SubsectionSlope-Intercept Form
Slope-Intercept Form
The form of a linear equation that you're probably most famililar with is called the slope-intercept form, which looks like
\begin{equation*}
y=mx+b.
\end{equation*}
If a linear equation is in this form, then \(m\text{,}\) the coefficient in front of \(x\text{,}\) is the slope of the line, and \(b\text{,}\) the constant term, corresponds to the \(y\)-intercept of \((0,b)\text{.}\)
For example, suppose we're given the linear equation \(3x-4y=8\text{.}\) Solving for \(y\text{,}\) we have
\begin{align*}
3x-4y\amp =8\\
-4y\amp = -3x+8\\
y\amp =\frac{-3x+8}{-4}\\
y\amp = \frac{-3x}{-4}+\frac{8}{-4}\\
y\amp = \frac{3}{4}x-2\text{.}
\end{align*}
From here, we can immediately see that the slope of this line is \(\frac{3}{4}\) and the \(y\)-intercept is at \((0,-2)\text{.}\) With this information, we can easily graph this line: we can first plot the \(y\)-intercept at \((0,-2)\text{,}\) and then use the slope to plot another point by going up \(3\) and over to the right \(4\text{.}\)
![Quadrants of the Plane](images/Slope4.png)
Example85
Put the linear equation \(3y+5x=18\) into slope-intercept form. Graph this equation, and identify the \(x\)-intercept.
SolutionTo put this equation into slope-intercept form, we need to solve for \(y\text{:}\)
\begin{equation*}
\begin{aligned}
3y + 5x \amp= 18 \\
3y \amp= -5x + 18 \\
y \amp= -\frac{5}{3}x + 6
\end{aligned}
\end{equation*}
From here, we see that the slope is \(-\frac{5}{3}\) and the \(y\)-intercept is \((0,6)\text{.}\) To graph this equation, we first plot the \(y\)-intercept. Using the fact that
\begin{equation*}
m=-\dfrac{5}{3}=\dfrac{-5}{3}=\dfrac{\text{rise}}{\text{run}},
\end{equation*}
we can find another point by starting from the \(y\)-intercept and moving down \(5\) units and right \(3\) units. Drawing the line passing through these two points, we get the line for this equation.
![Quadrants of the Plane](images/Slope6.png)
To determine the \(x\)-intercept, we substitute \(y=0\text{:}\)
\begin{equation*}
\begin{aligned}
3(\alert{0}) + 5x \amp = 18 \\
5x \amp= 18 \\
x \amp= \frac{18}{5}
\end{aligned}
\end{equation*}
Therefore, the \(x\)-intercept is \(\left(\frac{18}{5},0\right)\text{.}\) The general rule is to label all important points that cannot be clearly read from the graph.
![Quadrants of the Plane](images/Slope7.png)
Example86
Write a linear equation that defines the graph below, and find the \(x\)-intercept of the graph.
![](images/Slope9.png)
SolutionConsider the two points marked on the graph, \((0,3)\) and \((3,1)\text{.}\) From these points, we see that the slope is
\begin{equation*}
m=\dfrac{\text{rise}}{\text{run}}=\dfrac{-2}{3}
\end{equation*}
In addition, the \(y\)-intercept is \((0, 3)\) and thus \(b=3.\)Putting these together, we can write an equation in slope-intercept form:
\begin{equation*}
y=-\dfrac{2}{3}x+3
\end{equation*}
To find the \(x\)-intercept of the graph, we substitute \(y=0\) and solve for \(x\text{:}\)
\begin{equation*}
\begin{aligned}
\alert{0} \amp= -\dfrac{2}{3}x+3 \\
\dfrac{2}{3}x \amp= 3 \\
x \amp= 3\cdot\dfrac{3}{2} \\
x\amp= \dfrac{9}{2}
\end{aligned}
\end{equation*}
Thus the \(x\)-intercept is given by \(\left(\frac{9}{2},0\right)\text{.}\)
SubsectionPoint-Slope Form
The slope-intercept form of a line is perhaps the simplest form, but it has its limitations. For example, suppose we wanted to write an equation for the following line.
![](images/point-slope-1.png)
While we can certainly find the slope of this line, the \(y\)-intercept is not immediately clear. In cases like this where the \(y\)-intercept is not known, we can use the point-slope form of a line.
As the name suggests, point-slope form is useful when we know the slope and one point on a line. Suppose the slope of a line is \(m\) and we have a point \((x_1,y_1)\text{.}\) Let \((x,y)\) be a variable point. Then using the slope formula and rearranging, we have
\begin{equation*}
\begin{aligned}
m \amp= \dfrac{y-y_1}{x-x_1} \\
m(x-x_1) \amp = y - y_1
\end{aligned}
\end{equation*}
Therefore, the equation of a line with a slope of \(m\) through the point \((x_1,y_1)\) can be written in point-slope form:
\begin{equation*}
y-y_1=m(x-x_1)
\end{equation*}
Let's return to the graph above. On the graph, we can identify two points, \((-1,0)\) and \((2,1)\text{.}\) These two points give us a slope of \(m=\frac{1}{3}\text{.}\) Next, we can use either point to write an equation for this line. Using \((x_1,y_1)=(-1,0)\text{,}\) we get an equation of
\begin{equation*}
y-0=\dfrac{1}{3}(x-(-1))
\end{equation*}
Using \((x_1,y_1)=(2,1)\text{,}\) we get
\begin{equation*}
y-1=\dfrac{1}{3}(x-2)
\end{equation*}
It is not obvious that these are actually the same line; to verify this, we can simplify them both into slope-intercept form.
\begin{equation*}
\begin{aligned}
y-0 \amp= \dfrac{1}{3}(x-(-1)) \\
y \amp= \dfrac{1}{3}(x+1) \\
y \amp= \dfrac{1}{3}x+\dfrac{1}{3} \\
\\
y-1 \amp= \dfrac{1}{3}(x-2) \\
y-1 \amp= \dfrac{1}{3}x - \dfrac{2}{3} \\
y \amp= \dfrac{1}{3}x + \dfrac{1}{3}
\end{aligned}
\end{equation*}
Point-Slope Form
The point-slope form looks like
\begin{equation*}
y-y_1=m(x-x_1)
\end{equation*}
where \(m\) is the slope and \((x_1,y_1)\) is any point on the line. Sometimes, it is helpful to rewrite the formula as
\begin{equation*}
y=y_1+m(x-x_1).
\end{equation*}
Example88
Write an equation for the line passing through the point \((3,6)\) and \((-2,5)\text{.}\)
SolutionFirst, we compute the slope:
\begin{equation*}
m=\dfrac{6-5}{3-(-2)}=\dfrac{1}{5}
\end{equation*}
Next, we use either point to construct an equation in point-slope form. It does not matter which point we choose. Using \((3,6)\text{,}\) we have
\begin{equation*}
y-6=\dfrac{1}{5}(x-3)
\end{equation*}
Example89
Write an equation that defines the following graph.
![](images/point-slope-2.png)
SolutionFrom the two marked points on the graph, we see that the slope is
\begin{equation*}
m=\dfrac{\text{rise}}{\text{run}}=\dfrac{1}{6}
\end{equation*}
Next, we use either point to construct an equation in point-slope form. Using \((-4,-2)\text{,}\) we have
\begin{equation*}
\begin{aligned}
y-(-2) \amp= \dfrac{1}{6}(x-(-4)) \\
y+2 \amp= \dfrac{1}{6}(x+4)
\end{aligned}
\end{equation*}