**3.3 Find the Equation of a Line**

Topics covered in this section are:

- Find an equation of the line given the slope and $y$-intercept
- Find an equation of the line given the slope and a point
- Find an equation of the line given two points
- Find an equation of a line parallel to a given line
- Find an equation of a line perpendicular to a given line

How do online companies know that “you may also like” a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.

The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section, we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.

**3.3.1 Find an Equation of the Line Given the Slope and ***y*-Intercept

*y*-Intercept

We can easily determine the slope and intercept of a line if the equation is written in slope-intercept form, $y=mx+b$. Now we will do the reverse—we will start with the slope and $y$-intercept and use them to find the equation of the line.

**Example 1**

Find the equation of a line with slope $-9$ and $y$-intercept $(0, -4)$.

**Solution**

Since we are given the slope and $y$-intercept of the line, we can substitute the needed values into the slope-intercept form, $y=mx+b$.

Name the slope. | $m=\textcolor{red}{-9}$ |

Name the $y$-intercept. | $y$-intercept $(0, \textcolor{blue}{4})$ |

Substitute the values into $y=mx+b$. | $y=\textcolor{red}{m}x+\textcolor{blue}{b}$ $y=\textcolor{red}{-9}x+\textcolor{blue}{(-4)}$ $y=-9x-4$ |

Sometimes, the slope and intercept need to be determined from the graph.

**Example 2**

Find the equation of the line shown in the graph.

**Solution**

We need to find the slope and $y$-intercept of the line from the graph so we can substitute the needed values into the slope-intercept form, $y=mx+b$.

To find the slope, we choose two points on the graph.

The $y$-intercept is $(0, -4)$ and the graph passes through $(3, -2)$.

Find the slope, by counting the rise and run. | $m=\frac{rise}{run}$ |

$m=\textcolor{red}{\frac{2}{3}}$ | |

Find the $y$-intercept. | $y$-intercept $(0, \textcolor{blue}{-4})$ |

Substitute the value into $y=mx+b$. | $y=\textcolor{red}{m}x+\textcolor{blue}{b}$ $y=\textcolor{red}{\frac{2}{3}}x\textcolor{blue}{-4}$ |

**3.3.2 Find an Equation of the Line Given the Slope and a Point**

Finding an equation of a line using the slope-intercept form of the equation works well when you are given the slope and $y$-intercept or when you read them off a graph. But what happens when you have another point instead of the $y$-intercept?

We are going to use the slope formula to derive another form of an equation of the line.

Suppose we have a line that has slope $m$ and that contains some specific point $(x_1, y_1)$ and some other point, which we will just call $(x, y)$. We can write the slope of this line and then change it to a different form.

$m=\frac{y-y_1}{x-x_1}$ | |

Multiply both sides of the equation by $x-x_1$. | $m(x-x_1)=\frac{y-y_1}{x-x_1}(x-x_1)$ |

Simplify. | $m(x-x_1)=y-y_1$ |

Rewrite the equation with the $y$ terms on the left. | $y-y_1=m(x-x_1)$ |

This format is called the **point-slope form** of an equation of a line.

**POINT-SLOPE FORM OF AN EQUATION OF A LINE**

The **point-slope form** of an equation of a line with slope $m$ and containing the point $(x_1, y_1)$ is:

$y-y_1 = m(x-x_1)$

We can use the point-slope form of an equation to find an equation of a line when we know the slope and at least one point. Then, we will rewrite the equation in slope-intercept form. Most applications of linear equations use the the slope-intercept form.

**Example 3**

Find an equation of a line with slope $m=-\frac{1}{3}$ that contains the point $(6, −4)$. Write the equation in slope-intercept form.

**Solution**

Step 1. Identify the slope. | The slope is given. | $\textcolor{blue}{m=-\frac{1}{3}}$ |

Step 2. Identify the point. | The point is given. | $\begin{pmatrix}x_1 \ \ \ y_1 \\ 6, -4 \end{pmatrix}$ |

Step 3. Substitute the values into the point-slope form $y-y_1=m(x-x_1)$. | Simplify. | $\begin{align*} y-\textcolor{red}{y_1}&=\textcolor{blue}{m}(x-\textcolor{red}{x_1}) \\ y-\textcolor{red}{(-4)}&=\textcolor{blue}{-\frac{1}{3}}(x-\textcolor{red}{6}) \\ y+4 &=-\frac{1}{3}x+2 \end{align*}$ |

Step 4. Write the equation in slope-intercept form. | $y=-\frac{1}{3}x-2$ |

We list the steps for easy reference.

**HOW TO: find an equation of a line given the slope and a point.**

- Identify the slope.
- Identify the point.
- Substitute the values into the point-slope form, $y-y_1=m(x-x_1)$.
- Write the equation in slope-intercept form.

**Example 4**

Find an equation of a horizontal line that contains the point $(−2, −6)$. Write the equation in slope-intercept form.

**Solution**

Every horizontal line has slope 0. We can substitute the slope and points into the point-slope form, $y-y_1=m(x-x_1)$.

Identify the slope. | $m=\textcolor{blue}{0}$ |

Identify the point. | $\begin{pmatrix} \textcolor{red}{x_1} \ \ \ \ \textcolor{red}{y_1} \\ \textcolor{red}{-2,} \textcolor{red}{-6} \end{pmatrix}$ |

Substitute the values into $y-y_1=m(x-x_1)$. | $y-\textcolor{red}{y_1} = \textcolor{blue}{m}(x- \textcolor{red}{x_1})$ $y-\textcolor{red}{(-6)} = \textcolor{blue}{m}(x- \textcolor{red}{(-2)})$ |

Simplify. | $y+6=0$ |

Write in slope-intercept form. | It is in $y$-form, but could be written $y=0x-6$. |

Did we end up with the form of a horizontal line, $y=b$?

**3.3.3 Find an Equation of the Line Given Two Points**

When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.

So far, we have two options for finding an equation of a line: slope-intercept or point-slope. When we start with two points, it makes more sense to use the point-slope form.

But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.

**Example 5**

Find an equation of a line that contains the points $(−3, −1)$ and $(2, −2)$. Write the equation in slope-intercept form.

**Solution**

Step 1. Find the slope using the given points. | Find the slope of the line through $(-3, -1)$ and $(2, -2)$. | $\begin{align*} m&=\frac{y_2-y_1}{x_2-x_1} \\ m&=\frac{-2-(-1)}{2-(-3)} \\ m&=\frac{-1}{5} \\ &\textcolor{blue}{m=-\frac{1}{5}} \end{align*}$ |

Step 2. Choose one point. | Choose either point. | $\begin{pmatrix} \textcolor{red}{x_1} \ \ \ \ \textcolor{red}{y_1} \\ \textcolor{red}{2, } \textcolor{red}{-2} \end{pmatrix}$ |

Step 3. Substitute the values into the point-slope form $y-y_1=m(x-x_1)$. | Simplify. | $\begin{align*} y-\textcolor{red}{y_1}&=\textcolor{blue}{m}(x-\textcolor{red}{x_1}) \\ y-\textcolor{red}{(-2)}&=\textcolor{blue}{-\frac{1}{5}}(x-\textcolor{red}{2}) \\ y+2 &=-\frac{1}{5}x+\frac{2}{5} \end{align*}$ |

Step 4. Write the equation in slope-intercept form. | $y=-\frac{1}{5}x-\frac{8}{5}$ |

The steps are summarized here.

**HOW TO: Find an equation of a line given two points.**

- Find the slope using the given points. $m=\frac{y_2−y_1}{x_2−x_1}$
- Choose one point.
- Substitute the values into the point-slope form: $y−y_1=m(x−x_1)$.
- Write the equation in slope-intercept form.

**Example 6**

Find an equation of a line that contains the points $(−3, 5)$ and $(−3, 4)$. write the equation in slope-intercept form.

**Solution**

Again, the first step will be to find the slope.

Find the slope of the line through $(-3, 5)$ and $(-3, 4)$. | $m=\frac{y_2-y_1}{x_2-x_1}$ |

$m=\frac{4-5}{-3-(-3)}$ | |

$m=\frac{-1}{0}$ | |

The slope is undefined. |

This tells us it is a vertical line. Both of our points have an $x$-coordinate of $-3$. So our equation of the line is $x=−3$. Since there is no $y$, we cannot write it in slope-intercept form.

You may want to sketch a graph using the two given points. Does your graph agree with our conclusion that this is a vertical line?

We have seen that we can use either the slope-intercept form or the point-slope form to find an equation of a line. Which form we use will depend on the information we are given.

To Write an Equation of a Line | ||

If given: | Use: | Form: |

Slope and $y$-intercept | slope-intercept | $y=mx+b$ |

Slope and a point | point-slope | $y-y_1=m(x-x_1)$ |

Two points | point-slope | $y-y_1=m(x-x_1)$ |

**3.3.4 Find an Equation of a Line Parallel to a Given Line**

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point-slope equation.

First, let’s look at this graphically.

This graph shows $y=2x−3$. We want to graph a line parallel to this line and passing through the point $(−2, 1)$.

We know that parallel lines have the same slope. So the second line will have the same slope as $y=2x−3$. That slope is $m_\parallel =2$. We’ll use the notation $m_\parallel$ to represent the slope of a line parallel to a line with slope $m$. (Notice that the subscript $\parallel$ looks like two parallel lines.)

The second line will pass through $(−2, 1)$ and have $m=2$.

To graph the line, we start at $(−2, 1)$ and count out the rise and run.

With $m=2$ (or $m=\frac{2}{1}$), we count out the rise $2$ and the run $1$. We draw the line as shown in the graph.

Do the lines appear parallel? Does the second line pass through $(−2, 1)$?

We were asked to graph the line, now let’s see how to do this algebraically.

We can use either the slope-intercept form or the point-slope form to find an equation of a line. Here we know one point and can find the slope. So we will use the point-slope form.

**Example 7**

Find an equation of a line parallel to $y=2x−3$ that contains the point $(−2, 1)$. Write the equation in slope-intercept form.

**Solution**

Step 1. Find the slope of the given line. | The line is in slope-intercept form, $y=2x-3$. | $m=2$ |

Step 2. Find the slope of the parallel line. | Parallel lines have the same slope. | $m_\parallel = 2$ |

Step 3. Identify the point. | The given point is $(-2, 1)$. | $\begin{pmatrix} x_1 \ y_1 \\ -2, 1 \end{pmatrix}$ |

Step 4. Substitute the values into the point-slope form, $y-y_1=m(x-x_1)$. | Simplify. | $\begin{align*} y-y_1&=m(x-x_1) \\ y-1 &= 2(x-(-2)) \\ y-1 &=2(x+2) \\ y-1&=2x+4 \end{align*}$ |

Step 5. Write the equation in slope-intercept form. | $y=2x+5$ |

Look at graph with the parallel lines shown previously. Does this equation make sense? What is the $y$-intercept of the line? What is the slope?

**HOW TO: Find an equation of a line parallel to a given line.**

- Find the slope of the given line.
- Find the slope of the parallel line.
- Identify the point.
- Substitute the values into the point-slope form: $y−y_1=m(x−x_1)$.
- Write the equation in slope-intercept form.

**3.3.5 Find an Equation of a Line Perpendicular to a Given Line**

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point-slope equation, like we did with parallel lines.

This graph shows $y=2x−3$. Now, we want to graph a line perpendicular to this line and passing through $(-2, 1)$.

We know that perpendicular lines have slopes that are negative reciprocals.

We’ll use the notation $m_\perp$ to represent the slope of a line perpendicular to a line with slope $m$. (Notice that the subscript $\perp$ looks like the right angles made by two perpendicular lines.)

$y=2x-3$ | perpendicular line |

$m=2$ | $m_\perp=-\frac{1}{2}$ |

We now know the perpendicular line will pass through $(−2, 1)$ with $m_\perp =−\frac{1}{2}$.

To graph the line, we will start at $(−2, 1)$ and count out the rise $−1$ and the run $2$. Then we draw the line.

Do the lines appear perpendicular? Does the second line pass through $(−2, 1)$?

We were asked to graph the line, now, let’s see how to do this algebraically.

We can use either the slope-intercept form or the point-slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point-slope form.

**Example 8**

Find an equation of a line perpendicular to $y=2x−3$ that contains the point $(−2, 1)$. Write the equation in slope-intercept form.

**Solution**

Step 1. Find the slope of the given line. | The line is in slope-intercept form, $y=2x-3$. | $m=2$ |

Step 2. Find the slope of the perpendicular line. | The slopes of perpendicular lines are negative reciprocals. | $m_\perp=-\frac{1}{2}$ |

Step 3. Identify the point. | The given point is $(-2, 1)$. | $\begin{pmatrix} x_1 \ y_1 \\ -2, \ 1 \end{pmatrix}$ |

Step 4. Substitute the values in the the point-slope form, $y-y_1=m(x-x_1)$. | Simplify. | $\begin{align*}y-y_1&=m(x-x_1 \\ y-1&=-\frac{1}{2}(x-(-2)) \\ y-1&=-\frac{1}{2}(x+2) \\ y-1&=-\frac{1}{2}x-1 \end{align*}$ |

Step 5. Write the equation in slope-intercept form. | $y=-\frac{1}{2}x$ |

**HOW TO: Find an equation of a line perpendicular to a given line.**

- Find the slope of the given line.
- Find the slope of the perpendicular line.
- Identify the point.
- Substitute the values into the point-slope form, $y−y_1=m(x−x_1)$.
- Write the equation in slope-intercept form.

**Example 9**

Find an equation of a line perpendicular to $x=5$ that contains the point $(3, −2)$. Write the equation in slope-intercept form.

**Solution**

Again, since we know one point, the point-slope option seems more promising than the slope-intercept option. We need the slope to use this form, and we know the new line will be perpendicular to $x=5$. This line is vertical, so its perpendicular will be horizontal. This tells us the $m_\perp = 0$.

Identify the point. | $(3, -2)$ |

Identify the slope of the perpendicular line. | $m_\perp = 0$ |

Substitute the values into $y-y_1=m(x-x_1)$. | $\begin{align*}y-y_1&=m(x-x_1) \\ y-(-2)&=0(x-(-2))\end{align*}$ |

Simplify. | $\begin{align*} y+2&=0 \\ y&=-2 \end{align*}$ |

Sketch the graph of both lines. On your graph, do the lines appear to be perpendicular?

In Example 9, we used the point-slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to $x=5$ that contains the point $(3, −2)$. This graph shows us the line $x=5$ and the point $(3, -2)$.

We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through $(3, -2)$.

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have $y$-coordinates of $−2$. So, the equation of the line perpendicular to the vertical line $x=5$ is $y=-2$.

**Example 10**

Find an equation of a line that is perpendicular to $y=−3$ that contains the point $(−3, 5)$. Write the equation in slope-intercept form.

**Solution**

The line $y=−3$ is a horizontal line. Any line perpendicular to it must be vertical, in the form $x=a$. Since the perpendicular line is vertical and passes through $(−3, 5)$, every point on it has an $x$-coordinate of $−3$. The equation of the perpendicular line is $x=−3$.

You may want to sketch the lines. Do they appear perpendicular?

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*Marecek, L., & Mathis, A. H. (2020). Find the Equation of a Line. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/3-3-find-the-equation-of-a-line*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction*