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Section6.4The 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 Table 6.4.1. 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 \(2\text{.}\)

\(x\) \(y\)
\(-6\) \(0\)
\(-3\) \(2\)
\(0\) \(4\)
\(3\) \(6\)
\(6\) \(8\)
\(9\) \(10\)
Table6.4.1\(y=\frac{2}{3}x+4\)

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 \((0,7)\text{.}\)

Consider the line graphed in Figure 6.4.2. 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*}
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Figure6.4.2By observation \(y=-\frac{3}{4}x+2\)

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

Example6.4.3

Determine the slope-intercept equation of the line that passes through the two points indicated in Table 6.4.4.

\(x\) \(y\)
\(-7\) \(-8\)
\(14\) \(1\)
Table6.4.4Points on a line
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*}

Subsection6.4.1Exercises

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 \(x=-\frac{1}{5}y-7\text{.}\)

\begin{align*} x\amp=-\frac{1}{5}y-7\\ x\addright{7}\amp=-\frac{1}{5}y-7\addright{7}\\ x+7\amp=-\frac{1}{5}y\\ \multiplyleft{-5}(x+7)\amp=\multiplyleft{-5}-\frac{1}{5}y\\ -5x-35\amp=y \end{align*}

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 6.4.5.

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Figure6.4.5Determine the equation of the line
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!