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Section 4.5 Point-Slope Formula

Subsection Finding an Equation for a Line

The slope of a line is the same, no matter which points we use to compute it. Therefore, if we know the slope of a line and any one point on the line, then all other points on the line must satisfy the slope formula:

\begin{equation*} \blert{\dfrac{y-y_1}{x-x_1} = m} \end{equation*}

This is, in fact, an equation for the line.

Example 4.31.
  1. Graph the line that passes through the point \((1,3)\) and has slope \(-2\text{.}\)
  2. Find an equation for the line that passes through the point \((1,3)\) and has slope \(-2\text{.}\)
Solution
  1. We plot the point \((1,3)\text{,}\) then use the slope, \(-2\text{,}\) to find another point on the line. From the point \((1,3)\text{,}\) we move 2 units down and 1 unit to the right, arriving at \((2,1)\text{.}\) We draw the line through these two points. The graph is shown below.

    graph of line
  2. We use the formula

    \begin{equation*} \dfrac{y-y_1}{x-x_1} = m \end{equation*}

    with \(\dfrac{\Delta y}{\Delta x}=\dfrac{-2}{1}\) and \((x_1,y_1)=(\alert{1},\alert{3})\) to get

    \begin{equation*} \dfrac{-2}{1}=\dfrac{y-\alert{3}}{x-\alert{1}} \end{equation*}

    To simplify the equation, we cross-multiply.

    \begin{equation*} \begin{aligned} 1(y-3) \amp = -2(x-1) \amp\amp \blert{\text{Apply the distributive law.}}\\ y-3\amp = -2x+2 \amp\amp \blert{\text{Add 3 to both sides.}}\\ y \amp = -2x+5 \end{aligned} \end{equation*}

    You can verify on the graph that \(y=-2x+5\) is an equation for the line.

In Example 4.31 we used the slope formula in a new way: to find the equation of a line. We can simplify the formula by clearing the denominator.

\begin{equation*} \blert{(x-x_1)}\dfrac{y-y_1}{x-x_1} = m\blert{(x-x_1)} \end{equation*}

to get

\begin{equation*} y-y_1 = m(x-x_1) \end{equation*}

We call this the point-slope formula for linear equations.

Point-Slope Formula.

To find an equation for the line of slope \(m\) passing through the point \((x_1,y_1)\text{,}\) we use the point-slope formula

\begin{equation*} \blert{\dfrac{y-y_1}{x-x_1} = m} \end{equation*}

or

\begin{equation*} \blert{y-y_1 = m(x-x_1)} \end{equation*}
Look Closer.

What is the difference between the slope formula and the point-slope formula?

\begin{equation*} \begin{aligned} \text{the slope formula} \amp \amp\amp m=\dfrac{y_2-y_1}{x_2-x_1}\\ \text{the point-slope formula}\amp \amp\amp m=\dfrac{y-y_1}{x-x_1} \end{aligned} \end{equation*}

They are really the same formula, but they are used for different purposes:

  • We use the slope formula to calculate the slope of a line when we know two points on the line. That is, we know \((x_1,y_1)\) and \((x_2,y_2)\text{,}\) and we are looking for \(m\text{.}\)
  • We use the point-slope formula to find the equation of a line. That is, we know \((x_1,y_1)\) and \(m\text{,}\) and we are looking for \(y=mx+b\text{.}\)

Reading Questions Reading Questions

1.

Give two versions of the point-slope formula

2.

What is the point-slope formula used for?

Example 4.32.

Find an equation for the line that passes through the point \((1,4)\) and is perpendicular to the line \(4x-2y=6\text{.}\)

Solution

We know one point on the required line, but we also need to know its slope. Now, our line is perpendicular to the given line, and the slope of the given line is \(~m_1=2~\) (as you can check by putting its equation into slope-intercept form). So the slope of our line is

\begin{equation*} m_2=\dfrac{-1}{m_1} = \dfrac{-1}{2} \end{equation*}

Now we can use the point-slope formula with \(~m_1=\dfrac{-1}{2}\) and \((x_1,y_1) = (1,4)\text{.}\)

\begin{equation*} \begin{aligned} y-y_1 \amp = m(x-x_1) \\ y-\alert{4}\amp = \alert{\dfrac{-1}{2}}(x-\alert{1}) \amp\amp \blert{\text{Apply the distributive law.}}\\ y - 4 \amp = \dfrac{-1}{2}x + \dfrac{1}{2} \amp\amp \blert{\text{Add 4 to both sides.}}\\ y \amp = -\dfrac{-1}{2}x + \dfrac{9}{2} \amp\amp \blert{\dfrac{1}{2}+4=\dfrac{1}{2}+\dfrac{8}{2}= \dfrac{9}{2}} \end{aligned} \end{equation*}

The given line and the perpendicular line are shown in the figure.

perpendicular lines

Caution 4.33.

If we happen to know the \(y\)-intercept of a line, we can write its equation using the slope-intercept formula, \(y=mx+b\text{.}\) But if we don't know the \(y\)-intercept and instead know a different point, it is easier to use the point-slope formula.

Subsection The Line Through Two Points

How many lines pass through two given points? There is only one. We can use the point-slope formula to can find its equation.

Example 4.34.

Find an equation for the line that passes through \((2,-1)\) and \((-1,3)\text{.}\)

Solution

We solve this problem in two steps: First, we find the slope of the line, and then we use the point-slope formula.

Step 1 Let \((x_1,y_1) = (2,-1)\) and \((x_2,y_2) = (-1,3)\text{.}\) Using the slope formula, we find

\begin{equation*} \begin{aligned} m \amp =\dfrac{y_2-y_1}{x_2-x_1}\\ \amp =\dfrac{3-(-1)}{-1-2} = \dfrac{4}{-3} = \dfrac{-4}{3} \end{aligned} \end{equation*}

Step 2 We apply the point-slope formula with \(m=\dfrac{-4}{3}\) and \((x_1,y_1) = (2,-1)\text{.}\) (We can use either point to find the equation of the line.) Then

\begin{equation*} \dfrac{y-y_1}{x-x_1} = m~~~~~~\text{becomes} ~~~~~~ \dfrac{y-(-1)}{x-2} = \dfrac{-4}{3} \end{equation*}

We cross-multiply to find

\begin{equation*} \begin{aligned} 3(y+1)\amp = -4(x-2) \amp\amp \blert{\text{Apply the distributive law.}}\\ 3y+3 \amp = -4x+8 \amp\amp \blert{\text{Solve for}~y.}\\ 3y \amp = -4x+5 \\ y \amp = \dfrac{-4}{3}x+\dfrac{5}{3} \end{aligned} \end{equation*}

The graph of the line is shown in the figure.

line through two points

Reading Questions Reading Questions

3.

How many different lines pass through two given points?

4.

What formula do we use to find the equation of a line through two points?

To Fit a Line through Two Points.
  1. Compute the slope between the two points.
  2. Substitute the slope and either point into the point-slope formula.

Subsection Applications

Variables that increase or decrease at a constant rate can be described by linear equations.

Example 4.35.

In 1993, Americans drank 188.6 million cases of wine. Wine consumption increased at a constant rate over the next decade, and in 2003 we drank 258.3 million cases of wine. (Source: LA Times, Adams Beverage Group)

  1. Find a formula for wine consumption, \(W\text{,}\) in millions of cases, \(t\) years after 1990.
  2. State the slope as a rate of change. What does the slope tell us about this problem?
Solution
  1. We have two data points of the form \((t,W)\text{,}\) namely \((t_1,W_1)=(3,188.6)\) and \((t_2,W_2)=(13,258.3)\text{.}\) We use the point slope formula to fit a line through these two points. First we compute the slope.

    \begin{equation*} \dfrac{\Delta W}{\Delta t}=\dfrac{W_2-W_2}{t+2-t_1}=\dfrac{258.3-188.6}{13-3}=6.97 \end{equation*}

    Next, we substitute the slope \(m=6.97\) and either of the two data points into the point-slope formula.

    \begin{equation*} \begin{aligned} W \amp = W_1+m(t-t_1)\\ W \amp = 188.6+6.97(t-3)\\ W \amp = 167.69+6.97t \end{aligned} \end{equation*}

    Thus, \(W = 167.69+6.97t\text{.}\)

  2. The slope gives us the rate of change of \(W\) with respect to \(t\text{.}\) The units of the variables can help us interpret the slope in context.

    \begin{equation*} \dfrac{\Delta W}{\Delta t}=\dfrac{258.3-188.6~\text{millions of cases}}{13-3~\text{years}} = 6.97~ \text{millions of cases/years} \end{equation*}

    Over the ten years between 1993 and 2003, wine consumption in the US increased at a rate of 6.97 million cases per year.

Reading Questions Reading Questions

3.

What are the two steps to fit a line through two points?

Subsection Skills Warm-Up

Exercises Exercises

Use cross-multiplying to solve the proportion for \(y\) in terms of \(x\text{.}\)

1.
\(\dfrac{y}{x} = \dfrac{-5}{2}\)
2.
\(\dfrac{y-3}{4} = \dfrac{x}{2}\)
3.
\(\dfrac{x+1}{5} = \dfrac{y-1}{3}\)
4.
\(-2 = \dfrac{y+6}{x}\)
5.
\(\dfrac{y+2}{x-5} = \dfrac{3}{4}\)
6.
\(\dfrac{-1}{3} = \dfrac{4-y}{1-x}\)

Solutions Answers to Skills Warm-Up

Exercises Exercises

Exercises Homework 4.5

For Problems 1–6,

  1. Use the point-slope method to graph the line with the given slope and passing through the given point.
  2. Find an equation for the line. Write your equation in slope-intercept form.
  3. Find the \(x\)-intercept of the line.
1.
\(m=-2,~~(-3,4)\)
grid
2.
\(m=\dfrac{1}{2},~~(4,-3)\)
grid
3.
\(m=\dfrac{-2}{3},~~(-6,2)\)
grid
4.
\(m=0,~~(-3,5)\)
grid
5.
\(m=\dfrac{-5}{2},~~(6,-10)\)
grid
6.
\(m=3,~~(1,-1)\)
grid

For Problems 7–9, without doing any calculations, give the slope of each line and the coordinates of one point on the line.

7.
\(y=\dfrac{3}{5}-7\)
8.
\(y-2=3(x+5)\)
9.
\(y=\dfrac{4}{5}x)\)

For Problems 10–12, find an equation for the line passing through the given points. Write your answer in slope-intercept form.

10.
\((-2,4),~(1,7)\)
11.
\((3,5),~(-3,-5)\)
12.
\((6,4),~(-2,5)\)

For Problems 13–16, find the equation of the line shown in the graph.

13.
line through two points
14.
line through two points
15.
line through two points
16.
line through two points
17.
  1. Put the equation \(2y-3x=5\) into slope-intercept form, and graph the equation.

    grid
  2. What is the slope of any line that is parallel to \(2y-3x=5\text{?}\)
  3. On your graph for part (a), sketch by hand a line that is parallel to \(2y-3x=5\) and passes through the point \((4,2)\text{.}\)
  4. Use the point-slope formula to write an equation for the line that is parallel to the graph of \(2y-3x=5\) and passes through the point \((4,2)\text{.}\)
18.
  1. Put the equation \(x-2y=5\) into slope-intercept form, and graph the equation.

    graph
  2. What is the slope of any line that is perpendicular to \(x-2y=5\text{?}\)
  3. On your graph for part (a), sketch by hand a line that is perpendicular to \(x-2y=5\) and passes through the point \((4,-3)\text{.}\)
  4. Use the point-slope formula to write an equation for the line that is perpendicular to the graph of \(x-2y=5\) and passes through the point \((4,-3)\text{.}\)
19.
  1. What is the slope of the line \(y=2x+1\text{?}\)
  2. What is the slope of a line parallel to \(y=2x+1\text{?}\)
  3. A line is parallel to \(y=2x+1\) and passes through \((-2,1)\text{.}\) What is its equation?
20.
  1. What is the slope of the line \(y=-3x-2\text{?}\)
  2. What is the slope of a line parallel to \(y=-3x-2\text{?}\)
  3. A line is parallel to \(y=-3x-2\) and passes through \((1,3)\text{.}\) What is its equation?

For Problems 21–24, we'll find a linear model from two data points.

  1. Make a table showing the coordinates of two data points for the model.
  2. Find a linear equation in slope-intercept form relating the variables.
  3. Use your equation to answer the questions.
21.

Francine is driving into the mountains and stopping periodically to record the temperature, \(T\text{,}\) at various altitudes, \(h\text{.}\) The temperature at an altitude of 3200 feet is \(77 \degree\text{,}\) and the temperature at 8000 feet is \(65 \degree\text{.}\)

  1. What is the slope, including units? What does it tell you about the problem?
  2. What will the temperature be at 10,000 feet?
  3. What was the temperature at sea level?
22.

Envirotech is marketing a new line of microwave clothes dryers. It cost them $45,000 to produce the first 100 dryers. When they had produced 180 dryers, their total cost was up to $61,000. They would like to know the total cost, \(C\text{,}\) of producing \(x\) dryers.

  1. What is the slope, including units? What does it tell you about the problem?
  2. Envirotech has budgeted $100,000 for microwave clothes dryers this year. How many can they produce?
  3. How much did Envirotech invest in development before they made the first dryer?
23.

Flying lessons cost $645 for an 8-hour course and $1425 for a 20-hour course. Both prices include a fixed insurance fee. Express the cost, \(C\text{,}\) of flying lessons in terms of the length, \(h\text{,}\) of the course in hours.

  1. What is the slope, including units? What does it tell you about the problem?
  2. How much does a 10-hour course cost?
  3. How much is the fixed insurance fee?
24.

On an international flight a passenger may check two bags each weighing 70 kilograms, or 154 pounds, and one carryon bag weighing 50 kilograms, or 110 pounds. Express the weight, \(p\text{,}\) of a bag in pounds in terms of its weight, \(k\text{,}\) in kilograms.

  1. What is the slope, including units? What does it tell you about the problem?
  2. What does a 50-pound bag weigh in kilograms?
  3. Why is the constant term in your equation equal to zero?