# Equations of Lines

Now that we've learned how to plot points on a coordinate plane and graph linear equations, we can begin to analyze the equations of lines and evaluate the different characteristics of these lines. In this section, we'll learn about the commonly used forms for writing linear equations and the properties of lines that can be determined from their equations. For example, without creating a table of values, you will be able to match each equation below to its corresponding graph. You will also be able to explain the similarities and differences of each line, how they relate to each other, and why they behave that way. (a) [latex]y=3x+2[/latex] (b) [latex]y-4=-\frac{1}{2}(x+2)[/latex] (c) [latex]x=5[/latex] (d) [latex]y=-2[/latex] (e) [latex]3x=y+1[/latex] (f) [latex]2y-3x=6[/latex]## Writing Equations of Lines

### Slope-Intercept Form

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as [latex]y=mx+b[/latex], where [latex]m=\text{slope}[/latex] and [latex]b=y\text{-intercept}[/latex]. Let us begin with the slope.### The Slope of a Line

The**slope**of a line refers to the ratio of the vertical change in

*y*over the horizontal change in

*x*between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

### A General Note: The Slope of a Line

The slope of a line,*m*, represents the change in

*y*over the change in

*x.*Given two points, [latex]\left({x}_{1},{y}_{1}\right)[/latex] and [latex]\left({x}_{2},{y}_{2}\right)[/latex], the following formula determines the slope of a line containing these points:

### Example: Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points [latex]\left(2,-1\right)[/latex] and [latex]\left(-5,3\right)[/latex].Answer:
We substitute the *y-*values and the *x-*values into the formula.

#### Analysis of the Solution

It does not matter which point is called [latex]\left({x}_{1},{y}_{1}\right)[/latex] or [latex]\left({x}_{2},{y}_{2}\right)[/latex]. As long as we are consistent with the order of the*y*terms and the order of the

*x*terms in the numerator and denominator, the calculation will yield the same result.

### Try It

Find the slope of the line that passes through the points [latex]\left(-2,6\right)[/latex] and [latex]\left(1,4\right)[/latex].Answer: [latex]x=-5[/latex]

### Example: Identifying the Slope and *y-*intercept of a Line Given an Equation

Identify the slope and *y-*intercept, given the equation [latex]y=-\frac{3}{4}x - 4[/latex].

Answer: As the line is in [latex]y=mx+b[/latex] form, the given line has a slope of [latex]m=-\frac{3}{4}[/latex]. The *y-*intercept is [latex]b=-4[/latex].

#### Analysis of the Solution

The*y*-intercept is the point at which the line crosses the

*y-*axis. On the

*y-*axis, [latex]x=0[/latex]. We can always identify the

*y-*intercept when the line is in slope-intercept form, as it will always equal

*b.*Or, just substitute [latex]x=0[/latex] and solve for

*y.*

### The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.### A General Note: The Point-Slope Formula

Given one point and the slope, the point-slope formula will lead to the equation of a line:### Example: Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope [latex]m=-3[/latex] and passing through the point [latex]\left(4,8\right)[/latex]. Write the final equation in slope-intercept form.Answer:
Using the point-slope formula, substitute [latex]-3[/latex] for *m *and the point [latex]\left(4,8\right)[/latex] for [latex]\left({x}_{1},{y}_{1}\right)[/latex].

#### Analysis of the Solution

### Try It

Given [latex]m=4[/latex], find the equation of the line in slope-intercept form passing through the point [latex]\left(2,5\right)[/latex].Answer: [latex-display]y=4x - 3[/latex-display]

### Example: Finding the Equation of a Line Passing Through Two Given Points

Find the equation of the line passing through the points [latex]\left(3,4\right)[/latex] and [latex]\left(0,-3\right)[/latex]. Write the final equation in slope-intercept form.Answer: First, we calculate the slope using the slope formula and two points.

#### Analysis of the Solution

### Standard Form of a Line

Another way that we can represent the equation of a line is in**standard form**. Standard form is given as

*x-*and

*y-*terms are on one side of the equal sign and the constant term is on the other side.

### Example: Finding the Equation of a Line and Writing It in Standard Form

Find the equation of the line with [latex]m=-6[/latex] and passing through the point [latex]\left(\frac{1}{4},-2\right)[/latex]. Write the equation in standard form.Answer: We begin using the point-slope formula.

### Try It

Find the equation of the line in standard form with slope [latex]m=-\frac{1}{3}[/latex] and passing through the point [latex]\left(1,\frac{1}{3}\right)[/latex].Answer: [latex]x+3y=2[/latex]

### Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a**vertical line**is given as

*c*is a constant. The slope of a vertical line is undefined, and regardless of the

*y-*value of any point on the line, the

*x-*coordinate of the point will be

*c*. Suppose that we want to find the equation of a line containing the following points: [latex]\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)[/latex], and [latex]\left(-3,5\right)[/latex]. First, we will find the slope.

*x-*coordinates are the same and we find a vertical line through [latex]x=-3[/latex]. The equation of a

**horizontal line**is given as

*c*is a constant. The slope of a horizontal line is zero, and for any

*x-*value of a point on the line, the

*y-*coordinate will be

*c*. Suppose we want to find the equation of a line that contains the following set of points: [latex]\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)[/latex], and [latex]\left(5,-2\right)[/latex]. We can use the point-slope formula. First, we find the slope using any two points on the line.

*y*-intercept.

*y-*coordinates are the same.

*x*= −3 is a vertical line. The line

*y*= −2 is a horizontal line.

### Example: Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points: [latex]\left(1,-3\right)[/latex] and [latex]\left(1,4\right)[/latex].Answer: The *x-*coordinate of both points is 1. Therefore, we have a vertical line, [latex]x=1[/latex].

### Try It

Find the equation of the line passing through [latex]\left(-5,2\right)[/latex] and [latex]\left(2,2\right)[/latex].Answer: Horizontal line: [latex]y=2[/latex]

## Parallel and Perpendicular Lines

Parallel lines have the same slope and different*y-*intercepts. Lines that are

**parallel**to each other will never intersect. For example, the figure below shows the graphs of various lines with the same slope, [latex]m=2[/latex].

*y-*intercepts. Lines that are

**perpendicular**intersect to form a [latex]{90}^{\circ }[/latex] -angle. The slope of one line is the negative

**reciprocal**of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\cdot {m}_{2}=-1[/latex]. For example, the figure above shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\frac{1}{3}[/latex].

[latex]\begin{array}{l}\text{ }{m}_{1}\cdot {m}_{2}=-1\hfill \\ \text{ }3\cdot \left(-\frac{1}{3}\right)=-1\hfill \end{array}[/latex]

### Example: Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: [latex]3y=-4x+3[/latex] and [latex]3x - 4y=8[/latex].Answer: The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form. First equation:

[latex]\begin{array}{l}3y=-4x+3\hfill \\ y=-\frac{4}{3}x+1\hfill \end{array}[/latex]

Second equation:[latex]\begin{array}{l}3x - 4y=8\hfill \\ -4y=-3x+8\hfill \\ y=\frac{3}{4}x - 2\hfill \end{array}[/latex]

See the graph of both lines in the graph below. From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.[latex]\begin{array}{l}{m}_{1}=-\frac{4}{3}\hfill \\ {m}_{2}=\frac{3}{4}\hfill \\ {m}_{1}\cdot {m}_{2}=\left(-\frac{4}{3}\right)\left(\frac{3}{4}\right)=-1\hfill \end{array}[/latex]

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.### Try It

Graph the two lines and determine whether they are parallel, perpendicular, or neither: [latex]2y-x=10[/latex] and [latex]2y=x+4[/latex].Answer: Parallel lines: equations are written in slope-intercept form.

### Writing Equations of Parallel Lines

Suppose for example, we are given the following equation.[latex]f\left(x\right)=3x+1[\latex]

We know that the slope of the line formed by the function is 3. We also know that the*y-*intercept is (0, 1). Any other line with a slope of 3 will be parallel to

*f*(

*x*). So the lines formed by all of the following functions will be parallel to

*f*(

*x*).

[latex]\begin{cases}g\left(x\right)=3x+6\hfill \\ h\left(x\right)=3x+1\hfill \\ p\left(x\right)=3x+\frac{2}{3}\hfill \end{cases}[/latex]

Suppose then we want to write the equation of a line that is parallel to*f*and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for

*b*will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

[latex]\begin{cases}y-{y}_{1}=m\left(x-{x}_{1}\right)\hfill \\ y - 7=3\left(x - 1\right)\hfill \\ y - 7=3x - 3\hfill \\ \text{ }y=3x+4\hfill \end{cases}[/latex]

So [latex]g\left(x\right)=3x+4[/latex] is parallel to [latex]f\left(x\right)=3x+1[/latex] and passes through the point (1, 7).### How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

- Find the slope of the function.
- Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.
- Simplify.

### Example: Finding a Line Parallel to a Given Line

Find a line parallel to the graph of [latex]f\left(x\right)=3x+6[/latex] that passes through the point (3, 0).Answer:
The slope of the given line is 3. If we choose the slope-intercept form, we can substitute *m *= 3, *x *= 3, and *f*(*x*) = 0 into the slope-intercept form to find the *y-*intercept.

[latex]\begin{cases}g\left(x\right)=3x+b\hfill \\ \text{ }0=3\left(3\right)+b\hfill \\ \text{ }b=-9\hfill \end{cases}[/latex]

The line parallel to*f*(

*x*) that passes through (3, 0) is [latex]g\left(x\right)=3x - 9[/latex].

#### Analysis of the Solution

We can confirm that the two lines are parallel by graphing them. The graph below shows that the two lines will never intersect.### Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:[latex]f\left(x\right)=2x+4[/latex]

The slope of the line is 2, and its negative reciprocal is [latex]-\frac{1}{2}[/latex]. Any function with a slope of [latex]-\frac{1}{2}[/latex] will be perpendicular to*f*(

*x*). So the lines formed by all of the following functions will be perpendicular to

*f*(

*x*).

[latex]\begin{cases}g\left(x\right)=-\frac{1}{2}x+4\hfill \\ h\left(x\right)=-\frac{1}{2}x+2\hfill \\ p\left(x\right)=-\frac{1}{2}x-\frac{1}{2}\hfill \end{cases}[/latex]

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to*f*(

*x*) and passes through the point (4, 0). We already know that the slope is [latex]-\frac{1}{2}[/latex]. Now we can use the point to find the

*y*-intercept by substituting the given values into the slope-intercept form of a line and solving for

*b*.

[latex]\begin{cases}g\left(x\right)=mx+b\hfill \\ 0=-\frac{1}{2}\left(4\right)+b\hfill \\ 0=-2+b\hfill \\ 2=b\hfill \\ b=2\hfill \end{cases}[/latex]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a*y-*intercept of 2 is [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex]. So [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]f\left(x\right)=2x+4[/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

### Q & A

**A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?**

*No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.*

### How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

- Find the slope of the function.
- Determine the negative reciprocal of the slope.
- Substitute the new slope and the values for
*x*and*y*from the coordinate pair provided into [latex]g\left(x\right)=mx+b[/latex]. - Solve for
*b*. - Write the equation for the line.

### Example: Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to [latex]y=3x+3[/latex] that passes through the point (3, 0).Answer:
The original line has slope *m *= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\frac{1}{3}[/latex]. Using this slope and the given point, we can find the equation for the line.

[latex]\begin{cases}y_{2}=-\frac{1}{3}x+b\hfill \\ \text{ }0=-\frac{1}{3}\left(3\right)+b\hfill \\ \text{ }1=b\hfill \\ \text{ }b=1\hfill \end{cases}[/latex]

The line perpendicular to*y*that passes through (3, 0) is [latex]y_{2}=-\frac{1}{3}x+1[/latex].

#### Analysis of the Solution

A graph of the two lines is shown in the graph below.### Try It

Given the line [latex]y=2x - 4[/latex], write an equation for the line passing through (0, 0) that is- parallel to
*y* - perpendicular to
*y*

Answer:

- [latex]y=2x[/latex]is parallel
- [latex]y=-\frac{1}{2}x[/latex]is perpendicular

### How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

- Determine the slope of the line passing through the points.
- Find the negative reciprocal of the slope.
- Use the slope-intercept form or point-slope form to write the equation by substituting the known values.
- Simplify.

### Example: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point

A line passes through the points (–2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).Answer: From the two points of the given line, we can calculate the slope of that line.

[latex]\begin{cases}{m}_{1}=\frac{5 - 6}{4-\left(-2\right)}\hfill \\ =\frac{-1}{6}\hfill \\ =-\frac{1}{6}\hfill \end{cases}[/latex]

Find the negative reciprocal of the slope.[latex]\begin{cases}{m}_{2}=\frac{-1}{-\frac{1}{6}}\hfill \\ =-1\left(-\frac{6}{1}\right)\hfill \\ =6\hfill \end{cases}[/latex]

We can then solve for the*y-*intercept of the line passing through the point (4, 5).

[latex]\begin{cases}g\left(x\right)=6x+b\hfill \\ 5=6\left(4\right)+b\hfill \\ 5=24+b\hfill \\ -19=b\hfill \\ b=-19\hfill \end{cases}[/latex]

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is[latex]y=6x - 19[/latex]

### Try It

A line passes through the points, (–2, –15) and (2, –3). Find the equation of a perpendicular line that passes through the point, (6, 4).Answer: [latex]y=-\frac{1}{3}x+6[/latex]

### Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the**point-slope formula**to write the equation of the new line.

### How To: Given an equation for a line, write the equation of a line parallel or perpendicular to it.

- Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
- Use the slope and the given point with the point-slope formula.
- Simplify the line to slope-intercept form and compare the equation to the given line.

### Example: Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a [latex]5x+3y=1[/latex] and passing through the point [latex]\left(3,5\right)[/latex].Answer: First, we will write the equation in slope-intercept form to find the slope.

[latex]\begin{array}{l}5x+3y=1\hfill \\ 3y=-5x+1\hfill \\ y=-\frac{5}{3}x+\frac{1}{3}\hfill \end{array}[/latex]

The slope is [latex]m=-\frac{5}{3}[/latex]. The*y-*intercept is [latex]\frac{1}{3}[/latex], but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the

*y-*intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

[latex]\begin{array}{l}y - 5=-\frac{5}{3}\left(x - 3\right)\hfill \\ y - 5=-\frac{5}{3}x+5\hfill \\ y=-\frac{5}{3}x+10\hfill \end{array}[/latex]

The equation of the line is [latex]y=-\frac{5}{3}x+10[/latex].### Try It

Find the equation of the line parallel to [latex]5x=7+y[/latex] and passing through the point [latex]\left(-1,-2\right)[/latex].Answer: [latex]y=5x+3[/latex]

### Example: Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to [latex]5x - 3y+4=0\left(-4,1\right)[/latex].Answer: The first step is to write the equation in slope-intercept form.

[latex]\begin{array}{l}5x - 3y+4=0\hfill \\ -3y=-5x - 4\hfill \\ y=\frac{5}{3}x+\frac{4}{3}\hfill \end{array}[/latex]

We see that the slope is [latex]m=\frac{5}{3}[/latex]. This means that the slope of the line perpendicular to the given line is the negative reciprocal, or [latex]-\frac{3}{5}[/latex]. Next, we use the point-slope formula with this new slope and the given point.[latex]\begin{array}{l}y - 1=-\frac{3}{5}\left(x-\left(-4\right)\right)\hfill \\ y - 1=-\frac{3}{5}x-\frac{12}{5}\hfill \\ y=-\frac{3}{5}x-\frac{12}{5}+\frac{5}{5}\hfill \\ y=-\frac{3}{5}x-\frac{7}{5}\hfill \end{array}[/latex]

## Key Concepts

- Given two points, we can find the slope of a line using the slope formula.
- We can identify the slope and
*y*-intercept of an equation in slope-intercept form. - We can find the equation of a line given the slope and a point.
- We can also find the equation of a line given two points. Find the slope and use the point-slope formula.
- The standard form of a line has no fractions.
- Horizontal lines have a slope of zero and are defined as [latex]y=c[/latex], where
*c*is a constant. - Vertical lines have an undefined slope (zero in the denominator), and are defined as [latex]x=c[/latex], where
*c*is a constant. - Parallel lines have the same slope and different
*y-*intercepts. - Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical.

- A linear equation can be used to solve for an unknown in a number problem.

## Glossary

**slope**the change in

*y-*values over the change in

*x-*values

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