PCC SLC Math Resources

Section 7.4 The 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 \(y\text{.}\)

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.

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

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:

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 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*}

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

Exercises Exercises

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 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!