What if you were given two parallel lines? How could you find how far apart these two lines are? After completing this Concept, you'll be able to find the distance between parallel lines using the distance formula.
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CK12 Foundation: Chapter3DistanceBetweenParallelLinesA
James Sousa: Determining the Distance Between Two Parallel Lines
Guidance
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.
Example A
Find the distance between \begin{align*}x = 3\end{align*}
Any line with \begin{align*}x = a\end{align*}
You can use this same method with horizontal lines as well. For example, \begin{align*}y = 1\end{align*}
Example B
What is the shortest distance between \begin{align*}y=2x+4\end{align*}
Graph the two lines and determine the perpendicular slope, which is \begin{align*}\frac{1}{2}\end{align*}
\begin{align*}d & = \sqrt{(1+1)^2+(12)^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*}
Example C
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*}
Next, find two points on the lines. Let’s use the \begin{align*}y\end{align*}
\begin{align*}d & = \sqrt{(03)^2+(36)^2}\\
& = \sqrt{(3)^2+(9)^2}\\
& = \sqrt{9+81}\\
& = \sqrt{90} \approx 9.49 \ units\end{align*}
Watch this video for help with the Examples above.
CK12 Foundation: Chapter3DistanceBetweenParallelLinesB
Vocabulary
The distance formula tells us that the distance between two points \begin{align*}(x_1, y_1)\end{align*}
Guided Practice
1. Find the distance between \begin{align*}x = 7\end{align*}
2. Find the distance between \begin{align*}y = x+6\end{align*}
3. Find the distance between \begin{align*}y=5\end{align*}
Answers:
1. 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*}
2. Find the perpendicular slope: \begin{align*}m = 1\end{align*}
\begin{align*}d & = \sqrt{(04)^2 + (62)^2}\\
& = \sqrt{(4)^2 + (4)^2}\\
& = \sqrt{16+16}\\
& = \sqrt{32} = 5.66 \ units\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*}
Interactive Practice
Practice
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*}
x=5, x=1 
\begin{align*}y = 2, \ y = 7\end{align*}
y=−2, y=7 
\begin{align*}x = 4, \ x = 11\end{align*}
x=−4, x=11 
\begin{align*}y = 6, \ y = 4\end{align*}
y=−6, y=4 
\begin{align*}x = 22, \ x = 16\end{align*}
x=22, x=16 
\begin{align*}y = 14, \ y = 2\end{align*}
y=−14, y=2 
\begin{align*}y=x+5, \ y=x3\end{align*}
y=x+5, y=x−3 
\begin{align*}y=2x+1, \ y=2x4\end{align*}
y=2x+1, y=2x−4 
\begin{align*}y=\frac{1}{3}x+2, \ y=\frac{1}{3}x8\end{align*}
y=−13x+2, y=−13x−8 
\begin{align*}y=4x+9, \ y=4x8\end{align*}
y=4x+9, y=4x−8 
\begin{align*}y=\frac{1}{2}x, \ y=\frac{1}{2}x5\end{align*}
y=12x, y=12x−5