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Section 2.3 Lines

Before Starting this Section, Review

1 Evaluating algebraic expressions (Section P.1 , page 15)
2 Solving linear equations (Section 1.1 , page 83)
3 Graphs of equations (Section 2.2 , page 186)

Objectives

1 Find the slope of a line.
2 Write the point–slope form of the equation of a line.
3 Write the slope–intercept form of the equation of a line.
4 Recognize the equations of horizontal and vertical lines.
5 Recognize the general form of the equation of a line.
6 Find equations of parallel and perpendicular lines.
7 Model data using linear regression.

A Texan’s Tall Tale

Gunslinger Wild Bill Longley’s first burial took place on October 11, 1878, after he was hanged before a crowd of thousands in Giddings, Texas.

Rumors persisted that Longley’s hanging had been a hoax and that he had somehow faked his death and escaped execution. In 2001, Longley’s descendants had his grave opened to determine whether the remains matched Wild Bill’s description: a tall white male, age 27. Both the skeleton and some personal effects suggested that this was indeed Wild Bill. Modern science lent a hand, too: The DNA of Wild Bill’s sister’s descendant Helen Chapman was a perfect match.

Now the notorious gunman could be buried back in the Giddings cemetery—for the second time. How did the scientists conclude from the skeletal remains that Wild Bill was approximately 6 feet tall? (See Example 8.)

Slope of a Line

1 Find the slope of a line.

In this section, we study various forms of first-degree equations in two variables. Because the graphs of these equations are straight lines, they are called linear equations. Just as we measure weight or temperature by a number, we measure the “steepness” of a line by a number called its slope.

Consider two points P(x1, y1) $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ on a line, as shown in Figure 2.22. We say that the rise is the change in y-coordinates between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ and that the run is the corresponding change in the x-coordinates. A positive run indicates change to the right, as shown in Figure 2.22; a negative run indicates change to the left. Similarly, a positive rise indicates upward change; a negative rise indicates downward change. A zero run or rise means no change in the corresponding coordinate.

Side Note

The symbols $Δ y$ (read “delta y”) and $Δ x$ (read delta x”) are used to indicate a “change in y” and a “change in x,” respectively. So the slope is sometimes denoted by $m = \frac{Δ y}{Δ x}$ .

Since $m = \frac{rise}{run} = \frac{change in y -coordinates}{change in x -coordinates},$ if the change in x, $Δ x$ , is one unit to the right (x increases by 1 unit), then the change in y, $Δ y$ , equals the slope of the line. The slope of a line is therefore the change in y per unit change in x. In other words, the slope of a line measures the rate of change of y with respect to x. Roofs, staircases, graded landscapes, and mountainous roads all have slopes. For example, if the pitch (slope) of a section of the roof is 0.4, then for every horizontal distance of 10 feet in that section, the roof ascends 4 feet.

Example 1 Finding and Interpreting the Slope of a Line

Sketch the graph of the line that passes through $P (1, - 1)$ and Q(3, 3). Find and interpret the slope of the line.

Solution

The graph of the line is sketched in Figure 2.23. The slope m of this line is given by

$\begin{array}{l} m & = & \frac{rise}{run} = \frac{change in y -coordinates}{change in x -coordinates} \\ = & \frac{(y -coordinate of Q) - (y -coordinate of P)}{(x -coordinate of Q) - (x -coordinate of P)} \\ = & \frac{(3) - (- 1)}{(3) - (1)} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2 \end{array}$

Interpretation

A slope of 2 means that the value of y increases two units for every one unit increase in the value of x.

Practice Problem 1

Find and interpret the slope of the line containing the points $(- 7, 5)$ and $(6, - 3) .$

Figure 2.24 shows several lines passing through the origin. The slope of each line is labeled.

Warning

Be careful when using the formula for finding the slope of a line joining the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . Be sure to subtract coordinates in the same order.

You can use either

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} or m = \frac{y_{1} - y_{2}}{x_{1} - x_{2}},$

but it is incorrect to use

$m = \frac{y_{1} - y_{2}}{x_{2} - x_{1}} o r m = \frac{y_{2} - y_{1}}{x_{1} - x_{2}} .$

Figure 2.25 shows how to construct a line with given slope, $\pm \frac{2}{3}$ , with rise $= \pm 2$ and run $= 3$ .

Point–Slope Form

2 Write the point–slope form of the equation of a line.

We now find the equation of a line $ℓ$ that passes through the point $A (x_{1}, y_{1})$ and has slope m. Let P(x, y), with $x \neq x_{1},$ be any point in the plane. Then P(x, y) is on the line $ℓ$ if and only if the slope of the line passing through P(x, y) and $A (x_{1}, y_{1})$ is m. This is true if and only if

$\frac{y - y_{1}}{x - x_{1}} = m .$

See Figure 2.26.

Multiplying both sides by $x - x_{1}$ gives $y - y_{1} = m (x - x_{1})$ , which is also satisfied when $x = x_{1}$ and $y = y_{1} .$

Example 2 Finding an Equation of a Line with Given Point and Slope

Find the point–slope form of the equation of the line passing through the point $(1, - 2)$ with slope $m = 3 .$ Then solve for y.

Solution

Practice Problem 2

Find the point–slope form of the equation of the line passing through the point $(- 2, - 3)$ and with slope $- \frac{2}{3} .$ Then solve for y.

Example 3 Finding an Equation of a Line Passing Through Two Given Points

Find the point–slope form of the equation of the line $ℓ$ passing through the points $(- 2, 1)$ and (3, 7). Then solve for y.

Solution

We first find the slope m of the line $ℓ .$

$\begin{array}{l} m = \frac{7 - 1}{3 - (- 2)} = \frac{6}{3 + 2} = \frac{6}{5} & m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array}$

We use $m = \frac{6}{5}$ and either of the two given points when substituting into the point–slope form:

Practice Problem 3

Find the point–slope form of the equation of a line passing through the points $(- 3, - 4)$ and $(- 1, 6) .$ Then solve for y.

Side Note

You can avoid operations on fractions

$\begin{array}{rcl} y - 7 & = & \frac{6}{5} (x - 3) & \begin{array}{l} Multiply \\ by 5 . \end{array} \\ 5 (y - 7) & = & 6 (x - 3) \\ 5 y - 35 & = & 6 x - 18 \\ 5 y & = & 6 x + 17 \end{array}$

and at the end divide by 5

$y = \frac{6}{5} x + \frac{17}{5} .$

Slope–Intercept Form

3 Write the slope–intercept form of the equation of a line.

One of the most convenient forms of an equation describing a straight line is called the slope–intercept form, where we know the slope of the line and the given point is the y-intercept.

Example 4 Finding an Equation of a Line with a Given Slope and `y`-intercept

Find the point–slope form of the equation of the line with slope m and y-intercept b. Then solve for y.

Solution

Since the line has y-intercept b, the line passes through the point (0, b). See Figure 2.27.

$\begin{array}{rcll} y - y_{1} & = & m (x - x_{1}) & Point–slope form \\ y - b & = & m (x - 0) & Substitute x_{1} = 0 and y_{1} = b . \\ y - b & = & m x & Simplify. \\ y & = & m x + b & Solve for y . \end{array}$

Practice Problem 4

Find the point–slope form of the equation of the line with slope 2 and y-intercept $- 3 .$ Then solve for y.

The linear equation in the form $y = m x + b$ displays the slope m (the coefficient of x) and the y-intercept b (the constant term). The number m tells which way and how much the line is tilted; the number b tells where the line intersects the y-axis.

Example 5 Graph Using the Slope and `y`-intercept

Graph the line whose equation is $y = \frac{2}{3} x + 2 .$

Solution

The equation

$y = \frac{2}{3} x + 2$

is in the slope–intercept form with slope $\frac{2}{3}$ and y-intercept 2. To sketch the graph, find two points on the line and draw a line through the two points. Use the y-intercept as one of the points, and then use the slope to locate a second point. Since $m = \frac{2}{3},$ let 2 be the rise and 3 be the run. From the point (0, 2), move three units to the right (run) and two units up (rise). This gives $(0 + run, 2 + rise) = (0 + 3, 2 + 2) = (3, 4)$ as the second point.

The line we want joins the points (0, 2) and (3, 4) and is shown in Figure 2.28.

Figure 2.28

Locating a second point on a line

As an alternative solution, we can locate the second point on the line by choosing some value of x and finding the corresponding value of y directly from the given equation. Choosing $x = 3$ will give us the corresponding value of $y = \frac{2}{3} \cdot 3 + 2 = 4$ . This gives (3, 4) as a second point. Note that some choices of x are better than others. Choosing $x = 1$ would give us the point $(1, \frac{8}{3})$ , which would not be easy to plot accurately on a Cartesian grid.

Practice Problem 5

Graph the line with slope $- \frac{2}{3}$ that contains the point (0, 4).

Side Note

To graph a line with integer slope, such as 3, write $3 = \frac{3}{1}$ where the rise $= 3$ and run $= 1$ .

Equations of Horizontal and Vertical Lines

4 Recognize the equations of horizontal and vertical lines.

Consider the horizontal line through the point (h, k). The y-coordinate of every point on this line is k. So we can write its equation as $y = k .$ This line has slope $m = 0$ and y-intercept k. Similarly, an equation of the vertical line through the point (h, k) is $x = h .$ This line has undefined slope and x-intercept h. See Figure 2.29.

Figure 2.29

Horizontal and vertical lines

Example 6 Recognizing Horizontal and Vertical Lines

Discuss the graph of each equation in the xy-plane.

$y = 2$
$x = 4$

Solution

The equation $y = 2$ may be considered as an equation in two variables x and y by writing

$0 \cdot x + y = 2 .$

Any ordered pair of the form (x, 2) is a solution of this equation. Some solutions are $(- 1, 2), (0, 2), (2, 2),$ and (7, 2). It follows that the graph is a line parallel to the x-axis and two units above it, as shown in Figure 2.30. Its slope is 0.

Figure 2.30 Horizontal line
The equation $x = 4$ may be written as

$x + 0 \cdot y = 4 .$

Any ordered pair of the form (4, y) is a solution of this equation. Some solutions are $(4, - 5), (4, 0), (4, 2),$ and (4, 6). The graph is a line parallel to the y-axis and four units to the right of it, as shown in Figure 2.31. Its slope is undefined.

Figure 2.31

Vertical line

Practice Problem 6

Sketch the graphs of the lines $x = - 3$ and $y = 7 .$

Side Note

Sometimes it is easier to remember that

given the equation $y = 2$ , there is no change in y, since y is constant. The line has to be horizontal $(Δ y = 0)$ .
given the equation $x = 4$ , there is no change in x, since x is constant. The line has to be vertical $(Δ x = 0) .$

General Form of the Equation of a Line

5 Recognize the general form of the equation of a line.

An equation of the form

$a x + b y + c = 0,$

where a, b, and c are constants and a and b are not both zero, is called the general form of a linear equation. Consider two possible cases: $b \neq 0$ and $b = 0 .$

Suppose $b \neq 0$ : We can isolate y on one side and rewrite the equation in slope–intercept form.

$\begin{array}{rcll} a x + b y + c & = & 0 & Original equation \\ b y & = & - a x - c & Subtract a x + c from both sides. \\ y & = & - \frac{a}{b} x - \frac{c}{b} & Divide by b . \end{array}$

The result is an equation of a line in slope–intercept form with slope $m = - \frac{a}{b}$ and y-intercept equal to $- \frac{c}{b}$ .

Suppose $b = 0$ : We know that $a \neq 0$ , since not both a and b are zero, and we can solve for x.

$\begin{array}{rcll} a x + 0 \cdot y + c & = & 0 & Replace b with 0 . \\ a x + c & = & 0 & Simplify. \\ x & = & - \frac{c}{a} & Solve for x . \end{array}$

The graph of the equation $x = - \frac{c}{a}$ is a vertical line.

Example 7 Graphing a Linear Equation Using Intercepts

Find the slope, y-intercept, and x-intercept of the line with equation

$3 x - 4 y + 12 = 0 .$

Then sketch the graph.

Solution

First, solve for y to write the equation in slope–intercept form:

$\begin{array}{rcll} 3 x - 4 y + 12 & = & 0 & Original equation \\ 4 y & = & 3 x + 12 & Subtract 3 x + 12; multiply by - 1 . \\ y & = & \frac{3}{4} x + 3 & Divide by 4 . \end{array}$

This equation tells us that the slope m is $\frac{3}{4}$ and the y-intercept is 3. To find the x-intercept, set $y = 0$ in the original equation and obtain $3 x + 12 = 0$ , or $x = - 4 .$ So the x-intercept is $- 4 .$

We can sketch the graph of the equation if we can find two points on the graph. So we use the intercepts and sketch the line joining the points $(- 4, 0)$ and (0, 3). The graph is shown in Figure 2.32.

Figure 2.32

Graphing a line using intercepts

Alternatively, we can verify that the two-intercept form of the equation of the line with x-intercept a and y-intercept b is

$\frac{x}{a} + \frac{y}{b} = 1 .$

We can convert the general form of a linear equation such as $3 x - 4 y + 12 = 0$ to the two-intercept form by isolating the variables from the constant part and then dividing both sides of the equation by the constant.

$\begin{array}{rcll} 3 x - 4 y + 12 & = & 0 & Original equation \\ 3 x - 4 y & = & - 12 & Separate variables. \\ \frac{3 x}{- 12} + \frac{- 4 y}{- 12} & = & 1 & Divide by constant. \\ \frac{x}{- 4} + \frac{y}{3} & = & 1 & Simplify. \end{array}$

The last equation shows that the x-intercept is $- 4$ and the y-intercept is 3.

Practice Problem 7

Sketch the graph of the equation $3 x + 4 y = 24 .$

Forensic scientists use various clues to determine the identity of deceased individuals. Scientists know that certain bones serve as excellent sources of information about a person’s height. The length of the femur, the long bone that stretches from the hip (pelvis) socket to the kneecap (patella), can be used to estimate a person’s height. Knowing both the gender and race of the individual increases the accuracy of this relationship between the length of the bone and the height of the person.

Example 8 Inferring Height from the Femur

The height H of a human male is related to the length x of his femur by the formula

$H = 2.6 x + 65,$

where all measurements are in centimeters. The femur of Wild Bill Longley measured between 45 and 46 centimeters. Estimate the height of Wild Bill (in feet).

Solution

Substituting $x = 45$ and $x = 46$ into the preceding formula, we have possible heights

$\begin{array}{rcll} H_{1} & = & (2.6) (45) + 65 = 182 \\ and H_{2} & = & (2.6) (46) + 65 = 184.6 . \end{array}$

Therefore, Wild Bill’s height was between 182 and 184.6 centimeters.

Converting to feet, we conclude that his height was between $\frac{182}{30.48} \approx 5.97 ft$ and $\frac{184.6}{30.48} \approx 6.06 ft.$ He was approximately 6 feet tall.

Practice Problem 8

Suppose the femur of a person measures between 43 and 44 centimeters. Estimate the height of the person in meters.

Parallel and Perpendicular Lines

6 Find equations of parallel and perpendicular lines.

We can use slope to decide whether two lines are parallel, perpendicular, or neither.

In Exercises 135 and 136, you are asked to prove these assertions. Figure 2.33 shows examples of the slopes of a pair of parallel and a pair of perpendicular lines.

Procedure in Action

Example 9 Finding Equations of Parallel and Perpendicular Lines

OBJECTIVE	EXAMPLE
Let $ℓ : a x + b y + c = 0 .$ Find an equation of each line through the point $(x_{1},_{} y_{1}) :$ $ℓ_{1}$ parallel to $ℓ$ . $ℓ_{2}$ perpendicular to $ℓ$ .	Let $ℓ : 2 x - 3 y + 6 = 0 .$ Find a general form of the equation of each line through the point (2, 8): $ℓ_{1}$ parallel to $ℓ$ . $ℓ_{2}$ perpendicular to $ℓ$ .
Step 1 Find the slope `m` of $ℓ$ .	$\begin{array}{rcll} ℓ : 2 x - 3 y + 6 & = & 0 & Original equation \\ - 3 y & = & - 2 x - 6 & Subtract 2 x + 6 . \\ y & = & \frac{2}{3} x + 2 & Divide by - 3 . \end{array}$ The slope of $ℓ$ is $m = \frac{2}{3}$ .
Step 2 Write slope of $ℓ_{1}$ and $ℓ_{2} .$ The slope $m_{1}$ of $ℓ_{1}$ is `m`. The slope $m_{2}$ of $ℓ_{2}$ is $- \frac{1}{m} .$	$\begin{array}{l} m_{1} = \frac{2}{3} & ℓ_{1} parallel to ℓ \\ m_{2} = - \frac{3}{2} & ℓ_{2} perpendicular to ℓ \end{array}$
Step 3 Write equations of $ℓ_{1}$ and $ℓ_{2} .$ Use point–slope form to write equations of $ℓ_{1}$ and $ℓ_{2} .$ Simplify to write equations in the requested form.	$\begin{array}{rcll} ℓ_{1} : y - 8 & = & \frac{2}{3} (x - 2) & Point–slope form \\ 2 x - 3 y + 20 & = & 0 & Simplify. \\ ℓ_{2} : y - 8 & = & - \frac{3}{2} (x - 2) & Point–slope form \\ 3 x + 2 y - 22 & = & 0 & Simplify. \end{array}$

Practice Problem 9

Find the general form of the equation of the line
1. through $(- 2, 5)$ and parallel to the line containing (2, 3) and (5, 7).
2. through $(3, - 4)$ and perpendicular to the line $4 x + 5 y + 1 = 0 .$

Summary of Main Facts

Form	Equation	Slope	Other Information
General	$a x + b y + c = 0$ $a \neq 0,$ or $b \neq 0$	$- \frac{a}{b}$ if $b \neq 0$ Undefined if $b = 0$	`x`-intercept is $- \frac{c}{a}, a \neq 0$ . `y`-intercept is $- \frac{c}{b}, b \neq 0$ .
Point–slope	$y - y_{1} = m (x - x_{1})$	`m`	Line passes through $(x_{1}, y_{1}) .$
Slope–intercept	$y = m x + b$	`m`	`x`-intercept is $- \frac{b}{m}$ if $m \neq 0;$ `y`-intercept is `b`.
Two-intercept	$\frac{x}{a} + \frac{y}{b} = 1$	$- \frac{b}{a}$	`x`-intercept is a; `y`-intercept is b.
Vertical line through (`h`, `k`)	$x = h$	undefined	When $h \neq 0$ , no `y`-intercept; `x`-intercept is `h`. When $h = 0,$ the graph is the `y`-axis.
Horizontal line through (`h`, `k`)	$y = k$	0	When $k \neq 0,$ no `x`-intercept; `y`-intercept is `k`. When $k = 0,$ the graph is the `x`-axis.
Parallel and perpendicular lines: Two lines with slopes $m_{1}$ and $m_{2}$ are (a) parallel if $m_{1} = m_{2}$ and (b) perpendicular if $m_{1} \cdot m_{2} = - 1 .$

Modeling Data Using Linear Regression

7 Model data using linear regression.

Many relationships between variables can be modeled using linear equations. The success of modeling many phenomena by linear equations can be easily explained. On a local scale, many nonlinear equations may be reduced to linear equations by assuming that the variables change only to a small extent. We experience this in our daily life when we perceive the surface of Earth locally as flat. That means that any short-term data can be accurately modeled by linear models. A line that best fits the given data is called a regression line. To illustrate the method, we use Table 2.5, which gives the data for the number of registered motorcycles in the United States for the years 2006–2012.

Table 2.5

Year	2006	2007	2008	2009	2010	2011	2012
Registered motorcycles (millions)	6.7	7.1	7.8	7.9	8.4	8.8	9.5

The scatter diagram for the motorcycle data is shown in Figure 2.34(a), and the line used to model the data is added to the scatter diagram in Figure 2.34(b). The years 2006–2012 are represented by the numbers 0–6, respectively.

The line is called the regression line or the least-squares line. See Figure 2.35. The slope-intercept form of the equation for the line in Figure 2.35(b) is $y = 0.44 x + 6.70$ , where x is the number of years after 2006, and the slope and intercept values are rounded to the nearest hundredth. The equation is found by using either a graphing calculator or statistical formulas.

Figure 2.35

Least-squares fit of the curve $y = a x + b$ to 10 given data points

Side Note

Out of all possible linear fits, the least-squares regression line is the one that minimizes the sum of the squares made from the vertical offsets.

Example 10 Predictions Using Linear Regression

Use the equation $y = 0.44 x + 6.70$ to predict the number of registered motorcycles in the United States for 2015.

Solution

Because 2015 is nine years after 2006, we set $x = 9$ ; then $y = 0.44 (9) + 6.70 \approx 10.66$ .

So according to the regression prediction, 10.66 million motorcycles will be registered in the United States for 2015. The quality of a prediction based on linear regression depends on how close the relationship between the data quantities is to being linear.

Practice Problem 10

Repeat Example 10 to predict the number of registered motorcycles in the United States for 2016.

The following Technology Connection shows how a graphing calculator displays the scatter diagram, regression line, and regression line equation.

Technology Connection

You can use a graphing utility to investigate the linear regression model for data points that lie on or near a straight line. On the TI-84, the data for Example 10 are entered in five steps.

Step 1 Enter the data.

Press $STAT$ 1 for STAT Edit. Enter the values of the independent variable x in L1 and the corresponding values of the dependent variable y in L2.
Step 2 Turn on STAT Plot

Press $2^{nd} Y =$ 1 for Plot 1. Select On; then choose Type. $⋰ ⋰$

Choose X List: L1; Y list: L2; and mark: $\cdot$ .
Step 3 Graph the scatterplot. Press $Graph$ .
Step 4 Find the equation of the regression line. Press $STAT$ and highlight CALC. Press 4 for LinReg $(a x + b)$ . Press $ENTER$ .
Step 5 Graph the regression line with the scatterplot.

Press $Y =$ .

Press $VARS$ 5 for statistics. Highlight EQ and press $ENTER$ . (You should see the regression equation in the $Y = window .$ )

Press $ZOOM$ 9 for ZoomStat. (You should see the data graphed along with the regression line.)

Section 2.3 Exercises

Concepts and Vocabulary

The slope of a horizontal line is ; the slope of vertical line is .
The slope of the line passing through the points $P = (x_{1}, y_{1})$ and $Q = (x_{2}, y_{2})$ is given by the formula $m =$ .
Every line parallel to the line $y = 3 x - 2$ has slope, m, equal to .
Every line perpendicular to the line $y = 3 x - 2$ has slope, m, equal to .
True or False. The slope of the line $y = - \frac{1}{4} x + 5$ is equal to $\frac{1}{4} .$
True or False. The y-intercept of the line $y = 2 x - 3$ is equal to 3.
True or False. The graph of the line $y = 4$ is a horizontal line.
True or False. The graph of the line $x = - 5$ is a vertical line.

Building Skills

In Exercises 9–16, find the slope of the line through the given pair of points. Without plotting any points, state whether the line is rising, falling, horizontal, or vertical.

(1, 3), (4, 7)
(0, 4), (2, 0)
$(3, - 2), (6, - 2)$
$(- 3, 7), (- 3, - 4)$
$(0.5, 2), (3, - 3.5)$
$(3, - 2), (2, - 3)$
$(\sqrt{2}, 1), (1 + \sqrt{2}, 5)$
$(1 - \sqrt{3}, 0), (1 + \sqrt{3}, 3 \sqrt{3})$

In the figure, identify the line with the given slope m.

$m = 1$
$m = - 1$
$m = 0$
m is undefined.

In the figure below, find the slope of each line. (The scale is the same on both axes.)

$ℓ_{1}$
$ℓ_{2}$
$ℓ_{3}$
$ℓ_{4}$

In Exercises 25–32, find an equation in slope–intercept form of the line that passes through the given point and has slope m. Also sketch the graph of the line by locating the second point with the rise-and-run method.

$(0, 5); m = 3$
$(0, 9); m = - 2$
$(0, 4); m = \frac{1}{2}$
$(0, 4); m = - \frac{1}{2}$
$(2, 1); m = - \frac{3}{2}$
$(- 1, 0); m = \frac{2}{5}$
$(5, - 4); m = 0$
$(5, - 4); m$ is undefined.

In Exercises 33–54, use the given conditions to find an equation in slope–intercept form of each nonvertical line. Write vertical lines in the form $x = h .$

Passing through (0, 1) and (1, 0)
Passing through (0, 1) and (1, 3)
Passing through $(- 1, 3)$ and (3, 3)
Passing through $(- 5, 1)$ and (2, 7)
Passing through $(- 2, - 1)$ and (1, 1)
Passing through $(- 1, - 3)$ and $(6, - 9)$
Passing through $(\frac{1}{2}, \frac{1}{4})$ and (0, 2)
Passing through $(4, - 7)$ and (4, 3)
A vertical line through (5, 1.7)
A horizontal line through (1.4, 1.5)
A horizontal line through (0, 0)
A vertical line through (0, 0)
$m = 0; y -intercept = 14$
$m = 2; y -intercept = 5$
$m = - \frac{2}{3}; y -intercept = - 4$
$m = - 6; y -intercept = - 3$
$x -intercept = - 3; y -intercept = 4$
$x -intercept = - 5; y -intercept = - 2$
Parallel to $y = 5;$ passing through (4, 7)
Parallel to $x = 5;$ passing through (4, 7)
Perpendicular to $x = - 4;$ passing through $(- 3, - 5)$
Perpendicular to $y = - 4;$ passing through $(- 3, - 5)$

In Exercises 55–64, find the slope and intercepts from the equation of the line. Sketch the graph of each equation.

$y = 3 x - 2$
$y = - 2 x + 3$
$x + 2 y - 4 = 0$
$x = 3 y - 9$
$3 x - 2 y + 6 = 0$
$2 x = - 4 y + 15$
$x - 5 = 0$
$2 y + 5 = 0$
$x = 0$
$y = 0$

In Exercises 65–68, use the two-intercept form of the equation of a line.

Find an equation of the line whose x-intercept is 4 and y-intercept is 3.
Find an equation of the line whose x-intercept is $- 3$ and y-intercept is 2.
Find the x- and y-intercepts of the graph of the equation $2 x + 3 y = 6$ .
Repeat Exercise 67 for the equation $3 x - 4 y + 12 = 0 .$

In Exercises 69–72, find the x- and y-intercepts and sketch the graph of the equation

$2 x - 3 y = 12$
$3 x - 4 y = 12$
$- 5 x + 2 y = 10$
$- 4 x + 5 y = 20$
Find an equation of the line passing through the points (2, 4) and (7, 9). Use the equation to show that the three points (2, 4) and (7, 9), and $(- 1, 1)$ are on the same line (collinear).
Use the technique of Exercise 73 to check whether the points $(7, 2), (2, - 3),$ and (5, 1) are on the same line.
Write the slope–intercept form of the equation of the line that passes through the origin and is parallel to the red line shown in the figure.
Write the slope–intercept form of the equation of the line that passes through the origin and is perpendicular to the red line shown in the figure in Exercise 75.
Write the slope–intercept form of the equation of the blue line shown in the figure that is parallel to the red line.
Write the slope–intercept form of the equation of the green line shown in the figure that is perpendicular to the red line in Exercise 77.

In Exercises 79–88, determine whether each pair of lines is parallel, perpendicular, or neither.

$y = 3 x - 1$ and $y = 3 x + 2$
$y = 2 x + 2$ and $y = - 2 x + 2$
$y = 2 x - 4$ and $y = - \frac{1}{2} x + 4$
$y = 3 x + 1$ and $y = \frac{1}{3} x - 1$
$3 x + 8 y = 7$ and $5 x - 7 y = 0$
$10 x + 2 y = 3$ and $5 x + y = - 1$
$x = 4 y + 8$ and $y = - 4 x + 1$
$y = 3 x + 1$ and $6 y + 2 x = 0$
$x = - 1$ and $2 x + 7 = 9$
$2 x + 3 y = 7$ and $y = 2$

In Exercises 89–104, find the equation of the line in slope–intercept form satisfying the given conditions.

Parallel to a line with slope 3; passing through $(2, - 3) .$
Parallel to a line with slope $- 2;$ passing through $(- 1, 3) .$
Perpendicular to a line with slope $- \frac{1}{2}$ ; passing through $(- 1, 2) .$
Perpendicular to a line with slope $\frac{1}{3}$ ; passing through $(2, - 1) .$
Parallel to a line joining points $(1, - 2)$ and $(- 3, 2);$ passing through $(- 2, - 5) .$
Parallel to a line joining points $(- 2, 1)$ and (3, 5); passing through (1, 2).
Perpendicular to a line joining points $(- 3, 2)$ and $(- 4, - 1);$ passing through $(1, - 2) .$
Perpendicular to a line joining points (2, 1) and $(4, - 1);$ passing through $(- 1, 2) .$
Parallel to $y = 6 x + 5$ ; y-intercept 4.
Parallel to $y = - \frac{1}{2} x + 5$ ; y-intercept 2.
Perpendicular to $y = 6 x + 5$ ; y-intercept 4.
Perpendicular to $y = - \frac{1}{2} x + 5$ ; y-intercept $- 4 .$
Parallel to $x + y = 1$ ; passing through (1, 1).
Parallel to $- 2 x + 3 y = 7$ ; passing through (1, 0).
Perpendicular to $3 x - 9 y = 18$ ; passing through $(- 2, 4) .$
Perpendicular to $- 2 x + y = 14$ ; passing through (0, 2).

Applying the Concepts

Dance Class. A dance teacher runs a modern dance class for 12 weeks. The class meets each week in a rented dance studio and the teacher has to pay upfront all the costs associated with renting the studio. Each student pays the teacher a single fee for the course. The graph represents the profit the teacher makes depending on the number of students.
1. What does the y-intercept represent?
2. What does the x-intercept represent?
3. What is the teacher’s profit if there are 16 students in the class?
4. Write a linear equation for the profit P the teacher makes depending on the number of students n.
Prepaid Cellphone. A prepaid cellphone is loaded with a prepaid SIM card with initial airtime credit for calls. The company charges a flat fee per minute of calling within the continental United States. The graph represents the prepaid amount depending on usage of the cellphone.
1. What does the y-intercept represent?
2. What does the x-intercept represent?
3. Write a linear equation for the amount left on the SIM card P depending on the time the cellphone was used.
4. What is the cost per minute of calling from this cellphone?
Jet takeoff. Upon takeoff, a jet climbs to 4 miles as it passes over 40 miles of land below it. Find the slope of the jet’s climb.
Road gradient. Driving down the Smoky Mountains in Tennessee, Samantha descends 2000 feet in elevation while moving 4 miles horizontally away from a high point on the straight road. Find the slope (gradient) of the road $(1 mile = 5280 feet) .$
Trimming the bush. John noticed that it is springtime and that the hedge by his window has been growing. It grew 8 inches in two weeks. How often does John need to trim his hedge to be sure that it doesn’t grow more than 6 inches between trimmings?
Taking a bath. Miguel is preparing to take a bath. He sees that it takes 2 minutes to raise the bathtub water level by 5 inches. His bathtub is 36 inches deep, so he has 31 more inches left before water overflows. How much time does he have left before the water overflows?

In Exercises 111–115, write an equation of a line in slope–intercept form. To describe this situation, interpret (a) the variables x and y and (b) the meaning of the slope and the y-intercept.

Christmas savings account. You currently have $130 in a Christmas savings account. You deposit $7 per week into this account. (Ignore interest.)
Golf club charges. Your golf club charges a yearly membership fee of $1000 and $35 per session of golf.
Paying off a refrigerator. You bought a new refrigerator for $700. You made a down payment of $100 and promised to pay $15 per month. (Ignore interest.)
Converting currency. In 2012, according to the International Monetary Fund, 1 U.S. dollar equaled 50.5 Indian rupees. Convert currency from rupees to dollars.
Life expectancy. In 2010, the life expectancy of a female born in the United States was 80.8 years and was increasing at a rate of 0.17 year per year. (Assume that this rate of increase remains constant.)
Depreciating a tractor. The value $υ$ of a tractor purchased for $14,000 and depreciated linearly at the rate of $1400 per year is given by $υ = - 1400 t + 14, 000,$ where t represents the number of years since the purchase. Find the value of the tractor after (a) two years and (b) six years. When will the tractor have no value?
Data usage. Your monthly data limit on your cellphone plan is 1 gigabyte $(1 GB = 1024 MB) .$ By the end of June 13, you realized you have used up 435 MB of your quota. Assuming you plan to keep your usage patterns, will your limit be enough, or will you have to buy extra data by the end of June?
Car trip. You set out on a trip from your hometown to a vacation spot located 520 miles away. You started driving at 9 a.m. and by 12 p.m. you have realized you traveled 195 miles. Assuming your driving patterns continue, what time will you arrive at your destination?
Playing a concert. A famous band is considering playing a concert and charging $40,000 plus $5 per person attending the concert. Write an equation relating the income y of the band to the number x of people attending the concert.
Car rental. The U Drive car rental agency charges $30.00 per day and $0.25 per mile.
1. Find an equation relating the cost C of renting a car from U Drive for a one-day trip covering x miles.
2. Draw the graph of the equation in part (a).
3. How much does it cost to rent the car for one day covering 60 miles?
4. How many miles were driven if the one-day rental cost was $47.75?
Fahrenheit and Celsius. In the Fahrenheit temperature scale, water boils at $212 ° F$ and freezes at $32 ° F .$ In the Celsius scale, water boils at $100 ° C$ and freezes at $0 ° C .$ Assume that the Fahrenheit temperature F and Celsius temperature C are linearly related.
1. Find the equation in the slope–intercept form relating F and C, with C as the independent variable.
2. What is the meaning of slope in part (a)?
3. Find the Fahrenheit temperatures, to the nearest degree, corresponding to $40 ° C, 25 ° C, - 5 ° C,$ and $- 10 ° C .$
4. Find the Celsius temperatures, to the nearest degree, corresponding to $100 ° F, 90 ° F, 75 ° F, - 10 ° F,$ and $- 20 ° F .$
5. The normal body temperature in humans ranges from $97.6 ° F$ to $99.6 ° F .$ Convert this temperature range to degrees Celsius.
6. When is the Celsius temperature the same numerical value as the Fahrenheit temperature?
Cost. The cost of producing four modems is $210.20, and the cost of producing ten modems is $348.80.
1. Write a linear equation in slope–intercept form relating cost to the number of modems produced. Sketch a graph of the equation.
2. What is the practical meaning of the slope and the intercepts of the equation in part (a)?
3. Predict the cost of producing 12 modems.
Female prisoners. The number of females in a state’s prisons rose from 2425 in 2005 to 4026 in 2011.
1. Find a linear equation relating the number y of women prisoners to the year t. (Take $t = 0$ to represent 2005.)
2. Draw the graph of the equation from part (a).
3. How many women prisoners were there in 2008?
4. Predict the number of women prisoners in 2017.
Viewers of a TV show. After five months, the number of viewers of The Weekly Show on TV was 5.73 million. After eight months, the number of viewers of the show rose to 6.27 million. Assume that the linear model applies.
1. Write an equation expressing the number of viewers V after x months.
2. Interpret the slope and the intercepts of the line in part (a).
3. Predict the number of viewers of the show after 11 months.
Health insurance coverage. In Michigan, 11.7% of the population was not covered by health insurance in 2000. In 2005, the percentage of uninsured people rose to 12.7%. Use the linear model to predict the percentage of people in Michigan that were not covered in 2010.

(Source: Michigan Department of Community Health)
Oil consumption. The total world consumption of oil increased from 82.7 million barrels per day in 2004 to 84.2 million barrels per day in 2009. Create a linear model involving a relation between the year and the daily oil consumption in that year. Let $t = 0$ represent the year 2004.

(Source: www.cia.gov)
Newspaper sales. The table shows the number of newspapers (in thousands) that sell at a given price (in nickels) per copy for the college newspaper Midstate Oracle at Midstate College.

Price (nickels) 1 2 3 4 5

Number of Newspapers Sold (thousands) 11 8 6 4 3
1. Use a graphing utility to find the equation of the line of regression relating the variable x (price in nickels) and y (number sold in thousands).
2. Use a graphing utility to plot the points and graph the regression line.
3. Approximately how many papers will be sold if the price per copy is 35¢?
Sales and advertising. Bildwell Computers keeps a careful record of its weekly advertising expenses (in thousands of dollars) and the number of computers (in thousands) it sells the following week. The table shows the data the company collected.

Weekly Advertising Expenses (in thousands of dollars) 100 200 300 400 500 600

Sales the Following Week (in thousands) 21 29 36 49 58 67
1. Use a graphing utility to find the equation of the line of regression equation relating the variables x (weekly advertising expenses in thousands of dollars) and y (number of computers sold, in thousands).
2. Use a graphing utility to plot the points and graph the regression line.
3. Approximately how many thousands of computers will be sold if the company spends $700,000 on advertising?

Beyond the Basics

The graph of a line has slope $m = 3 .$ If $P (- 2, 3)$ and Q(1, c) are points on the graph, find c.
Find a real number c such that the line $3 x - c y - 2 = 0$ has a y-intercept of $- 4 .$

In Exercises 131–132, show that the three given points are collinear by using (a) slopes and (b) the distance formula.

(0, 1), (1, 3), and $(- 1, - 1)$
(1, 0.5), (2, 0), and (0.5, 0.75)
Show that the triangle with vertices $A (1, 1), B (- 1, 4),$ and C(5, 8) is a right triangle by using (a) slopes and (b) the Pythagorean Theorem.
Show that the four points $A (- 4, - 1),$ $B (1, 2), C (3, 1),$ and, $D (- 2, - 2)$ are the vertices of a parallelogram.

[Hint: Opposite sides of a parallelogram are parallel.]
In the accompanying figure, note that lines $ℓ_{1}$ and $ℓ_{2}$ have slopes $m_{1}$ and $m_{2}$ . Assume that $ℓ_{1}$ and $ℓ_{2}$ are parallel. Deduce that ABCD is a parallelogram and that triangles ABD and CDB are congruent, with $A B = C D .$ Deduce that $m_{1} = m_{2}$ .
In the accompanying figure, note that lines $ℓ_{1}$ and $ℓ_{2}$ have slopes $m_{1}$ and $m_{2}$ . Assume that $ℓ_{1}$ and $ℓ_{2}$ are perpendicular. Show that the right triangles OKA and BLO are similar and from that deduce that $m_{1} \cdot m_{2} = - 1$ .
Geometry. Use a coordinate plane to prove that if two sides of a quadrilateral are congruent (equal in length) and parallel, then so are the other two sides.
Geometry. Use a coordinate plane to prove that the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram.
Geometry. Find the coordinates of the point B of a line segment AB, given that $A = (2, 2), d (A, B) = 12.5,$ and the slope of the line segment AB is $\frac{4}{3} .$
Geometry. Show that an equation of a circle with $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ as endpoints of a diameter can be written in the form

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0 .$

[Hint: Use the fact that when the two endpoints of a diameter and any other point on the circle are used as vertices for a triangle, the angle opposite the diameter is a right angle.]
Geometry. Find an equation of the tangent line to the circle $x^{2} + y^{2} = 169$ at the point $(- 5, 12)$ . (See the accompanying figure.)
Geometry. Show that $x x_{1} + y y_{1} = a^{2}$ is an equation of the tangent line to the circle $x^{2} + y^{2} = a^{2}$ at the point $(x_{1}, y_{1})$ on the circle.

In Exercises 143–146, use a graphing device to graph the given lines on the same screen. Interpret your observations about the family of lines.

$y = 2 x + b$ for $b = 0, \pm 2, \pm 4$
$y = m x + 2$ for $m = 0, \pm 2, \pm 4$
$y = m (x - 1)$ for $m = 0, \pm 3, \pm 5$
$y = m (x + 1) - 2$ for $m = 0, \pm 3, \pm 5$

Critical Thinking / Discussion / Writing

1. The equation $y + 2 x + k = 0$ describes a family of lines for each value of k. Sketch the graphs of the members of this family for $k = - 3, 0,$ and 2. What is the common characteristic of the family?
2. Repeat part (a) for the family of lines
  
  $y + k x + 4 = 0 .$
Explain how you can determine the sign of the x-coordinate of the point of intersection of the lines $y = m_{1} x + b_{1}$ and $y = m_{2} x + b_{2}$ , if
1. $m_{1} > m_{2} > 0$ and $b_{1} > b_{2} .$
2. $m_{1} > m_{2} > 0$ and $b_{1} < b_{2} .$
3. $m_{1} < m_{2} < 0$ and $b_{1} > b_{2} .$
4. $m_{1} < m_{2} < 0$ and $b_{1} < b_{2} .$

Getting Ready for the Next Section

In Exercises 149–152, solve each equation for x.

$x^{2} - 4 = 0$
$1 - x^{2} = 0$
$x^{2} - x - 2 = 0$
$x^{2} + 2 x - 3 = 0$

In Exercises 153–156, simplify each expression.

$(3 (a + h) + 1) - (3 a + 1)$
$(2 {(a + h)}^{2} + 1) - (2 a^{2} + 1)$
$\frac{- {(a + h)}^{2} + a^{2}}{h}$
$\frac{1}{h} (\frac{1}{a + h} - \frac{1}{a})$

In Exercises 157–158, solve each inequality.

$(x - 1) (x - 3) < 0$
$x^{2} - 2 x - 3 \geq 0$

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Price (nickels)	1	2	3	4	5
Number of Newspapers Sold (thousands)	11	8	6	4	3

Weekly Advertising Expenses (in thousands of dollars)	100	200	300	400	500	600
Sales the Following Week (in thousands)	21	29	36	49	58	67

Table of Contents for Section 2.3 Lines

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Table of Contents for
Section 2.3 Lines