In Section 1.3, we developed the slope–intercept equation $y=mx+b$, or $f(x)=mx+b$. If we know the slope and the y-intercept of a line, we can find an equation of the line using the slope–intercept equation.

Another formula that can be used to determine an equation of a line is the point–slope equation. Suppose that we have a nonvertical line and that the coordinates of point $P_1$ on the line are $(x_1,y_1)$. We can think of $P_1$ as fixed and imagine another point P on the line with coordinates $(x,y)$. Thus the slope is given by

Multiplying by $x-x_1$ on both sides, we get the point–slope equation of the line:

If we know the slope of a line and the coordinates of one point on the line, we can find an equation of the line using either the point–slope equation,

or the slope–intercept equation,

Example 3

Find an equation of the line containing the points (2, 3) and (1, −4).

Solution

We first determine the slope:

$$m=\frac{-4-3}{1-2}=\frac{-7}{-1}=7.$$

Using the Point–Slope Equation: We substitute 7 for m and either of the points (2, 3) or (1, −4) for $(x_1,y_1)$ in the point–slope equation. In this case, we use $(2,3)$.

$$\begin{array}{rcll}y-y_1& =& m(x-x_1)& \mathbf{\text{Point-slope equation}}\\ y-3& =& 7(x-2)& \mathbf{\text{Substituting}}\\ y-3& =& 7x-14& \\ y& =& 7x-11,\text{or}& \\ f(x)& =& 7x-11& \end{array}$$

Using the Slope–Intercept Equation: We substitute 7 for m and either of the points (2, 3) or (1, −4) for (x, y) in the slope–intercept equation and solve for b. Here we use (1, −4).

$$\begin{array}{rcll}y& =& mx+b& \mathbf{\text{Slope-intercept equation}}\\ -4& =& 7\cdot 1+b& \mathbf{\text{Substituting}}\\ -4& =& 7+b& \\ -11& =& b& \mathbf{\text{Solving for }}\mathit{b}\end{array}$$

We substitute 7 for m and −11 for b in $y=mx+b$ to get

$$\begin{array}{rcl}y& =& 7x-11,\text{or}\\ f(x)& =& 7x-11.\end{array}$$

Now Try Exercise 19.

Can we determine whether the graphs of two linear equations are parallel without graphing them? Let’s look at three pairs of equations and their graphs.

If two different lines, such as $x=-4$ and $x=-2.5$, are vertical, then they are parallel. Thus two equations such as $x=a_1$ and $x=a_2$, where $a_1\ne a_2$, have graphs that are parallel lines. Two nonvertical lines, such as $y=2x+4$ and $y=2x-3$, or, in general, $y=mx+b_1$ and $y=mx+b_2$, where the slopes are the same and $b_1\ne b_2$, also have graphs that are parallel lines.

Can we examine a pair of equations to determine whether their graphs are perpendicular without graphing the equations? Let’s look at the following pairs of equations and their graphs.

If one line is vertical and another is horizontal, they are perpendicular. For example, the lines $x=5$ and $y=-3$ are perpendicular. Otherwise, how can we tell whether two lines are perpendicular? Consider a line $\overleftrightarrow{AB}$, as shown in the figure at left, with slope $a/b$. Then think of rotating the line ${90}^{°}$ to get a line $\overleftrightarrow{A_1{B}_1}$ perpendicular to $\overleftrightarrow{AB}$. For the new line, the rise and the run are interchanged, but the run is now negative. Thus the slope of the new line is $-b/a$, which is the opposite of the reciprocal of the slope of the first line. Also note that when we multiply the slopes, we get

This is the condition under which lines will be perpendicular.

If a line has slope $m_1$, the slope $m_2$ of a line perpendicular to it is $-1/m_1$. The slope of one line is the opposite of the reciprocal of the other:

When a real-world situation can be described in mathematical language, we have a mathematical model. For example, the natural numbers constitute a mathematical model for situations in which counting is essential. Situations in which algebra can be brought to bear often require the use of functions as models.

Mathematical models are abstracted from real-world situations. The mathematical model gives results that allow one to predict what will happen in that real-world situation. If the predictions are inaccurate or the results of experimentation do not conform to the model, the model must be changed or discarded.

Mathematical modeling can be an ongoing process. For example, finding a mathematical model that will provide an accurate prediction of population growth is not a simple problem. Any population model that one might devise would need to be reshaped as further information is acquired.

We will develop and use many kinds of mathematical models in this text. In this chapter, we have used linear functions as models. Other types of functions, such as quadratic, cubic, and exponential functions, can also model data. These functions are nonlinear.

In general, we try to find a function that fits, as well as possible, observations (data), theoretical reasoning, and common sense. We call this curve fitting; it is one aspect of mathematical modeling.

Let’s look at some data and related graphs, or scatterplots, and determine whether a linear function seems to fit the set of data.

Year, x	Gross Domestic Product (GDP) (in trillions)	Scatterplot
1990, 0	$6.0	It appears that the data points can be represented or modeled by a linear function. The graph is linear.
1995, 5	7.7
2000, 10	10.3
2005, 15	13.1
2010, 20	15.0
2011, 21	15.5
2012, 22	16.2
Sources: Bureau of Economic Analysis, U.S. Department of Commerce

Year, x	Estimated Number of Alternative-Fueled Vehicles (in thousands)	Scatterplot
1995, 0	247	It appears that the data points cannot be modeled accurately by a linear function. The graph is nonlinear.
1997, 2	280
1999, 4	322
2001, 6	425
2003, 8	534
2005, 10	592
2007, 12	696
2009, 14	826
2011, 16	1192
Source: U.S. Energy Information Administration

Looking at the scatterplots, we see that the data on gross domestic product seem to be rising in a manner to suggest that a linear function might fit, although a “perfect” straight line cannot be drawn through the data points. A linear function does not seem to fit the data on alternative-fueled vehicles.

Example 6

U.S. Gross Domestic Product. The gross domestic product (GDP) of a country is the market value of final goods and services produced. Market value depends on the quantity of goods and services and their price. Model the data in the table above on the U.S. Gross Domestic Product with a linear function. Then estimate the GDP in 2018.

Solution

We can choose any two of the data points to determine an equation. Note that the first coordinate is the number of years since 1990 and the second coordinate is the corresponding GDP in trillions of dollars. Let’s use (5, 7.7) and (21, 15.5).

We first determine the slope of the line:

$$m=\frac{15.5-7.7}{21-5}=\frac{7.8}{16}=\mathrm{0.4875.}$$

Then we substitute 0.4875 for m and either of the points (5, 7.7) or (21, 15.5) for $(x_1,y_1)$ in the point–slope equation. In this case, we use (5, 7.7). We get

$$\begin{array}{rcll}y-y_1& =& m(x-x_1)& \mathbf{\text{Point-slope equation}}\\ y-7.7& =& 0.4875(x-5),& \mathbf{\text{Substituting}}\end{array}$$

which simplifies to

$$y=0.4875x+5.2625,$$

where x is the number of years after 1990 and y is in trillions of dollars.

Next, we estimate the GDP in 2018 by substituting 28 $(2018-1990=28)$ for x in the model:

$$\begin{array}{rcll}y& =& 0.4875x+5.2625& \mathbf{\text{Model}}\\ & =& 0.4875(28)+5.2625& \mathbf{\text{Substituting}}\\ & =& 18.9125\approx \mathrm{18.91.}& \end{array}$$

We estimate that the gross domestic product will be $18.91 trillion in 2018.

Now Try Exercise 61.

In Example 6, if we were to use the data points $(0,6.0)$ and $(20,15.0)$, our model would be

and our estimate for the GDP in 2018 would be $18.60 trillion, about $0.31 trillion less than the estimate provided by the first model. This illustrates that a model and the estimates that it produces are dependent on the data points used.

Models that consider all the data points, not just two, are generally better models. The model that best fits the data can be found using a graphing calculator and a procedure called linear regression. This procedure is explained in the following Technology Connection.

The following scatterplots summarize the interpretation of a correlation coefficient.