### Distance Between Parallel Lines

The shortest distance between two parallel lines is the length of the perpendicular segment between them. It doesn’t matter which perpendicular line you choose, as long as the two points are on the lines. Recall that there are infinitely many perpendicular lines between two parallel lines.

Notice that all of the pink segments are the same length. So, when picking a perpendicular segment, be sure to pick one with endpoints that are integers.

#### Calculating the Distance Between Two Points

Find the distance between \begin{align*}x = 3\end{align*} and \begin{align*}x = -5\end{align*}.

Any line with \begin{align*}x = a\end{align*} number is a vertical line. In this case, we can just count the squares between the two lines. The two lines are \begin{align*}3-(-5)\end{align*} units apart, or 8 units.

You can use this same method with horizontal lines as well. For example, \begin{align*}y = -1\end{align*} and \begin{align*}y = 3\end{align*} are \begin{align*}3-(-1)\end{align*} units, or 4 units apart.

#### Finding the Shortest Distance Between Two Lines

What is the shortest distance between \begin{align*}y=2x+4\end{align*} and \begin{align*}y=2x-1\end{align*}?

Graph the two lines and determine the perpendicular slope, which is \begin{align*}-\frac{1}{2}\end{align*}. Find a point on \begin{align*}y=2x+4\end{align*}, let’s say (-1, 2). From here, use the slope of the perpendicular line to find the corresponding point on \begin{align*}y=2x-1\end{align*}. If you move down 1 from 2 and over to the right 2 from -1, you will hit \begin{align*}y=2x-1\end{align*} at (1, 1). Use these two points to determine the distance between the two lines.

\begin{align*}d & = \sqrt{(1+1)^2+(1-2)^2}\\ & = \sqrt{2^2+(-1)^2}\\ & = \sqrt{4+1}\\ & = \sqrt{5} \approx 2.24 \ units\end{align*}

The lines are about 2.24 units apart.

Notice that you could have used any two points, as long as they are on the same perpendicular line. For example, you could have also used (-3, -2) and (-1, -3) and you still would have gotten the same answer.

\begin{align*}d & = \sqrt{(-1+3)^2+(-3+2)^2}\\ & = \sqrt{2^2+(-1)^2}\\ & = \sqrt{4+1}\\ & = \sqrt{5} \approx 2.24 \ units\end{align*}

#### Calculating the Distance Between Parallel Lines

Find the distance between the two parallel lines below.

First you need to find the slope of the two lines. Because they are parallel, they are the same slope, so if you find the slope of one, you have the slope of both.

Start at the \begin{align*}y-\end{align*}intercept of the top line, 7. From there, you would go down 1 and over 3 to reach the line again. Therefore the slope is \begin{align*}-\frac{1}{3}\end{align*} and the perpendicular slope would be 3.

Next, find two points on the lines. Let’s use the \begin{align*}y-\end{align*}intercept of the bottom line, (0, -3). Then, rise 3 and go over 1 until your reach the second line. Doing this three times, you would hit the top line at (3, 6). Use these two points in the distance formula to find how far apart the lines are.

\begin{align*}d & = \sqrt{(0-3)^2+(-3-6)^2}\\ & = \sqrt{(-3)^2+(-9)^2}\\ & = \sqrt{9+81}\\ & = \sqrt{90} \approx 9.49 \ units\end{align*}

### Examples

#### Example 1

Find the distance between \begin{align*}x = 7\end{align*} and \begin{align*}x = -1\end{align*}.

These are vertical lines, so we can just count the squares between the two lines. The two lines are \begin{align*}7-(-1)\end{align*} units apart, or 8 units.

#### Example 2

Find the distance between \begin{align*}y = x+6\end{align*} and \begin{align*}y=x-2\end{align*}.

Find the perpendicular slope: \begin{align*}m = 1\end{align*}, so \begin{align*}m_\perp=-1\end{align*}. Then, find the \begin{align*}y-\end{align*}intercept of the top line, \begin{align*}y=x+6\end{align*}: (0, 6). Use the slope and count down 1 and to the right 1 until you hit \begin{align*}y=x-2\end{align*} at the point (4, 2). Use these two points in the distance formula to determine how far apart the lines are.

\begin{align*}d & = \sqrt{(0-4)^2 + (6-2)^2}\\ & = \sqrt{(-4)^2 + (4)^2}\\ & = \sqrt{16+16}\\ & = \sqrt{32} = 5.66 \ units\end{align*}

#### Example 3

Find the distance between \begin{align*}y=5\end{align*} and \begin{align*}y = -6\end{align*}.

3. These are horizontal lines, so we can just count the squares between the two lines. The two lines are \begin{align*}5-(-6)\end{align*} units apart, or 11 units.

### Interactive Practice

### Review

Use each graph below to determine how far apart each the parallel lines are. Round your answers to the nearest hundredth.

Determine the shortest distance between the each pair of parallel lines. Round your answer to the nearest hundredth.

- \begin{align*}x = 5, \ x = 1\end{align*}
- \begin{align*}y = -2, \ y = 7\end{align*}
- \begin{align*}x = -4, \ x = 11\end{align*}
- \begin{align*}y = -6, \ y = 4\end{align*}
- \begin{align*}x = 22, \ x = 16\end{align*}
- \begin{align*}y = -14, \ y = 2\end{align*}
- \begin{align*}y=x+5, \ y=x-3\end{align*}
- \begin{align*}y=2x+1, \ y=2x-4\end{align*}
- \begin{align*}y=-\frac{1}{3}x+2, \ y=-\frac{1}{3}x-8\end{align*}
- \begin{align*}y=4x+9, \ y=4x-8\end{align*}
- \begin{align*}y=\frac{1}{2}x, \ y=\frac{1}{2}x-5\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 3.11.