Find the Equation of a Line

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.

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

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.

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

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.

We list the steps for easy reference.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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.

The steps are summarized here.

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.

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 \({m}_{∥}=2.\) We’ll use the notation \({m}_{∥}\) to represent the slope of a line parallel to a line with slope m. (Notice that the subscript || 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.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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 \({m}_{⊥}\) to represent the slope of a line perpendicular to a line with slope m. (Notice that the subscript \(⊥\) looks like the right angles made by two perpendicular lines.)

We now know the perpendicular line will pass through \((-2,1)\) with \({m}_{⊥}=-\frac{1}{2}.\)

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\((-2,1)?\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

• How to find an equation of a line given the slope and a point.
  1. Identify the slope.
  2. Identify the point.
  3. Substitute the values into the point-slope form, \(y-{y}_{1}=m(x-{x}_{1}).\)
  4. Write the equation in slope-intercept form.
     
     
     
     
• How to find an equation of a line given two points.
  1. Find the slope using the given points. \(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)
  2. Choose one point.
  3. Substitute the values into the point-slope form: \(y-{y}_{1}=m(x-{x}_{1}).\)
  4. Write the equation in slope-intercept form.
     

     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	\(y-{y}_{1}=m(x-{x}_{1})\)

• How to find an equation of a line parallel to a given line.
  1. Find the slope of the given line.
  2. Find the slope of the parallel line.
  3. Identify the point.
  4. Substitute the values into the point-slope form: \(y-{y}_{1}=m(x-{x}_{1}).\)
  5. Write the equation in slope-intercept form
• How to find an equation of a line perpendicular to a given line.
  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line.
  3. Identify the point.
  4. Substitute the values into the point-slope form, \(y-{y}_{1}=m(x-{x}_{1})\)
  5. Write the equation in slope-intercept form.

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

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope-intercept form.

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope-intercept form.

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

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope-intercept form.

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope-intercept form.

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope-intercept form.

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope-intercept form.

Mixed Practice

In the following exercises, find the equation of each line. Write the equation in slope-intercept form.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

We can easily determine the slope and intercept of a line if the equation was 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.

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

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.

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

We can use the point–slope form of an equation to find an equation of a line when we are given the slope and one point. Then we will rewrite the equation in slope–intercept form. Most applications of linear equations use the the slope–intercept form.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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.

We have two options so far for finding an equation of a line: slope–intercept or point–slope. Since we will know two points, it will make 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.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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.

The graph shows the graph of \(y=2x-3\). 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\({m}_{∥}=2\). We’ll use the notation \({m}_{∥}\) to represent the slope of a line parallel to a line with slope \(m\). (Notice that the subscript \(∥\) 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.

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

Now, let’s see how to do this algebraically.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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.

The graph shows the graph of \(y=2x-3\). 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 \({m}_{\text{⊥}}\) to represent the slope of a line perpendicular to a line with slope \(m\). (Notice that the subscript ⊥ looks like the right angles made by two perpendicular lines.)

We now know the perpendicular line will pass through \((-2,1)\) with \({m}_{\text{⊥}}=-\frac{1}{2}\).

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 \((-2,1)\)?

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.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

• To Find an Equation of a Line Given the Slope and a Point
  1. Identify the slope.
  2. Identify the point.
  3. Substitute the values into the point-slope form, \(y-{y}_{1}=m(x-{x}_{1})\).
  4. Write the equation in slope-intercept form.

  
  
  

• To Find an Equation of a Line Given Two Points
  1. Find the slope using the given points.
  2. Choose one point.
  3. Substitute the values into the point-slope form, \(y-{y}_{1}=m(x-{x}_{1})\).
  4. Write the equation in slope-intercept form.

  
  
  

• To Write and Equation of a Line
  • If given slope and y-intercept, use slope–intercept form \(y=mx+b\).
  • If given slope and a point, use point–slope form \(y-{y}_{1}=m(x-{x}_{1})\).
  • If given two points, use point–slope form \(y-{y}_{1}=m(x-{x}_{1})\).

  
  
  

• To Find an Equation of a Line Parallel to a Given Line
  1. Find the slope of the given line.
  2. Find the slope of the parallel line.
  3. Identify the point.
  4. Substitute the values into the point-slope form, \(y-{y}_{1}=m(x-{x}_{1})\).
  5. Write the equation in slope-intercept form.

  
  
  

• To Find an Equation of a Line Perpendicular to a Given Line
  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line.
  3. Identify the point.
  4. Substitute the values into the point-slope form, \(y-{y}_{1}=m(x-{x}_{1})\).
  5. Write the equation in slope-intercept form.

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

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope–intercept form.

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope–intercept form.

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

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope–intercept form.

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope–intercept form.

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope–intercept form.

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

Condensed — the full section is in OpenStax Elementary Algebra 2e.