# 5.5: Families of Lines

**At Grade**Created by: CK-12

**Practice**Families of Lines

What if you were given the equation of a line like \begin{align*}y = -4x - 3\end{align*}

### Watch This

CK-12 Foundation: 0505S Equations of Parallel and Perpendicular Lines (H264)

### Guidance

We can use the properties of parallel and perpendicular lines to write an equation of a line parallel or perpendicular to a given line. You might be given a line and a point, and asked to find the line that goes through the given point and is parallel or perpendicular to the given line. Here’s how to do this:

- Find the slope of the given line from its equation. (You might need to re-write the equation in a form such as the slope-intercept form.)
- Find the slope of the parallel or perpendicular line—which is either the same as the slope you found in step 1 (if it’s parallel), or the negative reciprocal of the slope you found in step 1 (if it’s perpendicular).
- Use the slope you found in step 2, along with the point you were given, to write an equation of the new line in slope-intercept form or point-slope form.

#### Example A

*Find an equation of the line perpendicular to the line \begin{align*}y=-3x+5\end{align*} that passes through the point (2, 6).*

**Solution**

The slope of the given line is -3, so the perpendicular line will have a slope of \begin{align*}\frac{1}{3}\end{align*}.

Now to find the equation of a line with slope \begin{align*}\frac{1}{3}\end{align*} that passes through (2, 6):

Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.

Plug in the slope: \begin{align*}y=\frac{1}{3}x+b\end{align*}.

Plug in the point (2, 6) to find \begin{align*}b\end{align*}: \begin{align*}6=\frac{1}{3}(2)+b \Rightarrow b=6-\frac{2}{3} \Rightarrow b=\frac{16}{3} \to 5 \frac{1}{3}\end{align*}.

**The equation of the line is** \begin{align*}y=\frac{1}{3}x + 5\frac{1}{3}\end{align*}.

#### Example B

Find the equation of the line parallel to \begin{align*}6x-5y=12\end{align*} that passes through the point (-5, -3).

**Solution**

Rewrite the equation in slope-intercept form: \begin{align*}6x-5y=12 \Rightarrow 5y=6x-12 \Rightarrow y=\frac{6}{5}x-\frac{12}{5}\end{align*}.

The slope of the given line is \begin{align*}\frac{6}{5}\end{align*}, so we are looking for a line with slope \begin{align*}\frac{6}{5}\end{align*} that passes through the point (-5, -3).

Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.

Plug in the slope: \begin{align*}y=\frac{6}{5}x+b\end{align*}.

Plug in the point (-5, -3): \begin{align*}n-3=\frac{6}{5}(-5)+b \Rightarrow -3=-6+b \Rightarrow b=3\end{align*}

**The equation of the line is** \begin{align*}y=\frac{6}{5}x+3\end{align*}.

**Investigate Families of Lines**

A **family of lines** is a set of lines that have something in common with each other. Straight lines can belong to two types of families: one where the slope is the same and one where the \begin{align*}y-\end{align*}intercept is the same.

**Family 1:** Keep the slope unchanged and vary the \begin{align*}y-\end{align*}intercept.

The figure below shows the family of lines with equations of the form \begin{align*}y=-2x+b\end{align*}:

All the lines have a slope of –2, but the value of \begin{align*}b\end{align*} is different for each line.

Notice that in such a family all the lines are parallel. All the lines look the same, except that they are shifted up and down the \begin{align*}y-\end{align*}axis. As \begin{align*}b\end{align*} gets larger the line rises on the \begin{align*}y-\end{align*}axis, and as \begin{align*}b\end{align*} gets smaller the line goes lower on the \begin{align*}y-\end{align*}axis. This behavior is often called a **vertical shift.**

Family 2: Keep the \begin{align*}y-\end{align*}intercept unchanged and vary the slope.

The figure below shows the family of lines with equations of the form \begin{align*}y=mx+2\end{align*}:

All the lines have a \begin{align*}y-\end{align*}intercept of two, but the slope is different for each line. The steeper lines have higher values of \begin{align*}m\end{align*}.

#### Example C

*Write the equation of the family of lines satisfying the given condition.*

a) parallel to the \begin{align*}x-\end{align*}axis

b) through the point (0, -1)

c) perpendicular to \begin{align*}2x+7y-9=0\end{align*}

d) parallel to \begin{align*}x+4y-12=0\end{align*}

**Solution**

a) All lines parallel to the \begin{align*}x-\end{align*}axis have a slope of zero; the \begin{align*}y-\end{align*}intercept can be anything. So the family of lines is \begin{align*}y=0x+b\end{align*} or just \begin{align*}y=b\end{align*}.

b) All lines passing through the point (0, -1) have the same \begin{align*}y-\end{align*}intercept, \begin{align*}b = -1\end{align*}. The family of lines is: \begin{align*}y=mx-1\end{align*}.

c) First we need to find the slope of the given line. Rewriting \begin{align*}2x+7y-9=0\end{align*} in slope-intercept form, we get \begin{align*}y=-\frac{2}{7}x+\frac{9}{7}\end{align*}. The slope of the line is \begin{align*}-\frac{2}{7}\end{align*}, so we’re looking for the family of lines with slope \begin{align*}\frac{7}{2}\end{align*}.

**The family of lines is** \begin{align*}y=\frac{7}{2}x+b\end{align*}.

d) Rewrite \begin{align*}x+4y-12=0\end{align*} in slope-intercept form: \begin{align*}y=-\frac{1}{4}x+3\end{align*}. The slope is \begin{align*}-\frac{1}{4}\end{align*}, so that’s also the slope of the family of lines we are looking for.

**The family of lines is** \begin{align*}y=-\frac{1}{4}x+b\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Equations of Parallel and Perpendicular Lines

### Vocabulary

- A
**family of lines**is a set of lines that have something in common with each other. Straight lines can belong to two types of families: one where the slope is the same and one where the \begin{align*}y-\end{align*}intercept is the same.

- Notice that in such a family all the lines are parallel. All the lines look the same, except that they are shifted up and down the \begin{align*}y-\end{align*}axis. As \begin{align*}b\end{align*} gets larger the line rises on the \begin{align*}y-\end{align*}axis, and as \begin{align*}b\end{align*} gets smaller the line goes lower on the \begin{align*}y-\end{align*}axis. This behavior is often called a
**vertical shift.**

### Guided Practice

*Find the equation of the line perpendicular to \begin{align*}x-5y=15\end{align*} that passes through the point (-2, 5).*

**Solution**

Re-write the equation in slope-intercept form: \begin{align*}x-5y=15 \Rightarrow -5y=-x+15 \Rightarrow y=\frac{1}{5}x-3\end{align*}.

The slope of the given line is \begin{align*}\frac{1}{5}\end{align*}, so we’re looking for a line with slope -5.

Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.

Plug in the slope: \begin{align*}y=-5x+b\end{align*}.

Plug in the point (-2, 5): \begin{align*}5=-5(-2)+b \Rightarrow b=5-10 \Rightarrow b=-5\end{align*}

**The equation of the line is** \begin{align*}y=-5x-5\end{align*}.

### Practice

- Find the equation of the line parallel to \begin{align*}5x-2y=2\end{align*} that passes through point (3, -2).
- Find the equation of the line perpendicular to \begin{align*}y=-\frac{2}{5}x-3\end{align*} that passes through point (2, 8).
- Find the equation of the line parallel to \begin{align*}7y+2x-10=0\end{align*} that passes through the point (2, 2).
- Find the equation of the line perpendicular to \begin{align*}y+5=3(x-2)\end{align*} that passes through the point (6, 2).
- Line \begin{align*}S\end{align*} passes through the points (2, 3) and (4, 7). Line \begin{align*}T\end{align*} passes through the point (2, 5). If Lines \begin{align*}S\end{align*} and \begin{align*}T\end{align*} are parallel, name one more point on line \begin{align*}T\end{align*}. (
**Hint:**you don’t need to find the slope of either line.) - Lines \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} both pass through (-1, 5). Line \begin{align*}P\end{align*} also passes through (-3, -1). If \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} are perpendicular, name one more point on line \begin{align*}Q\end{align*}. (This time you will have to find the slopes of both lines.)
- Write the equation of the family of lines satisfying the given condition.
- All lines that pass through point (0, 4).
- All lines that are perpendicular to \begin{align*}4x+3y-1=0\end{align*}.
- All lines that are parallel to \begin{align*}y-3=4x+2\end{align*}.
- All lines that pass through the point (0, -1).

- Name two lines that pass through the point (3, -1) and are perpendicular to each other.
- Name two lines that are each perpendicular to \begin{align*}y=-4x-2\end{align*}. What is the relationship of those two lines to each other?
- Name two perpendicular lines that both pass through the point (3, -2). Then name a line parallel to one of them that passes through the point (-2, 5).

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### Image Attributions

Here you'll learn how to write the equation of a line that is parallel or perpendicular to a second line given that second line's equation and one of the points it passes through. You'll also investigate families of lines.