## Section4.6Point-Slope Form

Ā¶In SectionĀ 4.5, we learned that a linear equation can be written in slope-intercept form, \(y=mx+b\text{.}\) This section covers an alternative that can often be more useful depending on the application: point-slope form.

### Subsection4.6.1Point-Slope Definition

Sometimes, one problem with slope-intercept 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 is a slope-intercept equation for the population of the United States in year \(x\text{,}\) where \(x\) can be any year from 1990 and beyond, and \(y\) is the population measured in millions:

What can we say about the two numbers \(2.865\) and \(-5451\text{?}\) The slope is \(2.865\) with units \(\frac{y\text{-unit}}{x\text{-unit}}=\frac{\text{million}}{\text{year}}\text{.}\) OK, so there is meaning there: the population has been growing by \(2.865\) million people per year.

But what about \(-5451\text{?}\) This number tells us that the \(y\)-intercept is \((0,-5451)\text{,}\) but what practical use is that? It's nonsense to say that in the year 0, the population of the United States was \(-5451\) million. It doesn't make sense to have a negative population. It doesn't make sense to talk about the Unted 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 it says clearly that the model is for 1990 and beyond.

###### Remark4.6.2

If the \(x\)-value \(0\) is not an appropriate \(x\)-value to consider in a linear model, then the āinitial valueā \(b\) from the slope-intercept form is not meaningful in the context of the model. Its only value is to be part of the formula for calculations. It can still be used to mark a \(y\)-intercept on the \(y\)-axis, but if you are treating the equation as a mathematical model then you shouldn't be thinking too hard about the portion of the line near the \(y\)-axis, since the \(x\)-values near \(0\) are not relevant.

###### Example4.6.3

Since 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. But let's look for a way to write the equation other than slope-intercept form. Here are some things we know:

The slope equation is \(m=\frac{y_2-y_1}{x_2-x_1}\text{.}\)

The slope is \(m=2.865\) (million per year).

One point on the line is \((1990,253)\text{,}\) since 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 point-slope form.

###### Definition4.6.4Point-Slope 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 point-slope form of the line. It is called this because the slope and one point on the line are immediately discernable from the numbers in the equation.

###### Alternative Point-Slope Form

It is also common to define point-slope form as

by subtracting the \(y_0\) to the right side. Some exercises may appear using this form.

Consider the graph in FigureĀ 4.6.6.

###### Exercise4.6.7

In the previous exercise, the solution explains that each of the following are accepatable equations for the same line:

Are those two equations really equivalent? Let's distribute and simplify each of them to get slope-intercept formĀ (4.5.1).

So, yes. It didn't matter which point we focused on in the line in FigureĀ 4.6.6. We get different-looking equations that still represent the same line (which, by the way, has \(y\)-intercept at \((0,-7)\)).

Point-slope form is preferable when we know a line's slope and a point on it, but we don't know the \(y\)-intercept.

###### Example4.6.8

A spa chain has been losing customers at a roughly constant rate since the year 2010. In 2013, it had \(2975\) customers; in 2016, it had \(2585\) customers. Management estimated that the company will go out of business once its customer base decreases to \(1800\text{.}\) 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 slope-intercept 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 real-world meaning in this context.

For point-slope 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.

Shown is a graph that represents the scenario. Note that to make a graph of \(y=-130(x-2013)+2975\text{,}\) we must first find the point \((2013,2975)\) and from there use the slope of \(-130\) to draw the line.

###### Exercise4.6.10

### Subsection4.6.2Using 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.

###### Example4.6.11

A line passes through \((-6,0)\) and \((9,-10)\text{.}\) Find this line's equation in both point-slope and slope-intercept 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 point-slope equation is \(y=-\frac{2}{3}(x-x_0)+y_0\text{.}\) 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 point-slope equation into slope-intercept form:

###### Exercise4.6.12

### Subsection4.6.3More on Point-Slope Form

We can tell a lot about a linear equation now that we have learned both slope-intercept formĀ (4.5.1) and point-slope formĀ (4.6.1). For example, we can know that \(y=4x+2\) is in slope-intercept 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 \(y=-5(x-3)+2\) is in point-slope form because it looks like \(y=m(x-x_0)+y_0\text{,}\) it will graph as a line that has slope \(-5\) and will pass through the point \((3,2)\text{.}\)

###### Example4.6.13

For the equations below, state whether they are in slope-intercept form or point-slope form. Then identify the slope of the line and at least one point that the line will pass through.

\(y=-3x+2\)

\(y=9(x+1)-6\)

\(y=5-x\)

\(y=-\frac{12}{5}(x-9)+1\)

The equation \(y=-3x+2\) is in slope-intercept form. The slope is \(-3\) and the vertical intercept is \((0,2)\text{.}\)

The equation \(y=9(x+1)-6\) is in point-slope form. The slope is \(9\) and the line passes through the point \((-1,6)\text{.}\)

The equation \(y=5-x\) is almost in slope-intercept 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 \(y=-\frac{12}{5}(x-9)+1\) is in point-slope form. The slope is \(-\frac{12}{5}\) and the line passes through the point \((9,1)\text{.}\)

###### Example4.6.14

Consider the graph in FigureĀ 4.6.15. Find three equivalent equations that describe the line shown. Three integer valued points are shown for convenience.

We could imagine infinitely many equivalent equations for the graph in FigureĀ 4.6.15 if we could find infinitely many points on the graph. In this case, we only need three and there are three integer-coordinate points identified. To write *any* of these equations, we must first find the slope of the line. Fortunately, we have EquationĀ (4.4.3). This formula finds the slope between any two points and we will arbitrarily choose \((0,30)\) and \((-5,42)\) to find the slope. Inputting these points into the slope formula yieldsā¦

Thus the slope of the line is \(-\frac{12}{5}\text{.}\) This will be the case for any version of the equation of the line that we write.

Next, we need to write an equation based at each point shown. The first point we will draw attention to is \((0,30)\text{.}\) This is the vertical intercept. Thus we should be able to write an equation in slope intercept form, \(y=mx+b\text{.}\) We know that \(m=-\frac{12}{5}\text{,}\) and \(b=30\text{,}\) thus the equation is \(\highlight{y=-\frac{12}{5}x+30}\text{.}\)

The next point we will examine is \((20,-18)\text{.}\) Since this is not the vertical intercept, the only formula we can use is point-slope form, \(y=m\left(x-x_0\right)+y_0\text{.}\) We know that \(m=-\frac{12}{5}\text{,}\) \(x_0=20\text{,}\) and \(y_0=-18\text{.}\) Inputting these values into the formula, we achieve \(\highlight{y=-\frac{12}{5}\left(x-20\right)-18}\text{.}\)

The last point we have is \((-5,42)\text{.}\) Since this is not the vertical intercept, the only formula we can use is point-slope form, \(y=m\left(x-x_0\right)+y_0\text{.}\) We know that \(m=-\frac{12}{5}\text{,}\) \(x_0=-5\text{,}\) and \(y_0=42\text{.}\) Inputting these values into the formula, we achieve \(y=-\frac{12}{5}\left(x-(-5)\right)+42\text{.}\) Note that we can simplify this to be \(\highlight{y=-\frac{12}{5}\left(x+5\right)+42}\text{.}\)

### Subsection4.6.4Exercises

Point-Slope Form Basics

###### 1

A lineās equation is given in point-slope form:

\({y}={2\!\left(x-3\right)+5}\)

This lineās slope is .

A point on this line that is apparent from the given equation is .

###### 2

A lineās equation is given in point-slope form:

\({y}={2\!\left(x-1\right)-3}\)

This lineās slope is .

A point on this line that is apparent from the given equation is .

###### 3

A lineās equation is given in point-slope form:

\({y}={3\!\left(x+1\right)-4}\)

This lineās slope is .

A point on this line that is apparent from the given equation is .

###### 4

A lineās equation is given in point-slope form:

\({y}={-3\!\left(x+2\right)+10}\)

This lineās slope is .

A point on this line that is apparent from the given equation is .

###### 5

A lineās equation is given in point-slope form:

\(\displaystyle{ {y}={\left(-\frac{5}{2}\right)\!\left(x+2\right)+4} }\)

This lineās slope is .

A point on this line that is apparent from the given equation is .

###### 6

A lineās equation is given in point-slope form:

\(\displaystyle{ {y}={\frac{6}{7}\!\left(x+7\right) - 10} }\)

This lineās slope is .

A point on this line that is apparent from the given equation is .

Use given information to find a line's point-slope form equation.

###### 7

A line passes through the points \((1,9)\) and \((5,29)\text{.}\) Find this lineās equation in point-slope form.

Using the point \((1,9)\text{,}\) this lineās point-slope form equation is .

Using the point \((5,29)\text{,}\) this lineās point-slope form equation is .

###### 8

A line passes through the points \((3,25)\) and \((5,35)\text{.}\) Find this lineās equation in point-slope form.

Using the point \((3,25)\text{,}\) this lineās point-slope form equation is .

Using the point \((5,35)\text{,}\) this lineās point-slope form equation is .

###### 9

A line passes through the points \((5,-18)\) and \((-2,17)\text{.}\) Find this lineās equation in point-slope form.

Using the point \((5,-18)\text{,}\) this lineās point-slope form equation is .

Using the point \((-2,17)\text{,}\) this lineās point-slope form equation is .

###### 10

A line passes through the points \((-2,7)\) and \((5,-7)\text{.}\) Find this lineās equation in point-slope form.

Using the point \((-2,7)\text{,}\) this lineās point-slope form equation is .

Using the point \((5,-7)\text{,}\) this lineās point-slope form equation is .

###### 11

A line passes through the points \((18,{10})\) and \((27,{12})\text{.}\) Find this lineās equation in point-slope form.

Using the point \((18,{10})\text{,}\) this lineās point-slope form equation is .

Using the point \((27,{12})\text{,}\) this lineās point-slope form equation is .

###### 12

A line passes through the points \((-10,{3})\) and \((15,{18})\text{.}\) Find this lineās equation in point-slope form.

Using the point \((-10,{3})\text{,}\) this lineās point-slope form equation is .

Using the point \((15,{18})\text{,}\) this lineās point-slope form equation is .

###### 13

A lineās slope is \(3\text{.}\) The line passes through the point \((4,14)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 14

A lineās slope is \(4\text{.}\) The line passes through the point \((1,9)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 15

A lineās slope is \(-3\text{.}\) The line passes through the point \((1,-4)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 16

A lineās slope is \(-2\text{.}\) The line passes through the point \((4,-12)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 17

A lineās slope is \(1\text{.}\) The line passes through the point \((4,8)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 18

A lineās slope is \(1\text{.}\) The line passes through the point \((2,7)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 19

A lineās slope is \(-1\text{.}\) The line passes through the point \((-4,0)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 20

A lineās slope is \(-1\text{.}\) The line passes through the point \((3,-6)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 21

A lineās slope is \({{\frac{3}{5}}}\text{.}\) The line passes through the point \((15,{14})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 22

A lineās slope is \({{\frac{4}{7}}}\text{.}\) The line passes through the point \((-14,{-5})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 23

A lineās slope is \({-{\frac{5}{7}}}\text{.}\) The line passes through the point \((-7,{8})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

###### 24

A lineās slope is \({-{\frac{6}{5}}}\text{.}\) The line passes through the point \((-15,{15})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.

An equation for this line in point-slope form is: .

An equation for this line in slope-intercept form is: .

Convert to slope-intercept form.

###### 25

Change this equation from point-slope form to slope-intercept form.

\({y}={5\!\left(x-1\right)}\)

In slope-intercept form, this lineās equation would be .

###### 26

Change this equation from point-slope form to slope-intercept form.

\({y}={5\!\left(x+5\right)-23}\)

In slope-intercept form, this lineās equation would be .

###### 27

Change this equation from point-slope form to slope-intercept form.

\({y}={-2\!\left(x+1\right)}\)

In slope-intercept form, this lineās equation would be .

###### 28

Change this equation from point-slope form to slope-intercept form.

\({y}={-5\!\left(x-4\right)-25}\)

In slope-intercept form, this lineās equation would be .

###### 29

Change this equation from point-slope form to slope-intercept form.

\(\displaystyle{ {y}={{\frac{2}{7}}\!\left(x-14\right)+1} }\)

In slope-intercept form, this lineās equation would be .

###### 30

Change this equation from point-slope form to slope-intercept form.

\(\displaystyle{ {y}={{\frac{3}{4}}\!\left(x-4\right)+5} }\)

In slope-intercept form, this lineās equation would be .

###### 31

Change this equation from point-slope form to slope-intercept form.

\(\displaystyle{ {y}={-{\frac{4}{9}}\!\left(x+18\right)+4} }\)

In slope-intercept form, this lineās equation would be .

###### 32

Change this equation from point-slope form to slope-intercept form.

\(\displaystyle{ {y}={-{\frac{5}{6}}\!\left(x-6\right)-1} }\)

In slope-intercept form, this lineās equation would be .

Applications

###### 33

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 \(300\) minutes on the phone, and paid \({\$26.50}\text{.}\) In another month, you spent \(370\) minutes on the phone, and paid \({\$29.65}\text{.}\)

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.

This linear modelās slope-intercept equation is .

If you spent \(170\) minutes over the phone in a month, you would pay .

If in a month, you paid \({\$34.60}\) of cell phone bill, you must have spent minutes on the phone in that month.

###### 34

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 \(2003\text{,}\) there was still \({\$840{,}000}\) left in the fund. In \(2006\text{,}\) there was \({\$711{,}000}\) 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.

The linear modelās slope-intercept equation is .

In the year \(2010\text{,}\) there was left in the fund.

In the year , the fund will be empty.

###### 35

A biologist has been observing a treeās height. \(14\) months into the observation, the tree was \(19.52\) feet tall. \(16\) months into the observation, the tree was \(20.08\) 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.

This lineās slope-intercept equation is .

\(28\) months after the observations started, the tree would be feet in height.

months after the observation started, the tree would be \(31.56\) feet tall.

###### 36

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.

Six minutes since the experiment started, the gas had a mass of \(136.4\) grams.

Eighteen minutes since the experiment started, the gas had a mass of \(99.2\) 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.

This lineās slope-intercept equation is .

\(36\) 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.

###### 37

By your cell phone contract, you pay a monthly fee plus \({\$0.03}\) for each minute you spend on the phone. In one month, you spent \(270\) minutes over the phone, and had a bill totaling \({\$26.10}\text{.}\)

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.

This lineās slope-intercept equation is .

If you spend \(160\) minutes on the phone in a month, you would be billed .

If your bill was \({\$32.10}\) one month, you must have spent minutes on the phone in that month.

###### 38

A company set aside a certain amount of money in the year 2000. The company spent exactly \({\$26{,}000}\) from that fund each year on perks for its employees. In \(2002\text{,}\) there was still \({\$760{,}000}\) 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.

The linear modelās slope-intercept equation is .

In the year \(2010\text{,}\) there was left in the fund.

In the year , the fund will be empty.

###### 39

A biologist has been observing a treeās height. This type of tree typically grows by \(0.16\) feet each month. Twelve months into the observation, the tree was \(14.32\) 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.

This lineās slope-intercept equation is .

\(28\) months after the observations started, the tree would be feet in height.

months after the observation started, the tree would be \(21.52\) feet tall.

###### 40

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. Its mass is leaking by \(8.5\) grams each minute. Ten minutes since the experiment started, the remaining gas had a mass of \(289\) 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.

This lineās slope-intercept equation 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.