3.4: Slope-Intercept Form of a Line

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We start with the definition of the $y$-intercept of a line.

The $y$-intercept

The point $(0,b)$ where the graph of a line crosses the $y$-axis is called the $y$-intercept of the line.

We will now generate the slope-intercept formula for a line having $y$-intercept $(0,b)$ and $\text {Slope} = m$ (see Figure $\PageIndex{1}$). Let $(x,y)$ be an arbitrary point on the line.

fig 3.4.1.png — Figure $\PageIndex{1}$: Line with $y$-intercept at $(0,b)$ and $\text {Slope} = m$.

Start with the fact that the slope of the line is the rate at which the dependent variable is changing with respect to the independent variable.

\[\text {Slope} =\dfrac{\Delta y}{\Delta x} \qquad \color{Red} \text {Slope formula.} \nonumber \]

Substitute $m$ for the slope. To determine both the change in $y$ and the change in $x$, subtract the coordinates of the point $(0,b)$ from the point $(x,y)$.

\[\begin{aligned} m&=\frac{y-b}{x-0} \quad \color {Red} \text { Substitute } m \text { for the Slope. } \Delta y =y-b \quad \color {Red} \text { and } \Delta x=x-0 .\\ m&=\frac{y-b}{x} \quad \color {Red}\text { Simplify. } \end{aligned} \nonumber \]

Clear fractions from the equation by multiplying both sides by the common denominator.

\[\begin{aligned} mx &= \left[\dfrac{y-b}{x}\right] x \quad \color {Red} \text { Multiply both sides by } x \\ mx &= y-b \quad \color {Red} \text { Cancel. } \end{aligned} \nonumber \]

Solve for $y$.

\[\begin{aligned} mx+b&= y-b+b \quad \color {Red} \text { Add } b \text { to both sides. } \\ mx+b&= y \quad \color {Red} \text { Simplify. } \end{aligned} \nonumber \]

Thus, the equation of the line is $y = mx + b$.

The Slope-Intercept form of a line

The equation of the line having $y$ intercept $(0,b)$ and slope $m$ is: \[y = mx + b \nonumber \]Because this form of a line depends on knowing the slope $m$ and the intercept $(0,b)$, this form is called the slope-intercept form of a line.

Example $\PageIndex{1}$

Sketch the graph of the line having equation $y=\dfrac{3}{5} x+1$.

Solution

Compare the equation $y=\dfrac{3}{5} x+1$ with the slope-intercept form $y = mx + b$, and note that $m =3 /5$ and $b = 1$. This means that the slope is $3/5$ and the $y$-intercept is $(0,1)$. Start by plotting the $y$-intercept $(0,1)$, then move uward $3$ units and right $5$ units, arriving at the point $(5,4)$. Draw the line through the points $(0,1)$ and $(5,4)$, then label it with its equation $y=\dfrac{3}{5} x+1$.

fig 3.4.2.png — Figure $\PageIndex{2}$: Hand-drawn graph of $y=\dfrac{3}{5} x+1$.

fig 3.4.3.png — Figure $\PageIndex{3}$: Select **5:ZSquare** from the ZOOM menu to draw the graph of $y=\dfrac{3}{5} x+1$.

When we compare the calculator image in Figure $\PageIndex{3}$ with the hand-drawn graph in Figure $\PageIndex{2}$, we get a better match.

Exercise $\PageIndex{1}$

Sketch the graph of the line having equation $y=-\dfrac{4}{3} x-2$.

Answer: $Exercise 3.4.1.png$

Example $\PageIndex{2}$

Sketch the line with $y$-intercept $(0,2)$ and slope $−7/3$. Label the line with the slope-intercept form of its equation.

Solution

Plot the $y$-intercept $(0,2)$. Now use the slope $−7/3$. Start at $(0,2)$, then move down seven units, followed by a three unit move to the right to the point $(3,−5)$. Draw the line through the points $(0,2)$ and $(3,−5)$. (See Figure $\PageIndex{4}$).

Next, the $y$-intercept is $(0,2)$, so $b = 2$. Further, the slope is $−7/3$, so $m = −7/3$. Substitute these numbers into the slope-intercept form of the line.

\[\begin{aligned} y&= mx+b \quad \color {Red} \text { Slope-intercept form. } \\ y&= -\dfrac{7}{3} x+2 \quad \color {Red} \text { Substitute: }-7 / 3 \text { for } m, 2 \text { for } b . \end{aligned} \nonumber \]

Therefore, the slope-intercept form of the line is $y=-\dfrac{7}{3} x+2$. Label the line with this equation.

fig 3.4.4.png — Figure $\PageIndex{4}$: Hand-drawn graph of $y=-\dfrac{7}{3} x+2$

Check: To graph $y=-\dfrac{7}{3} x+2$, enter $-7 / 3 * X+2$ in Y1 in the Y= menu. Select 6:ZStandard from the ZOOM menu, followed by 5:ZSquare from the ZOOM menu to produce the graph shown in Figure $\PageIndex{6}$.

fig 3.4.5.png — Figure $\PageIndex{5}$: Hand-drawn graph of $y=-\dfrac{7}{3} x+2$

fig 3.4.6.png — Figure $\PageIndex{6}$: Select **6:ZStandard** from the ZOOM menu, followed by **5:ZSquare** from the ZOOM menu to produce the graph of the equation $y=-\dfrac{7}{3} x+2$

This provides a good match of the hand-drawn graph in Figure $\PageIndex{5}$ and our graphing calculator result in Figure $\PageIndex{6}$.

Exercise $\PageIndex{2}$

Sketch the line with $y$-intercept $(0,−3)$ and slope $5/2$. Label the line with the slope-intercept form of its equation.

Answer: $Exercise 3.4.2.png$

Example $\PageIndex{3}$

Use the graph of the line in the following figure to find the equation of the line.

$Example 3.4.3.png$

Solution

Note that the $y$-intercept of the line is $(0,−1)$ (See Figure $\PageIndex{7}$). Next, we try to locate a point on the line that passes directly through a lattice point, a point where a vertical and horizontal grid line intersect. In Figure $\PageIndex{7}$, we chose the point $(5,6)$. Now, starting at the $y$-intercept $(0,1)$, we move up $7$ units, then to the right $5$ units. Hence, the slope is $m=\Delta y / \Delta x$, or $m =7 /5$.

Note

We could also subtract the coordinates of point $(0,−1)$ from the coordinates of point $(5,6)$ to determine the slope. \[\dfrac{6-(-1)}{5-0}=\dfrac{7}{5} \nonumber \]

fig 3.4.7.png — Figure $\PageIndex{7}$: The line has $y$-intercept $(0,−1)$ and slope $7/5$.

Next, state the slope-intercept form, the substitute $7/5$ for $m$ and $−1$ for $b$.

\[\begin{aligned} y&= mx+b \quad \color {Red} \text { Slope-intercept form. } \\ y&= \dfrac{7}{5} x+(-1) \quad \color {Red} \text { Substitute: } 7 / 5 \text { for } m,-1 \text { for } b \end{aligned} \nonumber \]

Thus, the equation of the line is $y=\dfrac{7}{5} x-1$.

Check: This is an excellent situation to run a check on your graphing calculator.

fig 3.4.8.png — Figure $\PageIndex{8}$: Enter $y=\dfrac{7}{5} x-1$ in the Y= menu.

fig 3.4.9.png — Figure $\PageIndex{9}$: Select **6:ZStandard** followed by **5:ZSquare** (both from the ZOOM menu) to produce this graph.

When we compare the result in Figure $\PageIndex{9}$ with the original hand-drawn graph (see Figure $\PageIndex{7}$), we’re confident we have a good match.

Exercise $\PageIndex{3}$

Use the graph of the line in the figure below to find the equation of the line.

$Exercise 3.4.3.png$

Answer: $y=-\dfrac{3}{5} x+3$

Applications

Let’s look at a linear application.

Example $\PageIndex{4}$

Jason spots his brother Tim talking with friends at the library, located $300$ feet away. He begins walking towards his brother at a constant rate of $2$ feet per second ($2$ ft/s).

Express the distance $d$ between Jason and his brother Tim in terms of the time $t$.
At what time after Jason begins walking towards Tim are the brothers $200$ feet apart?

Solution

Because the distance between Jason and his brother is decreasing at a constant rate, the graph of the distance versus time is a line. Let’s begin by making a rough sketch of the line. In Figure $\PageIndex{10}$, note that we’ve labeled what are normally the $x$- and $y$-axes with the time $t$ and distance $d$, and we’ve included the units with our labels.

fig 3.4.10.png — Figure $\PageIndex{10}$: A plot of the distance $d$ separating the brothers versus time $t$.

Let $t = 0$ seconds be the time that Jason begins walking towards his brother Tim. At time $t = 0$, the initial distance between the brothers is $300$ feet. This puts the $d$-intercept (normally the $y$-intercept) at the point $(0,300)$ (see Figure $\PageIndex{10}$).

Because Jason is walking toward his brother, the distance between the brothers decreases at a constant rate of $2$ feet per second. This means the line must slant downhill, making the slope negative, so $m = −2$ ft/s. We can construct an accurate plot of distance versus time by starting at the point $(0,300)$, then descending $\Delta d=-300$, then moving to the right $\Delta t=150$. This makes the slope $\Delta d / \Delta t=-300 / 150=-2$ (See Figure $\PageIndex{10}$). Note that the slope is the rate at which the distance $d$ between the brothers is changing with respect to time $t$.

Finally, the equation of the line is $y = mx + b$, where $m$ is the slope of the line and $b$ is the $y$-coordinate (in this case the $d$-coordinate) of the point where the graph crosses the vertical axis. Thus, substitute $−2$ for $m$, and $300$ for $b$ in the slope-intercept form of the line.

\[\begin{aligned} y&= mx+b \quad \color {Red} \text { Slope-intercept form. } \\ y&= -2x+300 \quad \color {Red} \text { Substitute: }-2 \text { for } m, 300 \text { for } b \end{aligned} \nonumber \]

One problem remains. The equation $y = −2x + 300$ gives us $y$ in terms of $x$.

The question required that we express the distance $d$ in terms of the time $t$. So, to finish the solution, replace $y$ with $d$ and $x$ with $t$ (check the axes labels in Figure $\PageIndex{10}$) to obtain a solution: \[d=-2t+300 \nonumber \]
Now that our equation expresses the distance between the brothers in terms of time, let’s answer part (b), “At what time after Jason begins walking towards Tim are the brothers $200$ feet apart?” To find this time, substitute $200$ for $d$ in the equation $d =−2t + 300$, then solve for $t$.\[\begin{aligned} d &= -2t+300 \quad \color {Red} \text { Distance equation } \\ 200 &= -2t+300 \quad \color {Red} \text { Substitute } 200 \text { for } d \end{aligned} \nonumber \]Solve this last equation for the time $t$.\[\begin{aligned} 200-300&= -2t+300-300 \quad \color {Red} \text { Subtract } 300 \text { from both sides. } \\ -100&= -2t \quad \color {Red} \text { Simplify both sides. } \\ \dfrac{-100}{-2}&= \dfrac{-2 t}{-2} \quad \color {Red} \text { Divide both sides by }-2 \\ 50&= t \quad \color {Red} \text { Simplify both sides. } \end{aligned} \nonumber \]Thus, it takes Jason $50$ seconds to close the distance between the brothers to $200$ feet.

Exercise $\PageIndex{4}$

A swimmer is $50$ feet from the beach, and then begins swimming away from the beach at a constant rate of $1.5$ feet per second ($1.5$ ft/s). Express the distance $d$ between the swimmer and the beach in terms of the time $t$.

Answer: $d=1.5 t+50$

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