Find the Equation of a 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.

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

We can easily determine the slope and intercept of a line if the equation is written in slope-intercept form,

y = m x + 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.

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})

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.

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.

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.

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.

To Write an Equation of a Line
If given:	Use:	Form:
Slope and y-intercept	slope-intercept	$y = m x + b$
Slope and a point	point-slope	$y - y_{1} = m (x - x_{1})$
Two points	point-slope	$y - y_{1} = m (x - x_{1})$

We want to graph a line parallel to this line and passing through the point

(−2, 1) .

This figure has a graph of a straight line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The point (negative 2, 1) is plotted. The line does not go through the point (negative 2, 1).

We know that parallel lines have the same slope. So the second line will have the same slope as

y = 2 x - 3 .

to represent the slope of a line parallel to a line with slope m. (Notice that the subscript \|\| looks like two parallel lines.)

), we count out the rise 2 and the run 1. We draw the line, as shown in the graph.

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.

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.

Now, we want to graph a line perpendicular to this line and passing through

(−2, 1) .

to represent the slope of a line perpendicular to a line with slope m. (Notice that the subscript

⊥

This figure has a graph of two perpendicular straight lines on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The first line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The points (negative 2, 1) and (0, 0) are plotted. A right triangle is drawn connecting the points (negative 2, 1), (negative 2, 0), and (0, 0). The second line goes through the points (negative 2, 1) and (0, 0).

Do the lines appear perpendicular? Does the second line pass through

(−2, 1) ?

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.

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

This figure has a graph of a straight vertical line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (5, 0), (5, 1), and (5, 2). The point (3, negative 2) is plotted. The line does not go through the point (3, negative 2).

We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through

(3, −2) .

If we look at a few points on this horizontal line, we notice they all have y-coordinates of

−2 .

Key Concepts

To Write an Equation of a Line
If given:	Use:	Form:
Slope and y-intercept	slope-intercept	$y = m x + b$
Slope and a point	point-slope	$y - y_{1} = m (x - x_{1})$
Two points	point-slope	$y - y_{1} = m (x - x_{1})$

Practice Makes Perfect

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.

slope 3 and* * *

y

-intercept $(0, 5)$

y = 3 x + 5

slope 8 and* * *

y-intercept $(0, −6)$

slope $−3$

and* * *

y

-intercept $(0, −1)$

y = −3 x - 1

slope $−1$

and* * *

y

-intercept $(0, 3)$

slope $\frac{1}{5}$

and* * *

y

-intercept $(0, −5)$

y = \frac{1}{5} x - 5

slope $- \frac{3}{4}$

and* * *

y

-intercept $(0, −2)$

slope 0 and* * *

y

-intercept $(0, −1)$

y = −1

slope $−4$

and* * *

y

-intercept $(0, 0)$

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

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 5), (1, negative 2), and (2, 1).

y = 3 x - 5

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 4), (1, 2), and (2, 0).

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 3), (2, negative 2), and (6, 0).

y = \frac{1}{2} x - 3

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 2), (4, 5), and (8, 8).

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 3), (3, negative 1), and (6, negative 5).

y = - \frac{4}{3} x + 3

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 1), (2, negative 4), and (4, negative 7).

This figure has a graph of a horizontal straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 2), (1, negative 2), and (2, negative 2).

y = −2

This figure has a graph of a horizontal straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 6), (1, 6), and (2, 6).

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.

m = \frac{5}{8},

point $(8, 3)$

y = \frac{5}{8} x - 2

m = \frac{5}{6},

point $(6, 7)$

m = - \frac{3}{5},

point $(10, −5)$

y = - \frac{3}{5} x + 1

m = - \frac{3}{4},

point $(8, −5)$

m = - \frac{3}{2},

point $(−4, −3)$

y = - \frac{3}{2} x + 9

m = - \frac{5}{2},

point $(−8, −2)$

m = −7,

point $(−1, −3)$

y = −7 x - 10

m = −4,

point $(−2, −3)$

Horizontal line containing $(−2, 5)$

y = 5

Horizontal line containing $(−2, −3)$

Horizontal line containing $(−1, −7)$

y = −7

Horizontal line containing $(4, −8)$

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.

(2, 6)

and $(5, 3)$

y = − x + 8

(4, 3)

and $(8, 1)$

(−3, −4)

and $(5 - 2) .$

y = \frac{1}{4} x - \frac{13}{4}

(−5, −3)

and $(4, −6) .$

(−1, 3)

and $(−6, −7) .$

y = 2 x + 5

(−2, 8)

and $(−4, −6) .$

(0, 4)

and $(2, −3) .$

y = - \frac{7}{2} x + 4

(0, −2)

and $(−5, −3) .$

(7, 2)

and $(7, −2) .$

x = 7

(−2, 1)

and $(−2, −4) .$

(3, −4)

and $(5, −4) .$

y = −4

(−6, −3)

and $(−1, −3)$

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.

line $y = 4 x + 2,$

point $(1, 2)$

y = 4 x - 2

line $y = −3 x - 1,$

point $(2, −3) .$

line $2 x - y = 6,$

point $(3, 0) .$

y = 2 x - 6

line $2 x + 3 y = 6,$

point $(0, 5) .$

line $x = −4,$

point $(−3, −5) .$

x = −3

line $x - 2 = 0,$

point $(1, −2)$

line $y = 5,$

point $(2, −2)$

y = −2

line $y + 2 = 0,$

point $(3, −3)$

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.

line $y = −2 x + 3,$

point $(2, 2)$

y = \frac{1}{2} x + 1

line $y = − x + 5,$

point $(3, 3)$

line $y = \frac{3}{4} x - 2,$

point $(−3, 4)$

y = - \frac{4}{3} x

line $y = \frac{2}{3} x - 4,$

point $(2, −4)$

line $2 x - 3 y = 8,$

point $(4, −1)$

y = - \frac{3}{2} x + 5

line $4 x - 3 y = 5,$

point $(−3, 2)$

line $2 x + 5 y = 6,$

point $(0, 0)$

y = \frac{5}{2} x

line $4 x + 5 y = −3,$

point $(0, 0)$

line $x = 3,$

point $(3, 4)$

y = 4

line $x = −5,$

point $(1, −2)$

line $x = 7,$

point $(−3, −4)$

y = −4

line $x = −1,$

point $(−4, 0)$

line $y - 3 = 0,$

point $(−2, −4)$

x = −2

line $y - 6 = 0,$

point $(−5, −3)$

line y-axis,* * *

point $(3, 4)$

y = 4

line y-axis,* * *

point $(2, 1)$

Mixed Practice

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

Containing the points $(4, 3)$

and $(8, 1)$

y = - \frac{1}{2} x + 5

Containing the points $(−2, 0)$

and $(−3, −2)$

m = \frac{1}{6},

containing point $(6, 1)$

y = \frac{1}{6} x

m = \frac{5}{6},

containing point $(6, 7)$

Parallel to the line $4 x + 3 y = 6,$

containing point $(0, −3)$

y = - \frac{4}{3} x - 3

Parallel to the line $2 x + 3 y = 6,$

containing point $(0, 5)$

m = - \frac{3}{4},

containing point $(8, −5)$

y = - \frac{3}{4} x + 1

m = - \frac{3}{5},

containing point $(10, −5)$

Perpendicular to the line $y - 1 = 0,$

point $(−2, 6)$

x = −2

Perpendicular to the line y-axis, point $(−6, 2)$

Parallel to the line $x = −3,$

containing point $(−2, −1)$

x = −2

Parallel to the line $x = −4,$

containing point $(−3, −5)$

Containing the points $(−3, −4)$

and $(2, −5)$

y = - \frac{1}{5} x - \frac{23}{5}

Containing the points $(−5, −3)$

and $(4, −6)$

Perpendicular to the line $x - 2 y = 5,$

point $(−2, 2)$

y = −2 x - 2

Perpendicular to the line $4 x + 3 y = 1,$

point $(0, 0)$

Writing Exercises

Why are all horizontal lines parallel?

Answers will vary.

Explain in your own words why the slopes of two perpendicular lines must have opposite signs.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Name the slope.
Name the y-intercept.
Substitute the values into $y = m x + b .$

Find the slope, by counting the rise and run.

Find the y-intercept.
Substitute the values into $y = m x + b .$

Identify the slope.
Identify the point.
Substitute the values into $y - y_{1} = m (x - x_{1}) .$



Simplify.

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

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

Key Concepts

Practice Makes Perfect

Writing Exercises

Self Check

Glossary