Suppose you were tasked with finding a line parallel or perpendicular to

### 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 y=−3x+5 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

Now to find the equation of a line with slope

Start with the slope-intercept form:

Plug in the slope:

Plug in the point (2, 6) to find

**The equation of the line is**

#### Example B

Find the equation of the line parallel to

**Solution**

Rewrite the equation in slope-intercept form:

The slope of the given line is

Start with the slope-intercept form:

Plug in the slope:

Plug in the point (-5, -3):

**The equation of the line is**

**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

**Family 1:** Keep the slope unchanged and vary the

The figure below shows the family of lines with equations of the form

All the lines have a slope of –2, but the value of

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 **vertical shift.**

Family 2: Keep the

The figure below shows the family of lines with equations of the form

All the lines have a

#### Example C

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

a) parallel to the

b) through the point (0, -1)

c) perpendicular to

d) parallel to

**Solution**

a) All lines parallel to the

b) All lines passing through the point (0, -1) have the same

c) First we need to find the slope of the given line. Rewriting

**The family of lines is**

d) Rewrite

**The family of lines is**

Watch this video for help with the Examples above.

CK-12 Foundation: Equations of Parallel and Perpendicular Lines

### Guided Practice

*Find the equation of the line perpendicular to x−5y=15 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*}.

### Explore More

- 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).

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.5.