ORCCA Point-Slope Form

Section4.6Point-Slope Form

ObjectivesPCC Course Content and Outcome Guide

C.1.1:8.c
C.1.1:8.h
C.1.1:8.l
C.1.1:8.m
C.1.1:9.b
C.1.1:9.c
C.1.1:9.d

In Section 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.

Figure4.6.1Alternative Video Lesson

Subsection4.6.1Point-Slope Definition

Sometimes, one problem with slope-intercept form (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:

\begin{equation*} y=2.865x-5451\text{.} \end{equation*}

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 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 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 “initial value” $b$ 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

\begin{equation*} \frac{y-253}{x-1990}=2.865\text{.} \end{equation*}

There is good reason¹It will help us to see that $y$ (population) depends on $x$ (whatever year it is). to want to isolate $y$ in this equation:

\begin{align*} \frac{y-253}{x-1990}\amp=2.865\\ y-253\amp=2.865\multiplyright{(x-1990)}\amp\amp\text{(could distribute, but not going to)}\\ y\amp=2.865(x-1990)\addright{253} \end{align*}

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

\begin{equation} y=m\left(x-x_0\right)+y_0\label{equation-point-slope-form}\tag{4.6.1} \end{equation}

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 discernible from the numbers in the equation.

Figure4.6.5

Consider the graph in Figure 4.6.6.

Figure4.6.6

Checkpoint4.6.7

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

\begin{align*} y\amp=3(x-3)+2\amp y\amp=3(x-2)-1 \end{align*}

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

\begin{align*} y\amp=3(x-3)+2\amp y\amp=3(x-2)-1\\ y\amp=3x-9+2\amp y\amp=3x-6-1\\ y\amp=3x-7\amp y\amp=3x-7 \end{align*}

So, yes. It didn't matter which point we focused on in the line in Figure 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 formula (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:

\begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{2585-2975}{2016-2013}\\ \amp=\frac{-390}{3}\\ \amp=-130 \end{align*}

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:

\begin{equation*} y=-130(x-x_0)+y_0\text{.} \end{equation*}

Now we can pick one of those two given points, say $(2013,2975)\text{,}$ and get the equation

\begin{equation*} y=-130(x-2013)+2975\text{.} \end{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.

\begin{align*} y\amp=-130(x-2013)+2975\\ \substitute{1800}\amp=-130(x-2013)+2975\\ 1800\subtractright{2975}\amp=-130(x-2013)\\ -1175\amp-130(x-2013)\\ \divideunder{-1175}{-130}\amp=x-2013\\ 9.038\amp\approx x-2013\\ 9.038\addright{2013}\amp\approx x\\ 2022\amp\approx x \end{align*}

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.

Figure4.6.9A Graph of $y=-130(x-2013)+2975$

Checkpoint4.6.10

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.

Solution

We will use the slope formula (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:

\begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-10-0}{9-(-6)}\\ \amp=\frac{-10}{15}\\ \amp=-\frac{2}{3} \end{align*}

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:

\begin{align*} y\amp=-\frac{2}{3}(x-x_0)+y_0\\ y\amp=-\frac{2}{3}(x-9)+(-10)\\ y\amp=-\frac{2}{3}(x-9)-10 \end{align*}

Next, we will change the point-slope equation into slope-intercept form:

\begin{align*} y\amp=-\frac{2}{3}(x-9)-10\\ y\amp=-\frac{2}{3}x+6-10\\ y\amp=-\frac{2}{3}x-4 \end{align*}

Checkpoint4.6.12

Subsection4.6.3More on Point-Slope Form

We can tell a lot about a linear equation now that we have learned both slope-intercept form (4.5.1) and point-slope form (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.

$y=-3x+2$
$y=9(x+1)-6$
$y=5-x$
$y=-\frac{12}{5}(x-9)+1$

Solution

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 Figure 4.6.15. Find three equivalent equations that describe the line shown. Three integer valued points are shown for convenience.

Figure4.6.15Find Three Equations that Describe this Line

Solution

We could imagine infinitely many equivalent equations for the graph in Figure 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 Equation (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 yields…

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_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 $-\frac{12}{5}\text{.}$ 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 $\highlight{y=-\frac{12}{5}\left(x+5\right)+42}\text{.}$

SubsectionExercises

Point-Slope Form Basics

1

A line’s equation is given in point-slope form:

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

This line’s slope is .

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

2

A line’s equation is given in point-slope form:

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

This line’s slope is .

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

3

A line’s equation is given in point-slope form:

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

This line’s slope is .

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

4

A line’s equation is given in point-slope form:

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

This line’s slope is .

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

5

A line’s equation is given in point-slope form:

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

This line’s slope is .

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

6

A line’s equation is given in point-slope form:

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

This line’s 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 line’s equation in point-slope form.

Using the point $(1,9)\text{,}$ this line’s point-slope form equation is .

Using the point $(5,29)\text{,}$ this line’s point-slope form equation is .

8

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

Using the point $(3,25)\text{,}$ this line’s point-slope form equation is .

Using the point $(5,35)\text{,}$ this line’s point-slope form equation is .

9

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

Using the point $(5,-18)\text{,}$ this line’s point-slope form equation is .

Using the point $(-2,17)\text{,}$ this line’s point-slope form equation is .

10

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

Using the point $(-2,7)\text{,}$ this line’s point-slope form equation is .

Using the point $(5,-7)\text{,}$ this line’s point-slope form equation is .

11

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

Using the point $(18,{10})\text{,}$ this line’s point-slope form equation is .

Using the point $(27,{12})\text{,}$ this line’s point-slope form equation is .

12

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

Using the point $(-10,{3})\text{,}$ this line’s point-slope form equation is .

Using the point $(15,{18})\text{,}$ this line’s point-slope form equation is .

13

A line’s 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: .