## Section7.4The Slope-Intercept Form of the Equation of a Line

The standard form of the linear equation in two variables is useful in enabling quick computations of both intercepts. It is also useful when solving systems of linear equations — a topic you may or may not have yet studied. In many other situations, however, there is a much more powerful form of a linear equation in two variables. That form is called the slope-intercept form of a linear equation.

Consider the line with equation $-2x+3y=12\text{.}$ If we isolate $y\text{,}$we get the equivalent equation $y=\frac{2}{3}x+4\text{.}$ Right away, we get a benefit from this new form of the equation. Rather than having to replace $x$ or $y$ with a value and solve for the other variable, we have a direct formula into which we can substitute a value for $x$ and subsequently directly evaluate the value of $y\text{.}$ For example, if we let $x=3\text{,}$ we can directly calculate $y\text{.}$

\begin{align*} y\amp=\frac{2}{3} \cdot 3+4\\ \amp=2+4\\ \amp=6 \end{align*}

This tells us that an ordered pair that satisfies both $y=\frac{2}{3}x+4$ and its parent equation, $-2x+3y=12\text{,}$ is $(3,6)\text{.}$

It's no accident that I chose $3$ as my value for $x\text{.}$ I can tell from the coefficient on $x$ in the equation $y=\frac{2}{3}x+4$ that if the value I choose for $x$ is evenly divisible by $3\text{,}$ then the resultant value for $y$ will be an integer. By inference, isolating $y$ in a linear equation with two variables can help identify points on a line where both coordinates of the point are integers. This can be especially useful when producing a graph of the line.

Several ordered pairs that satisfy the equation $y=\frac{2}{3}x+4$ are shown in Figure 7.4.1. Note that the third ordered pair in the table is the $y$-intercept of the line. Also note that as you read from one row to the next, the value of $x$ increases by $3$ and the value of $y$ increases by $2\text{.}$

These values correspond to runs of $3$ and rises of $2\text{.}$ From this we get

\begin{align*} m\amp=\frac{\text{rise}}{\text{run}}\\ \amp=\frac{2}{3} \end{align*}

Crikey! The coefficient on $x$ in the equation $y=\frac{2}{3}x+4$ is the slope of the line. That observation coupled with the more apparent fact (via direct substitution) that the $y$-intercept of the line is $(0,4)$ leads us to the following generalization.

The equation of the line with a slope of $m$ and the $y$-intercept $(0,b)$ can be written as:

\begin{equation*} y=mx+b\text{.} \end{equation*}

An equation of a line in the form $y=mx+b\text{,}$ where $m$ is the slope of the line and $(0,b)$ is the $y$-intercept of the line, is called the slope-intercept form of the equation of the line.

For example, when presented with the equation $y=-2x+7$ we can immediately conclude that the slope of the line is $-2$ and that the $y$-intercept of the line is $(0,7)\text{.}$

###### Example7.4.3.

Determine the slope of the line whose equation is $3x-5y=12\text{.}$

Solution

If we isolate $y$ the coefficient on $x$ will be the elope of the line and the constant term will be the $y$-coordinate of the $y$-intercept. Let's do it.

\begin{align*} 3x-5y\amp=12\\ 3x-5y\subtractright{3x}\amp=12\subtractright{3x}\\ -5y\amp=-3x+12\\ \divideunder{-5y}{-5}\amp=\divideunder{-3x+12}{-5}\\ y\amp=\frac{3}{5}x-\frac{12}{5} \end{align*}

The slope of the line is $\frac{3}{5}$ and the $y$-intercept of the line is $\left(0,-\frac{12}{5}\right)\text{.}$

Consider the line graphed in Figure 7.4.5. We can see that the $y$-intercept of the line is $(0,2)\text{.}$ From the indicated slope-triangle we have

\begin{align*} m\amp=\frac{\text{rise}}{\text{run}}\\ \amp=\frac{-3}{4}\text{.} \end{align*} Figure 7.4.5. By observation $y=-\frac{3}{4}x+2$

So we can conclude that the slope-intercept equation for the line is $y=-\frac{3}{4}x+2\text{.}$

###### Example7.4.6.

Determine the slope-intercept equation of the line that passes through the two points indicated in Figure 7.4.7.

Solution

The first thing we need to determine is the slope of the line. Letting $(x_1,y_1)$ be $(-7,-8)$ and $(x_2,y_2)$ be $(14,1)$we have:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{1-(-8)}{14-(-7))}\\ \amp=\frac{9}{21}\\ \amp=\frac{3}{7} \end{align*}

So we now know that the equation of the line is $y=\frac{3}{7}x+b$ where $(0,b)$ is the unknown $y$-intercept of the line. We can use the ordered pair $(14,1)$ to determine the value of $b\text{.}$

Letting $x=14$ and $y=1$ in the equation $y=\frac{3}{7}x+b$ we have the following.

\begin{align*} 1\amp=\frac{3}{7} \cdot 14+b\\ 1\amp=6+b\\ 1\subtractright{6}\amp=6+b\subtractright{6}\\ -5\amp=b \end{align*}

So we can conclude that the slope-intercept equation of the line is $y=\frac{3}{7}x+(-5)$ which we simplify to

\begin{equation*} y=\frac{3}{7}x-5\text{.} \end{equation*}

### ExercisesExercises

Determine information about a line given the equation of the line.

###### 1.

Determine the slope and $y$-intercept of the line with equation $2x-7y=21\text{.}$

Solution

We begin by isolating $y$ in the equation $2x-7y=21\text{.}$

\begin{align*} 2x-7y\amp=21\\ 2x-7y\subtractright{2x}\amp=21\subtractright{2x}\\ -7y\amp=-2x+21\\ \multiplyleft{-\frac{1}{7}}(7y)\amp=\multiplyleft{-\frac{1}{7}}(-2x+21)\\ y\amp=\frac{2}{7}x-3 \end{align*}

Now that the equation is in the form $y=mx+b\text{,}$ we can see that the slope of the line is $\frac{2}{7}$ and that the $y$-intercept is $(0,-3)\text{.}$

###### 2.

Determine the slope and $y$-intercept of the line with equation $x=-\frac{1}{5}y-7\text{.}$

Solution

We begin by isolating $y$ in the equation $x=-\frac{1}{5}y-7\text{.}$

The last equation is equivalent to the equation $y=-5x-35\text{.}$ Now that the equation is in the form $y=mx+b\text{,}$ we can see that the slope of the line is $-5$ and that the $y$-intercept is $(0,-35)\text{.}$

Determine the equation of a line given information about the line.

###### 3.

Determine the slope-intercept equation or the line shown in Figure 7.4.9.

Solution

By inspection, the slope of the line is $2$ and the $y$-intercept of the line is $(0,-3)\text{.}$ So fitting the form $y=mx+b\text{,}$ we know that the equation of the line is $y=2x-3\text{.}$

###### 4.

Determine the slope-intercept equation of the line that passes through the points $(-9,7)$ and $(18,-5)\text{.}$

Solution

We begin by determining the slope of the line.

\begin{equation*} (x_1,y_1) \text{ is } (-9,7) \text{ and } (x_2,y_2) \text{ is } (18,-5) \end{equation*}
\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-5-7}{18-(-9)}\\ \amp=\frac{-12}{27}\\ \amp=-\frac{4}{9} \end{align*}

We now know that the equation has form $y=-\frac{4}{9}x+b\text{.}$ Let's use the point $(-9,7)$ to determine $b\text{.}$

\begin{align*} 7\amp=-\frac{4}{9} \cdot -9+b\\ 7\amp=4+b\\ 7\subtractright{4}\amp=4+b\subtractright{4}\\ 3\amp=b \end{align*}

So the equation of the line is $y=-\frac{4}{9}x+3\text{.}$ (Both ordered pairs check in the equation.)

###### 5.

Determine the slope-intercept equation of the line that passes through the points $(-11,8)$ and $(-19,8)\text{.}$

Solution

We begin by determining the slope of the line.

\begin{equation*} (x_1,y_1) \text{ is } (-11,8) \text{ and } (x_2,y_2) \text{ is } (-19,8) \end{equation*}
\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{8-8}{-19-(-11)}\\ \amp=\frac{0}{-8}\\ \amp=0 \end{align*}

We now know that the equation has form $y=0x+b\text{.}$ We can use the point $(-11,8)$ to determine $b\text{.}$

\begin{align*} 8\amp=0 \cdot -11+b\\ 8\amp=0+b\\ 8\amp=b \end{align*}

So the equation of the line is $y=0x+8$ or, simplified, just $y=8\text{.}$ Looking again at the two points ... makes sense!