### Distance Between Parallel Lines

All vertical lines are in the form \begin{align*}x = a\end{align*}, where a is the \begin{align*}x\text{-} \!\end{align*}intercept. To find the distance between two vertical lines, count the squares between the two lines. You can use this method for horizontal lines as well. All horizontal lines are in the form \begin{align*}y = b\end{align*}, where \begin{align*}b\end{align*} is the \begin{align*}y\text{-} \!\end{align*}intercept.

In general, the shortest distance between two parallel lines is the length of a perpendicular segment between them. There are infinitely many perpendicular segments between two parallel lines, but they will all be the same length.

Remember that distances are always positive!

### Examples

#### Example 1

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

The two lines are 3 – (-5) units apart, or 8 units apart.

#### Example 2

Find the distance between \begin{align*}x = \text{-}5\end{align*} and \begin{align*}x = \text{-}10.\end{align*}

The two lines are -5 – (-10) units apart, or 5 units apart.

#### Example 3

Find the distance between \begin{align*}y = 5\end{align*} and \begin{align*}y = \text{-}8.\end{align*}

The two lines are 5 – (-8) units apart, or 13 units apart.

#### Example 4

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

Step 1: Find the perpendicular slope.

\begin{align*}m = 1\end{align*}, so \begin{align*}m_\perp=\text{-}1.\end{align*}

Step 2: Find the \begin{align*}y\text{-} \!\end{align*}intercept of the top line, \begin{align*}y=x+6.\end{align*}

The intercept is (0, 6).

Step 3: Use the slope and count down 1 and to the right 1 until you hit \begin{align*}y=x-2.\end{align*}

**Always rise/run the same amount for \begin{align*}\mathbf{m = 1} \ \mathbf{ or } \ \mathbf{m = \text{-}1}.\end{align*}**

Step 4: 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 \ \mathrm{units}\end{align*}

#### Example 5

Find the distance between \begin{align*}y=\text{-}x-1, \text{ and } y=\text{-}x-3.\end{align*}

Step 1: Find the perpendicular slope.

\begin{align*}m = \text{-}1, \text{ so } m\perp \; = 1.\end{align*}

Step 2: Find the \begin{align*}y\text{-}\!\end{align*}intercept of the top line, \begin{align*}y=\text{-}x-1\end{align*}.

The intercept is (0, -1).

Step 3: Use the slope and count down 1 and to the *left* 1 until you hit \begin{align*}y=x-3.\end{align*}

Step 4: Use these two points in the distance formula to determine how far apart the lines are.

\begin{align*}d & =\sqrt{(0-(\text{-}1))^2 + (\text{-}1-(\text{-}2))^2}\\ & = \sqrt{(1)^2+(1)^2}\\ & = \sqrt{1+1}\\ & = \sqrt{2} = 1.41 \ \text{units}\end{align*}

### Review

Use each graph below to determine how far apart each pair of parallel lines is.

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 = -6, y = 4\end{align*}
- \begin{align*}y = 3, y = 15\end{align*}
- \begin{align*}x = -10, x = -1\end{align*}
- \begin{align*}x = 8, x = 0\end{align*}
- \begin{align*}y = 7, y = -12\end{align*}

Find the distance between the given parallel lines.

- \begin{align*}y=x-3, \ y=x+11\end{align*}
- \begin{align*}y=-x+4, \ y=-x\end{align*}
- \begin{align*}y=-x-5, \ y = -x+1\end{align*}
- \begin{align*}y = x+12 , \ y=x-6\end{align*}

### Review (Answers)

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

### Resources