx+b y=mx+b if we replace −a b −a b with m m and constant c b c b with b b . (Note: The fact that we let b= c b b= c b is unfortunate and occurs beacuse of the letters we have chosen to use in the general form. The letter b b occurs on both sides of the equal sign and may not represent the same value at all. This problem is one of the historical convention and, fortunately, does not occur very often.)

The following examples illustrate this procedure.

Example 1

Solve 3x+2y=6 3x+2y=6 for y y .

3x+2y = 6 Subtract 3x  from both sides. 2y = −3x+6 Divide both sides by 2. y = − 3 2 x+3 3x+2y = 6 Subtract 3x  from both sides. 2y = −3x+6 Divide both sides by 2. y = − 3 2 x+3

This equation is of the form y=mx+b y=mx+b . In this case, m=- 3 2 m=- 3 2 and b=3 b=3 .

Example 2

Solve -15x+5y=20 -15x+5y=20 for y y .

−15x+5y = 20 5y = 15x+20 y = 3x+4 −15x+5y = 20 5y = 15x+20 y = 3x+4

This equation is of the form y=mx+b y=mx+b . In this case, m=3 m=3 and b=4 b=4 .

Example 3

Solve 4x-y=0 4x-y=0 for y y .

4x−y = 0 −y = −4x y = 4x 4x−y = 0 −y = −4x y = 4x

This equation is of the form y=mx+b y=mx+b . In this case, m=4 m=4 and b=0 b=0 . Notice that we can write y=4x y=4x as y=4x+0 y=4x+0 .

The Slope-Intercept Form of a Line

The Slope-Intercept Form of a Line y=mx+by=mx+b

A linear equation in two variables written in the form y=mx+b y=mx+b is said to be in slope-intercept form.

Sample Set A

The following equations are in slope-intercept form:

Example 4

y=6x−7. In this case m=6 and b=−7. y=6x−7. In this case m=6 and b=−7.

Example 5

y=−2x+9. In this case m=−2 and b=9. y=−2x+9. In this case m=−2 and b=9.

Example 6

y= 1 5 x+4.8 In this case m= 1 5  and b=4.8. y= 1 5 x+4.8 In this case m= 1 5  and b=4.8.

Example 7

y=7x. In this case m=7 and b=0 since we can write y=7x as y=7x+0. y=7x. In this case m=7 and b=0 since we can write y=7x as y=7x+0.

The following equations are not in slope-intercept form:

Example 8

2y=4x−1. The coefficient of y is 2.  To be in slope-intercept form, the coefficient of y must be 1. 2y=4x−1. The coefficient of y is 2.  To be in slope-intercept form, the coefficient of y must be 1.

Example 9

y+4x=5. The equation is not solved for y.  The x and y appear on the same side of the equal sign. y+4x=5. The equation is not solved for y.  The x and y appear on the same side of the equal sign.

Example 10

y+1=2x. The equation is not solved for y. y+1=2x. The equation is not solved for y.

Practice Set A

The following equation are in slope-intercept form. In each case, specify the slope and y-intercept y-intercept .

Exercise 1

y=2x+7; m=           b=           y=2x+7; m=           b=          

Solution

m=2, b=7 m=2, b=7

Exercise 2

y=−4x+2; m=           b=           y=−4x+2; m=           b=          

Solution

m=−4,b=2 m=−4,b=2

Exercise 3

y=−5x−1; m=           b=           y=−5x−1; m=           b=          

Solution

m=−5,b=−1 m=−5,b=−1

Exercise 4

y= 2 3 x−10; m=           b=           y= 2 3 x−10; m=           b=          

Solution

m= 2 3 , b= −10 m= 2 3 , b= −10

Exercise 5

y= −5 8 x+ 1 2 ; m=           b=           y= −5 8 x+ 1 2 ; m=           b=          

Solution

m= −5 8 ,  b= 1 2 m= −5 8 ,  b= 1 2

Exercise 6

y=−3x; m=           b=           y=−3x; m=           b=          

Solution

m=−3, b=0 m=−3, b=0

Slope and Intercept

When the equation of a line is written in slope-intercept form, two important properties of the line can be seen: the slope and the intercept. Let's look at these two properties by graphing several lines and observing them carefully.

Sample Set B

Example 11

Graph the line y=x−3 y=x−3 .

Table 1
x x y y ( x, y ) ( x, y )
0 −3 −3 (0, −3) (0, −3)
4 1 (4,  1) (4,  1)
−2 −2 −5 −5 (−2,  −5) (−2,  −5)

A graph of a line passing through three points with coordinates zero, negative three; four, one , and negative two, negative five.

Looking carefully at this line, answer the following two questions.

Problem 1

At what number does this line cross the y-axis y-axis ? Do you see this number in the equation?

Solution

The line crosses the y-axis y-axis at -3 -3 .

Problem 2

Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?

Solution

After moving horizontally one unit to the right, we must move exactly one vertical unit up. This number is the coefficient of x x .

Example 12

Graph the line y= 2 3 x+1 y= 2 3 x+1 .

Table 2
x x y y ( x, y ) ( x, y )
0 1 (0,  1) (0,  1)
3 3 (3,  3) (3,  3)
−3 −3 −1 −1 (−3,  −1) (−3,  −1)

A graph of a line passing through three points with coordinates zero, one; three, three, and negative three, negative one.

Looking carefully at this line, answer the following two questions.

Problem 1

At what number does this line cross the y-axis y-axis ? Do you see this number in the equation?

Solution

The line crosses the y-axis y-axis at +1 +1 .

Problem 2

Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?

Solution

After moving horizontally one unit to the right, we must move exactly 2 3 2 3 unit upward. This number is the coefficient of x x .

Practice Set B

Example 13

Graph the line y=−3x+4 y=−3x+4 .

Table 3
x x y y (x, y) (x, y)
0
3
2

An xy coordinate plane with gridlines, labeled negative five and five on the both axes.

Looking carefully at this line, answer the following two questions.

Exercise 7

At what number does the line cross the y-axis y-axis ? Do you see this number in the equation?

Solution

The line crosses the y-axis y-axis at +4 +4 . After moving horizontally 1 unit to the right, we must move exactly 3 units downward.

Exercise 8

Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?

Solution

A graph of a line passing through two points with coordinates zero,four; two, negative two ,and three, negative five.

In the graphs constructed in Sample Set B and Practice Set B, each equation had the form y=mx+b y=mx+b . We can answer the same questions by using this form of the equation (shown in the diagram).

A graph of a line sloped up and to the right in a first quadrant labeled with the equation y equal to mx plus b and intersecting y axis at point with coordinates zero, b. Lines illustrating an upward change of m units and a horizontal change of one unit to the right.

yy-Intercept

Exercise 9

At what number does the line cross the y-axis y-axis ? Do you see this number in the equation?

Solution

In each case, the line crosses the y-axis y-axis at the constant b b . The number b b is the number at which the line crosses the y-axis y-axis , and it is called the y-intercept y-intercept . The ordered pair corresponding to the y-intercept y-intercept is (0, b). (0, b).

Exercise 10

Place your pencil at any point on the line. Move your pencil exactly one unit horizontally to the right. Now, how many units straight up or down must you move your pencil to get back on the line? Do you see this number in the equation?

Solution

To get back on the line, we must move our pencil exactly m m vertical units.

Slope

The number m m is the coefficient of the variable x x . The number m m is called the slope of the line and it is the number of units that y y changes when x x is increased by 1 unit. Thus, if x x changes by 1 unit, y y changes by m m units.
Since the equation y=mx+b y=mx+b contains both the slope of the line and the y-intercept y-intercept , we call the form y=mx+b y=mx+b the slope-intercept form.

The Slope-Intercept Form of the Equation of a Line

The slope-intercept form of a straight line is
 y=mx+b y=mx+b
The slope of the line is m m , and the y-intercept y-intercept is the point (0 , b) (0 , b) .

The Slope is a Measure of the Steepness of a Line

The word slope is really quite appropriate. It gives us a measure of the steepness of the line. Consider two lines, one with slope 12 1 2 and the other with slope 3. The line with slope 3 is steeper than is the line with slope 12 1 2 . Imagine your pencil being placed at any point on the lines. We make a 1-unit increase in the x x -value by moving the pencil one unit to the right. To get back to one line we need only move vertically 12 1 2 unit, whereas to get back onto the other line we need to move vertically 3 units.

A graph of a line sloped up and to the right with lines illustrating an upward change of three units and a horizontal change of one unit to the right.

A graph of a line sloped up and to the right with lines illustrating an upward change of one half unit and a horizontal change of one unit to the right.

Sample Set C

Find the slope and the y y -intercept of the following lines.

Example 14

y=2x+7. y=2x+7.

The line is in the slope-intercept form y=mx+b. y=mx+b. The slope is m m , the coefficient of x x . Therefore, m=2. m=2. The y-intercept y-intercept is the point (0, b). (0, b). Since b=7 b=7 , the y-intercept y-intercept is (0, 7). (0, 7).

Slope:2 y-intercept:(0, 7) Slope:2 y-intercept:(0, 7)

Example 15

y=−4x+1. y=−4x+1.

The line is in slope-intercept form y=mx+b. y=mx+b. The slope is m m , the coefficient of x x . So, m=−4. m=−4. The y-intercept y-intercept is the point (0, b). (0, b). Since b=1 b=1 , the y-intercept y-intercept is (0, 1). (0, 1).

Slope:−4 y-intercept:(0, 1) Slope:−4 y-intercept:(0, 1)

Example 16

3x+2y=5. 3x+2y=5.

The equation is written in general form. We can put the equation in slope-intercept form by solving for y y .

3x+2y = 5 2y = −3x+5 y = − 3 2 x+ 5 2 3x+2y = 5 2y = −3x+5 y = − 3 2 x+ 5 2

Now the equation is in slope-intercept form.

Slope: − 3 2 y-intercept:  ( 0,  5 2 ) Slope: − 3 2 y-intercept:  ( 0,  5 2 )

Practice Set C

Exercise 11

Find the slope and y-intercept y-intercept of the line 2x+5y=15. 2x+5y=15.

Solution

Solving for y y we get y= −2 5 x+3. y= −2 5 x+3. Now, m= −2 5 m= −2 5 and b=3. b=3.

The Formula for the Slope of a Line

We have observed that the slope is a measure of the steepness of a line. We wish to develop a formula for measuring this steepness.

It seems reasonable to develop a slope formula that produces the following results:

Steepness of line 1> 1> steepness of line 2.

A graph of two lines sloped up and to the right in the first quadrant. Line with the lable 'Line one' has a steepness greater than the line with the lable 'Line two'.

Consider a line on which we select any two points. We’ll denote these points with the ordered pairs ( x 1,  y 1  ) ( x 1,  y 1  ) and ( x 2,  y 2  ) ( x 2,  y 2  ) . The subscripts help us to identify the points.

( x 1,  y 1  ) ( x 1,  y 1  ) is the first point. Subscript 1 indicates the first point.
 ( x 2,  y 2  ) ( x 2,  y 2  ) is the second point. Subscript 2 indicates the second point.

A graph of a line sloped up and to the right in the first quadrant passing through two points with coordinates x-one, y-one and x-two, y-two.

The difference in x x values ( x 2 − x 1 ) ( x 2 − x 1 ) gives us the horizontal change, and the difference in y y values ( y 2 − y 1 ) ( y 2 − y 1 ) gives us the vertical change. If the line is very steep, then when going from the first point to the second point, we would expect a large vertical change compared to the horizontal change. If the line is not very steep, then when going from the first point to the second point, we would expect a small vertical change compared to the horizontal change.

A graph of a line sloped up and to the right in a first quadrant. Lines illustrating an upward change of y-two minus y-one and a horizontal change x-two minus x-one. Horzontal change is small as compared to vertical change.

A graph of a line sloped up and to the right in a first quadrant. Lines illustrating an upward change of y-two minus y-one and a horizontal change x-two minus x-one. Vertical change is small as compared to horzontal change

We are comparing changes. We see that we are comparing

The vertical change to the horizontal change The change in y to the change in x y 2 − y 1 to x 2 − x 1 The vertical change to the horizontal change The change in y to the change in x y 2 − y 1 to x 2 − x 1

This is a comparison and is therefore a ratio. Ratios can be expressed as fractions. Thus, a measure of the steepness of a line can be expressed as a ratio.

The slope of a line is defined as the ratio

 Slope= change in y change in x Slope= change in y change in x

Mathematically, we can write these changes as

 Slope= y 2 − y 1 x 2 − x 1 Slope= y 2 − y 1 x 2 − x 1

Finding the Slope of a Line

The slope of a nonvertical line passing through the points ( x 1,  y 1  ) ( x 1,  y 1  ) and ( x 2,  y 2  ) ( x 2,  y 2  ) is found by the formula
 m= y 2 − y 1 x 2 − x 1 m= y 2 − y 1 x 2 − x 1

Sample Set D

For the two given points, find the slope of the line that passes through them.

Example 17

( 0, 1 ) ( 0, 1 ) and ( 1, 3 ) ( 1, 3 ) .

Looking left to right on the line we can choose ( x 1,  y 1  ) ( x 1,  y 1  ) to be ( 0, 1 ) ( 0, 1 ) , and ( x 2,  y 2  ) ( x 2,  y 2  ) to be ( 1, 3 ) . ( 1, 3 ) . Then,

 m= y 2 − y 1 x 2 − x 1 = 3−1 1−0 = 2 1 =2 m= y 2 − y 1 x 2 − x 1 = 3−1 1−0 = 2 1 =2

A graph of a line passing through two points with coordinates zero, one, and one, three with the upward change of two units and a horizontal change of one unit to the right.

This line has slope 2. It appears fairly steep. When the slope is written in fraction form, 2= 2 1 2= 2 1 , we can see, by recalling the slope formula, that as x x changes 1 unit to the right (because of the +1 +1 ) y y changes 2 units upward (because of the +2 +2 ).

 m= change in y change in x = 2 1 m= change in y change in x = 2 1

Notice that as we look left to right, the line rises.

Example 18

(2, 2 )  (2, 2 )  and (4, 3 )  (4, 3 )  .

Looking left to right on the line we can choose ( x 1,  y 1  ) ( x 1,  y 1  ) to be (2, 2 ) (2, 2 ) and ( x 2,  y 2  ) ( x 2,  y 2  ) to be (4, 3 ). (4, 3 ). Then,

 m= y 2 − y 1 x 2 − x 1 = 3−2 4−2 = 1 2 m= y 2 − y 1 x 2 − x 1 = 3−2 4−2 = 1 2

A graph of a line passing through two points with coordinates two, two, and four, three.

This line has slope 1 2 1 2 . Thus, as x x changes 2 units to the right (because of the +2 +2 ), y y changes 1 unit upward (because of the +1 +1 ).

 m= change in y change in x = 1 2 m= change in y change in x = 1 2

Notice that in examples 1 and 2, both lines have positive slopes, +2 +2 and + 1 2 + 1 2 , and both lines rise as we look left to right.

Example 19

(−2, 4) (−2, 4) and (1, 1) (1, 1) .

Looking left to right on the line we can choose ( x 1,  y 1  ) ( x 1,  y 1  ) to be (−2, 4) (−2, 4) and ( x 2,  y 2  ) ( x 2,  y 2  ) to be (1, 1) (1, 1) . Then,

 m= y 2 − y 1 x 2 − x 1 = 1−4 1−( −2 ) = −3 1+2 = −3 3 =−1 m= y 2 − y 1 x 2 − x 1 = 1−4 1−( −2 ) = −3 1+2 = −3 3 =−1

A graph of a line passing through two points with coordinates negative two, four, and one, one with a downward change of one unit and a horizontal change of one unit to the right.

This line has slope −1. −1.

When the slope is written in fraction form, m=−1= −1 +1 m=−1= −1 +1 , we can see that as x x changes 1 unit to the right (because of the +1 +1 ), y y changes 1 unit downward (because of the −1 −1 ).
Notice also that this line has a negative slope and declines as we look left to right.

Example 20

( 1, 3 ) ( 1, 3 ) and ( 5, 3 ) ( 5, 3 ) .

 m= y 2 − y 1 x 2 − x 1 = 3−3 5−1 = 0 4 =0 m= y 2 − y 1 x 2 − x 1 = 3−3 5−1 = 0 4 =0

A graph of a line parallel to x axis and passing through two points with coordinates one, three, and five, three.

This line has 0 slope. This means it has no rise and, therefore, is a horizontal line. This does not mean that the line has no slope, however.

Example 21

( 4, 4 ) ( 4, 4 ) and ( 4, 0 ) ( 4, 0 ) .

This problem shows why the slope formula is valid only for nonvertical lines.

m= y 2 − y 1 x 2 − x 1 = 0−4 4−4 = −4 0 m= y 2 − y 1 x 2 − x 1 = 0−4 4−4 = −4 0

A graph of a line parallel to y axis and passing through two points with coordinates four,zero and four, four.

Since division by 0 is undefined, we say that vertical lines have undefined slope. Since there is no real number to represent the slope of this line, we sometimes say that vertical lines have undefined slope, or no slope.

Practice Set D

Exercise 12

Find the slope of the line passing through ( 2, 1 ) ( 2, 1 ) and ( 6, 3 ) ( 6, 3 ) . Graph this line on the graph of problem 2 below.

Solution

m= 3−1 6−2 = 2 4 = 1 2 . m= 3−1 6−2 = 2 4 = 1 2 .

Exercise 13

Find the slope of the line passing through (3, 4 ) (3, 4 ) and (5, 5 ) (5, 5 ) . Graph this line.

 An xy-plane with gridlines

Solution

The line has slope 12 1 2 .

Exercise 14

Compare the lines of the following problems. Do the lines appear to cross? What is it called when lines do not meet (parallel or intersecting)? Compare their slopes. Make a statement about the condition of these lines and their slopes.

Solution

The lines appear to be parallel. Parallel lines have the same slope, and lines that have the same slope are parallel.

 A graph of two parallel lines. One of the lines passes through two points with coordinates two, one and six, three. Another straight line passes through two points with coordinates three, four and five, five.

Before trying some problems, let’s summarize what we have observed.

Exercise 15

The equation y=mx+b y=mx+b is called the slope-intercept form of the equation of a line. The number m m is the slope of the line and the point (0, b ) (0, b ) is the y-intercept y-intercept .

Exercise 16

The slope, m, m, of a line is defined as the steepness of the line, and it is the number of units that y y changes when x x changes 1 unit.

Exercise 17

The formula for finding the slope of a line through any two given points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

m= y 2 - y 1 x 2 - x 1 m= y 2 - y 1 x 2 - x 1

Exercise 18

The fraction y 2 - y 1 x 2 - x 1 y 2 - y 1 x 2 - x 1 represents the Change in y Change in x . Change in y Change in x .

Exercise 19

As we look at a graph from left to right, lines with positive slope rise and lines with negative slope decline.

Exercise 20

Parallel lines have the same slope.

Exercise 21

Horizontal lines have 0 slope.

Exercise 22

Vertical lines have undefined slope (or no slope).

Exercises

For the following problems, determine the slope and y-intercept y-intercept of the lines.

Exercise 23

y=3x+4 y=3x+4

Solution

slope=3;  y-intercept=( 0,4 ) slope=3;  y-intercept=( 0,4 )

Exercise 24

y=2x+9 y=2x+9

Exercise 25

y=9x+1 y=9x+1

Solution

slope=9;  y-intercept=( 0,1 ) slope=9;  y-intercept=( 0,1 )

Exercise 26

y=7x+10 y=7x+10

Exercise 27

y=−4x+5 y=−4x+5

Solution

slope=−4;  y-intercept=( 0,5 ) slope=−4;  y-intercept=( 0,5 )

Exercise 28

y=−2x+8 y=−2x+8

Exercise 29

y=−6x−1 y=−6x−1

Solution

slope=−6;  y-intercept=( 0,−1 ) slope=−6;  y-intercept=( 0,−1 )

Exercise 30

y=−x−6 y=−x−6

Exercise 31

y=−x+2 y=−x+2

Solution

slope=−1;  y-intercept=( 0,2 ) slope=−1;  y-intercept=( 0,2 )

Exercise 32

2y=4x+8 2y=4x+8

Exercise 33

4y=16x+20 4y=16x+20

Solution

slope=4;  y-intercept=( 0,5 ) slope=4;  y-intercept=( 0,5 )

Exercise 34

−5y=15x+55 −5y=15x+55

Exercise 35

−3y=12x−27 −3y=12x−27

Solution

slope=−4;  y-intercept=( 0,9 ) slope=−4;  y-intercept=( 0,9 )

Exercise 36

y= 3 5 x−8 y= 3 5 x−8

Exercise 37

y= 2 7 x−12 y= 2 7 x−12

Solution

slope= 2 7 ;  y-intercept=( 0,−12 ) slope= 2 7 ;  y-intercept=( 0,−12 )

Exercise 38

y= −1 8 x+ 2 3 y= −1 8 x+ 2 3

Exercise 39

y= −4 5 x− 4 7 y= −4 5 x− 4 7

Solution

slope=− 4 5 ;  y-intercept=( 0,− 4 7 ) slope=− 4 5 ;  y-intercept=( 0,− 4 7 )

Exercise 40

−3y=5x+8 −3y=5x+8

Exercise 41

−10y=−12x+1 −10y=−12x+1

Solution

slope= 6 5 ;  y-intercept=( 0,− 1 10 ) slope= 6 5 ;  y-intercept=( 0,− 1 10 )

Exercise 42

−y=x+1 −y=x+1

Exercise 43

−y=−x+3 −y=−x+3

Solution

slope=1;  y-intercept=( 0,−3 ) slope=1;  y-intercept=( 0,−3 )

Exercise 44

3x−y=7 3x−y=7

Exercise 45

5x+3y=6 5x+3y=6

Solution

slope=− 5 3 ;  y-intercept=( 0,2 ) slope=− 5 3 ;  y-intercept=( 0,2 )

Exercise 46

−6x−7y=−12 −6x−7y=−12

Exercise 47

−x+4y=−1 −x+4y=−1

Solution

slope= 1 4 ;  y-intercept=( 0,− 1 4 ) slope= 1 4 ;  y-intercept=( 0,− 1 4 )

For the following problems, find the slope of the line through the pairs of points.

Exercise 48

(1, 6), (4,9) (1, 6), (4,9)

Exercise 49

(1, 3), (4,7) (1, 3), (4,7)

Solution

m= 4 3 m= 4 3

Exercise 50

(3, 5), (4,7) (3, 5), (4,7)

Exercise 51

(6, 1), (2,8) (6, 1), (2,8)

Solution

m=− 7 4 m=− 7 4

Exercise 52

(0, 5), (2,-6) (0, 5), (2,-6)

Exercise 53

(-2, 1), (0,5) (-2, 1), (0,5)

Solution

m=2 m=2

Exercise 54

(3, -9), (5,1) (3, -9), (5,1)

Exercise 55

(4, -6), (-2,1) (4, -6), (-2,1)

Solution

m=− 7 6 m=− 7 6

Exercise 56

(-5, 4), (-1,0) (-5, 4), (-1,0)

Exercise 57

(-3, 2), (-4,6) (-3, 2), (-4,6)

Solution

m=−4 m=−4

Exercise 58

(9, 12), (6,0) (9, 12), (6,0)

Exercise 59

(0, 0), (6,6) (0, 0), (6,6)

Solution

m=1 m=1

Exercise 60

(-2, -6), (-4,-1) (-2, -6), (-4,-1)

Exercise 61

(-1, -7), (-2,-9) (-1, -7), (-2,-9)

Solution

m=2 m=2

Exercise 62

(-6, -6), (-5,-4) (-6, -6), (-5,-4)

Exercise 63

(-1, 0), (-2,-2) (-1, 0), (-2,-2)

Solution

m=2 m=2

Exercise 64

(-4, -2), (0,0) (-4, -2), (0,0)

Exercise 65

(2, 3), (10,3) (2, 3), (10,3)

Solution

m=0 ( horizontal line y=3 ) m=0 ( horizontal line y=3 )

Exercise 66

(4, -2), (4,7) (4, -2), (4,7)

Exercise 67

(8, -1), (8,3) (8, -1), (8,3)

Solution

No slope ( vertical line at x=8 ) No slope ( vertical line at x=8 )

Exercise 68

(4, 2), (6,2) (4, 2), (6,2)

Exercise 69

(5, -6), (9,-6) (5, -6), (9,-6)

Solution

m=0 ( horizontal line at y=−6 ) m=0 ( horizontal line at y=−6 )

Exercise 70

Do lines with a positive slope rise or decline as we look left to right?

Exercise 71

Do lines with a negative slope rise or decline as we look left to right?

Solution

decline

Exercise 72

Make a statement about the slopes of parallel lines.

Use a calculator. Calculator Problems

For the following problems, determine the slope and y-intercept y-intercept of the lines. Round to two decimal places.

Exercise 73

3.8x+12.1y=4.26 3.8x+12.1y=4.26

Solution

slope=−0.31 y−intercept=( 0,0.35 ) slope=−0.31 y−intercept=( 0,0.35 )

Exercise 74

8.09x+5.57y=−1.42 8.09x+5.57y=−1.42

Exercise 75

10.813x−17.0y=−45.99 10.813x−17.0y=−45.99

Solution

slope=0.64 y−intercept=( 0,2.71 ) slope=0.64 y−intercept=( 0,2.71 )

Exercise 76

−6.003x−92.388y=0.008 −6.003x−92.388y=0.008

For the following problems, find the slope of the line through the pairs of points. Round to two decimal places.

Exercise 77

(5.56, 9.37), (2.16, 4.90) (5.56, 9.37), (2.16, 4.90)

Solution

m=1.31 m=1.31

Exercise 78

(33.1, 8.9), (42.7, −1.06) (33.1, 8.9), (42.7, −1.06)

Exercise 79

(155.89, 227.61), (157.04,227.61) (155.89, 227.61), (157.04,227.61)

Solution

m=0 ( horizontal line at y=227.61 ) m=0 ( horizontal line at y=227.61 )

Exercise 80

(0.00426, −0.00404), (−0.00191, −0.00404) (0.00426, −0.00404), (−0.00191, −0.00404)

Exercise 81

(88.81, −23.19), (88.81, −26.87) (88.81, −23.19), (88.81, −26.87)

Solution

No slope ( vertical line x=88.81 ) No slope ( vertical line x=88.81 )

Exercise 82

(−0.0000567, −0.0000567), (−0.00765, 0.00764) (−0.0000567, −0.0000567), (−0.00765, 0.00764)

Exercises for Review

Exercise 83

((Reference)) Simplify ( x 2 y 3 w 4 ) 0 ( x 2 y 3 w 4 ) 0 .

Solution

1 if xyw≠0 1 if xyw≠0

Exercise 84

((Reference)) Solve the equation 3x−4(2−x)−3(x−2)+4=0 3x−4(2−x)−3(x−2)+4=0 .

Exercise 85

((Reference)) When four times a number is divided by five, and that result is decreased by eight, the result is zero. What is the original number?

Solution

10

Exercise 86

((Reference)) Solve −3y+10=x+2  −3y+10=x+2  if x=−4 x=−4 .

Exercise 87

((Reference)) Graph the linear equation x+y=3 x+y=3 .
An xy coordinate plane with gridlines, labeled negative five and five on the both axes.

Solution

A graph of a line passing through two points with coordinates three, zero and zero, three.

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