Section 4.6 PointSlope Form
Ā¶In SectionĀ 4.5, we learned that a linear equation can be written in slopeintercept form, \(y=mx+b\text{.}\) This section covers an alternative that can often be more useful depending on the application: pointslope form.
Subsection 4.6.1 PointSlope Motivation and Definition
Starting in 1990, the population of the United States has been growing by about \(2.865\) million per year. Also, back in 1990, the population was \(253\) million. Since the rate of growth has been roughly constant, a linear model is appropriate. Let's try to write an equation to model this.
We consider using slopeintercept formĀ (4.5.1), but we would need to know the \(y\)intercept, and nothing in the background tells us that. We'd need to know the population of the United States in the year 0, before there even was a United States.
We could do some side work to calculate the \(y\)intercept, but let's try something else. Here are some things we know:
The slope equation is \(m=\frac{y_2y_1}{x_2x_1}\text{.}\)
The slope is \(m=2.865\) (million per year).
One point on the line is \((1990,253)\text{,}\) because in 1990, the population was \(253\) million.
If we use the generic \((x,y)\) to represent a point somewhere on this line, then the rate of change between \((1990,253)\) and \((x,y)\) has to be \(2.865\text{.}\) So
There is good reason^{ā1ā}It will help us to see that \(y\) (population) depends on \(x\) (whatever year it is). to want to isolate \(y\) in this equation:
This is a good place to stop. We have isolated \(y\text{,}\) and three meaningful numbers appear in the population: the rate of growth, a certain year, and the population in that year. This is a specific example of pointslope form. Before we look deeper at pointslope form, let's continue reducing the line equation into slopeintercept form.
One concern with slopeintercept formĀ (4.5.1) is that it uses the \(y\)intercept, which might be somewhat meaningless in the context of an application. For example, here we have found that the \(y\)intercept is at \((0,5448.35)\text{,}\) but what practical use is that? It's nonsense to say that in the year 0, the population of the United States was \(5448.35\) million. It doesn't make sense to have a negative population. It doesn't make sense to talk about the United States population before there even was a United States. And it doesn't make sense to use this model for years earlier than 1990 because the background information says clearly that the rate of change we have applies to years 1990 and later.
For all these reasons, we prefer the equation when it was in the form
Definition 4.6.2 PointSlope Form
When \(x\) and \(y\) have a linear relationship where \(m\) is the slope and \((x_0,y_0)\) is some specific point that the line passes through, one equation for this relationship is
and this equation is called the pointslope form of the line. It is called this because the slope and one point on the line are immediately discernible from the numbers in the equation.
Remark 4.6.4 Alternative PointSlope Form
It is also common to define pointslope form as
by subtracting \(y_0\) from each side. Some exercises may appear using this form.
Checkpoint 4.6.5
In CheckpointĀ 4.6.5, the solution explains that each of the following are acceptable equations for the same line:
The first uses \((3,2)\) as a point on the line, and the second uses \((2,1)\text{.}\) Are those two equations really equivalent? Let's distribute and simplify each of them to get slopeintercept formĀ (4.5.1).
So, yes. It didn't matter which point we used to write a pointslope equation. We get differentlooking equations that still represent the same line.
Pointslope form is preferable when we know a line's slope and a point on it, but we don't know the \(y\)intercept.
Example 4.6.6
A spa chain has been losing customers at a roughly constant rate since the year 2010. In 2013, it had \(2{,}975\) customers; in 2016, it had \(2{,}585\) customers. Management estimated that the company will go out of business once its customer base decreases to \(1{,}800\text{.}\) If this trend continues, when will the company close?
The given information tells us two points on the line: \((2013,2975)\) and \((2016,2585)\text{.}\) The slope formulaĀ (4.4.3) will give us the slope. After labeling those two points as \((\overset{x_1}{2013},\overset{y_1}{2975})\) and \((\overset{x_2}{2016},\overset{y_2}{2585})\text{,}\) we have:
And considering units, this means they are losing \(130\) customers per year.
Let's note that we could try to make an equation for this line in slopeintercept form, but then we would need to calculate the \(y\)intercept, which in context would correspond to the number of customers in year \(0\text{.}\) We could do it, but we'd be working with numbers that have no realworld meaning in this context.
For pointslope form, since we calculated the slope, we know at least this much:
Now we can pick one of those two given points, say \((2013,2975)\text{,}\) and get the equation
Note that all three numbers in this equation have meaning in the context of the spa chain.
We're ready to answer the question about when the chain might go out of business. Substitute \(y\) in the equation with \(1800\) and solve for \(x\text{,}\) and we will get the answer we seek.
And so we find that at this rate, the company is headed toward a collapse in 2022.
Checkpoint 4.6.8
Subsection 4.6.2 Using Two Points to Build a Linear Equation
Since two points can determine a line's location, we can calculate a line's equation using just the coordinates from any two points it passes through.
Example 4.6.9
A line passes through \((6,0)\) and \((9,10)\text{.}\) Find this line's equation in both pointslope and slopeintercept form.
We will use the slope formulaĀ (4.4.3) to find the slope first. After labeling those two points as \((\overset{x_1}{6},\overset{y_1}{0}) \text{ and } (\overset{x_2}{9},\overset{y_2}{10})\text{,}\) we have:
The pointslope equation is \(y=\frac{2}{3}(xx_0)+y_0\text{.}\) Next, we will use \((9,10)\) and substitute \(x_0\) with \(9\) and \(y_0\) with \(10\text{,}\) and we have:
Next, we will change the pointslope equation into slopeintercept form:
Remark 4.6.10
Note that many other resources use the alternate pointslope formĀ (4.6.2) to write their equations. Those equations will always be equivalent to those created using our pointslope form. In ExampleĀ 4.6.9, we found the pointslope form \(y=\frac{2}{3}(x9)10\text{.}\) The alternate pointslope form equation^{ā2ā}khanacademy.org/math/algebra/twovarlinearequations/pointslope/a/pointslopeformreviewwould have given us \(y+10=\frac{2}{3}(x9)\text{.}\) If you solve this equation for \(y\) and simplify, you should still get \(y=\frac{2}{3}x4\text{,}\) as we did earlier.
Checkpoint 4.6.11
Subsection 4.6.3 More on PointSlope Form
We can tell a lot about a linear equation now that we have learned both slopeintercept formĀ (4.5.1) and pointslope formĀ (4.6.1). For example, we can know that \(y=4x+2\) is in slopeintercept form because it looks like \(y=mx+b\text{.}\) It will graph as a line with slope \(4\) and vertical intercept \((0,2)\text{.}\) Likewise, we know that the equation \(y=5(x3)+2\) is in pointslope form because it looks like \(y=m(xx_0)+y_0\text{.}\) It will graph as a line that has slope \(5\) and will pass through the point \((3,2)\text{.}\)
Example 4.6.12
For the equations below, state whether they are in slopeintercept form or pointslope form. Then identify the slope of the line and at least one point that the line will pass through.
\(y=3x+2\)
\(y=9(x+1)6\)
\(y=5x\)
\(y=\frac{12}{5}(x9)+1\)
The equation \(y=3x+2\) is in slopeintercept form. The slope is \(3\) and the vertical intercept is \((0,2)\text{.}\)
The equation \(y=9(x+1)6\) is in pointslope form. The slope is \(9\) and the line passes through the point \((1,6)\text{.}\)
The equation \(y=5x\) is almost in slopeintercept form. If we rearrange the right hand side to be \(y=x+5\text{,}\) we can see that the slope is \(1\) and the vertical intercept is \((0,5)\text{.}\)
The equation \(y=\frac{12}{5}(x9)+1\) is in pointslope form. The slope is \(\frac{12}{5}\) and the line passes through the point \((9,1)\text{.}\)
Remark 4.6.13
Again, we should note that the alternate pointslope formĀ (4.6.2) can be used to identify equations. For example, the equation \(y+10=\frac{2}{3}(x9)\) matches the alternate pointslope form equation^{ā3ā}en.wikipedia.org/wiki/Linear_equation#Pointāslope_form with slope \(\frac{2}{3}\) and the line passes through the point \((9,10)\text{.}\) Note that both coordinates are the opposite of what they appear to be in the equation with this form.
Example 4.6.14
Consider the graph in FigureĀ 4.6.15.
Find three equations that describe the line shown written in pointslope form. Three integervalued points are shown for convenience.
Determine the slopeintercept form of the equation of this line.

To write any of the equations representing this line in pointslope form, we must first find the slope of the line and we can use the slope formulaĀ (4.4.3) to do so. We will arbitrarily choose \((0,30)\) and \((5,42)\) as the two points. Inputting these points into the slope formula yields:
\begin{align*} m\amp=\frac{y_2y_1}{x_2x_1}\\ \amp=\frac{\substitute{42}\substitute{30}}{\substitute{5}\substitute{0}}\\ \amp=\frac{12}{5}\\ \amp=\frac{12}{5} \end{align*}Thus the slope of the line is \(\frac{12}{5}\text{.}\)
Next, we need to write an equation in pointslope form based on each point shown. Using the point \((0,30)\text{,}\) we have:
\begin{equation*} y=\frac{12}{5}(x0)+30 \end{equation*}(This simplifies to \(y=\frac{12}{5}x+30\text{.}\))
The next point is \((20,18)\text{.}\) Using this point, we can write an equation for this line as:
\begin{equation*} y=\frac{12}{5}(x20)18 \end{equation*}Finally, we can also use the point \((5,42)\) to write an equation for this line:
\begin{equation*} y=\frac{12}{5}(x(5))+42 \end{equation*}which can also be written as:
\begin{equation*} y=\frac{12}{5}(x+5)+42 \end{equation*} As \((0,30)\) is the vertical intercept, we can write the equation of this line in slopeintercept form as \(y=\frac{12}{5}x+30\text{.}\) It's important to note that each of the equations that were written in pointslope form simplify to this, making all four equations equivalent.
Subsection 4.6.4 Exercises
Review and Warmup
1
Evaluate \({5C7b}\) for \(C = 6\) and \(b = 7\text{.}\)
2
Evaluate \({a+3A}\) for \(a = 2\) and \(A = 7\text{.}\)
3
Evaluate
for \(x_1 = 14\text{,}\) \(x_2 = 10\text{,}\) \(y_1 = 19\text{,}\) and \(y_2 = 10\text{:}\)
4
Evaluate
for \(x_1 = 10\text{,}\) \(x_2 = 17\text{,}\) \(y_1 = 2\text{,}\) and \(y_2 = 19\text{:}\)
PointSlope Form
5
A lineās equation is given in pointslope form:
\({y}={5\!\left(x5\right)+28}\)
This lineās slope is .
A point on this line that is apparent from the given equation is .
6
A lineās equation is given in pointslope form:
\({y}={2\!\left(x1\right)+5}\)
This lineās slope is .
A point on this line that is apparent from the given equation is .
7
A lineās equation is given in pointslope form:
\({y}={2\!\left(x+2\right)+5}\)
This lineās slope is .
A point on this line that is apparent from the given equation is .
8
A lineās equation is given in pointslope form:
\({y}={3\!\left(x+4\right)+7}\)
This lineās slope is .
A point on this line that is apparent from the given equation is .
9
A lineās equation is given in pointslope form:
\(\displaystyle{ {y}={\frac{8}{3}\!\left(x+9\right)  23} }\)
This lineās slope is .
A point on this line that is apparent from the given equation is .
10
A lineās equation is given in pointslope form:
\(\displaystyle{ {y}={\frac{9}{8}\!\left(x+24\right)  29} }\)
This lineās slope is .
A point on this line that is apparent from the given equation is .
11
A line passes through the points \((2,9)\) and \((1,7)\text{.}\) Find this lineās equation in pointslope form.
Using the point \((2,9)\text{,}\) this lineās pointslope form equation is .
Using the point \((1,7)\text{,}\) this lineās pointslope form equation is .
12
A line passes through the points \((4,10)\) and \((1,4)\text{.}\) Find this lineās equation in pointslope form.
Using the point \((4,10)\text{,}\) this lineās pointslope form equation is .
Using the point \((1,4)\text{,}\) this lineās pointslope form equation is .
13
A line passes through the points \((3,17)\) and \((0,8)\text{.}\) Find this lineās equation in pointslope form.
Using the point \((3,17)\text{,}\) this lineās pointslope form equation is .
Using the point \((0,8)\text{,}\) this lineās pointslope form equation is .
14
A line passes through the points \((1,2)\) and \((3,14)\text{.}\) Find this lineās equation in pointslope form.
Using the point \((1,2)\text{,}\) this lineās pointslope form equation is .
Using the point \((3,14)\text{,}\) this lineās pointslope form equation is .
15
A line passes through the points \((6,{5})\) and \((6,{25})\text{.}\) Find this lineās equation in pointslope form.
Using the point \((6,{5})\text{,}\) this lineās pointslope form equation is .
Using the point \((6,{25})\text{,}\) this lineās pointslope form equation is .
16
A line passes through the points \((7,{5})\) and \((21,{17})\text{.}\) Find this lineās equation in pointslope form.
Using the point \((7,{5})\text{,}\) this lineās pointslope form equation is .
Using the point \((21,{17})\text{,}\) this lineās pointslope form equation is .
17
A lineās slope is \(4\text{.}\) The line passes through the point \((5,22)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
18
A lineās slope is \(5\text{.}\) The line passes through the point \((2,11)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
19
A lineās slope is \(2\text{.}\) The line passes through the point \((2,2)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
20
A lineās slope is \(5\text{.}\) The line passes through the point \((4,18)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
21
A lineās slope is \(1\text{.}\) The line passes through the point \((5,1)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
22
A lineās slope is \(1\text{.}\) The line passes through the point \((2,1)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
23
A lineās slope is \(1\text{.}\) The line passes through the point \((2,1)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
24
A lineās slope is \(1\text{.}\) The line passes through the point \((3,2)\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
25
A lineās slope is \({{\frac{6}{5}}}\text{.}\) The line passes through the point \((10,{8})\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
26
A lineās slope is \({{\frac{7}{4}}}\text{.}\) The line passes through the point \((12,{22})\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
27
A lineās slope is \({{\frac{8}{9}}}\text{.}\) The line passes through the point \((9,{13})\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
28
A lineās slope is \({{\frac{9}{5}}}\text{.}\) The line passes through the point \((10,{20})\text{.}\) Find an equation for this line in both pointslope and slopeintercept form.
An equation for this line in pointslope form is: .
An equation for this line in slopeintercept form is: .
PointSlope and SlopeIntercept
29
Change this equation from pointslope form to slopeintercept form.
\({y}={2\!\left(x4\right)+5}\)
In slopeintercept form:
30
Change this equation from pointslope form to slopeintercept form.
\({y}={2\!\left(x+2\right)}\)
In slopeintercept form:
31
Change this equation from pointslope form to slopeintercept form.
\({y}={4\!\left(x3\right)10}\)
In slopeintercept form:
32
Change this equation from pointslope form to slopeintercept form.
\({y}={4\!\left(x+4\right)+12}\)
In slopeintercept form:
33
Change this equation from pointslope form to slopeintercept form.
\(\displaystyle{ {y}={{\frac{5}{8}}\!\left(x16\right)+15} }\)
In slopeintercept form:
34
Change this equation from pointslope form to slopeintercept form.
\(\displaystyle{ {y}={{\frac{6}{5}}\!\left(x5\right)+11} }\)
In slopeintercept form:
35
Change this equation from pointslope form to slopeintercept form.
\(\displaystyle{ {y}={{\frac{7}{3}}\!\left(x+3\right)+6} }\)
In slopeintercept form:
36
Change this equation from pointslope form to slopeintercept form.
\(\displaystyle{ {y}={{\frac{8}{7}}\!\left(x+21\right)+29} }\)
In slopeintercept form:
PointSlope Form and Graphs
37
Determine the pointslope form of the linear equation from its graph.
38
Determine the pointslope form of the linear equation from its graph.
39
Determine the pointslope form of the linear equation from its graph.
40
Determine the pointslope form of the linear equation from its graph.
41
Determine the pointslope form of the linear equation from its graph.
42
Determine the pointslope form of the linear equation from its graph.
43
Determine the pointslope form of the linear equation from its graph.
44
Determine the pointslope form of the linear equation from its graph.
45
Determine the pointslope form of the linear equation from its graph.
46
Determine the pointslope form of the linear equation from its graph.
47
Determine the pointslope form of the linear equation from its graph.
48
Determine the pointslope form of the linear equation from its graph.
49
Graph the linear equation \(y=\frac{8}{3}(x4)5\) by identifying the slope and one point on this line.
50
Graph the linear equation \(y=\frac{5}{7}(x+3)+2\) by identifying the slope and one point on this line.
51
Graph the linear equation \(y=\frac{3}{4}(x+2)+1\) by identifying the slope and one point on this line.
52
Graph the linear equation \(y=\frac{5}{2}(x1)5\) by identifying the slope and one point on this line.
53
Graph the linear equation \(y=3(x9)+4\) by identifying the slope and one point on this line.
54
Graph the linear equation \(y=7(x+3)10\) by identifying the slope and one point on this line.
55
Graph the linear equation \(y=8(x+12)20\) by identifying the slope and one point on this line.
56
Graph the linear equation \(y=5(x20)70\) by identifying the slope and one point on this line.
Applications
57
By your cell phone contract, you pay a monthly fee plus \({\$0.04}\) for each minute you spend on the phone. In one month, you spent \(230\) minutes over the phone, and had a bill totaling \({\$22.20}\text{.}\)
Let \(x\) be the number of minutes you spend on the phone in a month, and let \(y\) be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone.
A pointslope equation to model this is .
If you spend \(160\) minutes on the phone in a month, you would be billed .
If your bill was \({\$30.60}\) one month, you must have spent minutes on the phone in that month.
58
A company set aside a certain amount of money in the year 2000. The company spent exactly \({\$31{,}000}\) from that fund each year on perks for its employees. In \(2003\text{,}\) there was still \({\$872{,}000}\) left in the fund.
Let \(x\) be the number of years since 2000, and let \(y\) be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.
A pointslope equation to model this is .
In the year \(2010\text{,}\) there was left in the fund.
In the year , the fund will be empty.
59
A biologist has been observing a treeās height. This type of tree typically grows by \(0.19\) feet each month. Ten months into the observation, the tree was \(17.4\) feet tall.
Let \(x\) be the number of months passed since the observations started, and let \(y\) be the treeās height at that time, in feet. Use a linear equation to model the treeās height as the number of months pass.
A pointslope equation to model this is .
\(28\) months after the observations started, the tree would be feet in height.
months after the observation started, the tree would be \(25.76\) feet tall.
60
Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(3\) grams. Seven minutes since the experiment started, the remaining gas had a mass of \(117\) grams.
Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
A pointslope equation to model this is .
\(35\) minutes after the experiment started, there would be grams of gas left.
If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
61
A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In \(2004\text{,}\) there was still \({\$783{,}000}\) left in the fund. In \(2006\text{,}\) there was \({\$701{,}000}\) left.
Let \(x\) be the number of years since 2000, and let \(y\) be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.
A pointslope equation to model this is .
In the year \(2010\text{,}\) there was left in the fund.
In the year , the fund will be empty.
62
By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent \(230\) minutes on the phone, and paid \({\$27.65}\text{.}\) In another month, you spent \(320\) minutes on the phone, and paid \({\$32.60}\text{.}\)
Let \(x\) be the number of minutes you talk over the phone in a month, and let \(y\) be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone.
A pointslope equation to model this is .
If you spent \(150\) minutes over the phone in a month, you would pay .
If in a month, you paid \({\$41.95}\) of cell phone bill, you must have spent minutes on the phone in that month.
63
Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way.
Nine minutes since the experiment started, the gas had a mass of \(106.6\) grams.
Eighteen minutes since the experiment started, the gas had a mass of \(83.2\) grams.
Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
A pointslope equation to model this is .
\(35\) minutes after the experiment started, there would be grams of gas left.
If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
64
A biologist has been observing a treeās height. \(10\) months into the observation, the tree was \(18.6\) feet tall. \(18\) months into the observation, the tree was \(19.4\) feet tall.
Let \(x\) be the number of months passed since the observations started, and let \(y\) be the treeās height at that time, in feet. Use a linear equation to model the treeās height as the number of months pass.
A pointslope equation to model this is .
\(27\) months after the observations started, the tree would be feet in height.
months after the observation started, the tree would be \(23.5\) feet tall.