Think about the members of your family. You probably all have some things in common, but you're definitely not all identical. The same is true of a family of lines. What could a family of lines have in common? What might be different?

### Families of Lines

A straight line has two very important properties, its slope and its \begin{align*}y-\end{align*}intercept. The slope tells us how steeply the line rises or falls, and the \begin{align*}y-\end{align*}intercept tells us where the line intersects the \begin{align*}y-\end{align*}axis. In this Concept, we will look at two 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: where the slope is the same and where the \begin{align*}y-\end{align*}intercept is the same.

#### Family 1: The slope is the same

Remember that lines with the same slope are parallel. Each line on the Cartesian plane below has an identical slope with different \begin{align*}y-\end{align*}intercepts. All the lines look the same but they are shifted up and down the \begin{align*}y-\end{align*}axis as shown in the graph below. 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**.

#### Let's write the equation for the red line in the image above:

We can see from the graph that the equation has a \begin{align*}y\end{align*}-intercept of 1. Since all the lines have the same slope, we can look at any line to determine the slope, so the slope is \begin{align*}-2\end{align*}. Therefore, the equation of the red line is:

\begin{align*}y=-2x+1.\end{align*}

#### Family 2: The \begin{align*}y-\end{align*}intercept is the same

The graph below shows several lines with the same \begin{align*}y-\end{align*}intercept but with varying slopes.

#### Let's write the equation for the brown line in the image above:

All the lines share the same \begin{align*}y\end{align*}-intercept, which is 2. Looking at the graph, the slope is -1. Thus, the equation is:

\begin{align*}y=-x+2.\end{align*}

#### Now, let's write a general equation for each family of lines shown in the images in this Concept.

- For family 1, the red line has the equation \begin{align*}y=-2x+1.\end{align*} Since all the lines share the same slope, we keep the slope of -2. But they all have different \begin{align*}y\end{align*}-intercepts, so we will use \begin{align*}b\end{align*}:

\begin{align*}y=-2x+b.\end{align*}

- For family 2, the brown line has the equation \begin{align*}y=-x+2.\end{align*} Since all the lines share the same \begin{align*}y\end{align*}-intercept but have different slopes:

\begin{align*}y=mx+2.\end{align*}

### Examples

#### Example 1

Earlier, you were asked what a family of lines could have in common and what could be different.

As shown in this concept, there are two important parts of a line, the \begin{align*}y-\end{align*}intercept and the slope, that a family of lines can have in common. A family of lines does not need to have both the \begin{align*}y-\end{align*}intercept and slope in common, just one.

#### Example 2

Write the equation of the family of lines perpendicular to \begin{align*}6x+2y=24\end{align*}.

First we must find the slope of \begin{align*}6x+2y=24\end{align*}:

\begin{align*}slope=-\frac{6}{2}=-3.\end{align*}

Now we find the slope of any line perpendicular to our original line:

\begin{align*}-3\cdot m=-1\end{align*}

\begin{align*}\frac{-3\cdot m}{-3}=\frac{-1}{-3}\end{align*}

\begin{align*} m=\frac{1}{3}\end{align*}

The family of lines perpendicular to \begin{align*}6x+2y=24\end{align*} will have a slope of \begin{align*} m=\frac{1}{3}\end{align*}. They will all have different \begin{align*}y\end{align*}-intercepts:

\begin{align*}y=\frac{1}{3}x+b.\end{align*}

### Review

- What is a family of lines?
- Find the equation of the line parallel to \begin{align*}5x-2y=2\end{align*} that passes through the point (3, –2).
- Find the equation of the line perpendicular to \begin{align*}y=-\frac{2}{5}x-3\end{align*} that passes through the 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).
- Find the equation of the line through (2, –4) perpendicular to \begin{align*}y=\frac{2}{7} x+3\end{align*}.
- Find the equation of the line through (2, 3) parallel to \begin{align*}y=\frac{3}{2} x+5\end{align*}.

In 8–11, write the equation of the family of lines satisfying the given condition.

- All lines pass through point (0, 4).
- All lines are perpendicular to \begin{align*}4x+3y-1=0\end{align*}.
- All lines are parallel to \begin{align*}y-3=4x+2\end{align*}.
- All lines pass through point (0, –1).
- Write an equation for a line parallel to the equation graphed below.
- Write an equation for a line perpendicular to the equation graphed below and passing through the point (0, –1).

#### Quick Quiz

1. Write an equation for a line with a slope of \begin{align*}\frac{4}{3}\end{align*} and a \begin{align*}y-\end{align*}intercept of (0, 8).

2. Write an equation for a line containing (6, 1) and (7, –3).

3. A plumber charges $75 for a 2.5-hour job and $168.75 for a 5-hour job.

Assuming the situation is linear, write an equation to represent the plumber’s charge and use it to predict the cost of a 1-hour job.

4. Rewrite in standard form: \begin{align*}y=\frac{6}{5} x+11\end{align*}.

5. Sasha took tickets for the softball game. Student tickets were $3.00 and adult tickets were $3.75. She collected a total of $337.50 and sold 75 student tickets. How many adult tickets were sold?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.9.