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:
\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 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
\begin{equation*} \frac{y-253}{x-1990}=2.865\text{.} \end{equation*}There is good reasonā1āIt 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\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 discernable from the numbers in the equation.
Note: it is also common to define point-slope form as
\begin{equation} y-y_0=m\left(x-x_0\right)\tag{4.6.2} \end{equation}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:
\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 formĀ (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 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:
\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.
Exercise4.6.9
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.
Example4.6.10
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 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:
\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*}Now the point-slope equation looks like \(y=-\frac{2}{3}(x-x_0)+y_0\text{.}\) 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*}Exercise4.6.11
Subsection4.6.3Exercises
Exercises on Point-Slope Form Equations
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Exercises on Finding Linear Equation by Two Points