# 5.5: Families of Lines

Difficulty Level: At Grade Created by: CK-12
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### Families of Lines

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:

1. 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.)
2. 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).
3. 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.

#### Finding an Equation of a Line

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

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*}.

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

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*}.

#### Writing an Equation of a Family of Lines

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

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

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) through the point (0, -1)

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

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

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*}.

### Example

#### Example 1

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

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*}.

### Review

1. Find the equation of the line parallel to \begin{align*}5x-2y=2\end{align*} that passes through point (3, -2).
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).
3. Find the equation of the line parallel to \begin{align*}7y+2x-10=0\end{align*} that passes through the point (2, 2).
4. Find the equation of the line perpendicular to \begin{align*}y+5=3(x-2)\end{align*} that passes through the point (6, 2).
5. 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.)
6. 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.)
7. Write the equation of the family of lines satisfying the given condition.
1. All lines that pass through point (0, 4).
2. All lines that are perpendicular to \begin{align*}4x+3y-1=0\end{align*}.
3. All lines that are parallel to \begin{align*}y-3=4x+2\end{align*}.
4. All lines that pass through the point (0, -1).
8. Name two lines that pass through the point (3, -1) and are perpendicular to each other.
9. 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?
10. 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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