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3.11: Distance Between Parallel Lines

Difficulty Level: At Grade Created by: CK-12
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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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CK-12 Foundation: Chapter3DistanceBetweenParallelLinesA

James Sousa: Determining the Distance Between Two 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.

Example A

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.

Example B

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*}

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*}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*}

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter3DistanceBetweenParallelLinesB


The distance formula tells us that the distance between two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*} can be defined as \begin{align*}d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\end{align*}.

Guided Practice

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

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

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


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*} units apart, or 8 units.

2. 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*}

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


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.

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

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Distance Formula The distance between two points (x_1, y_1) and (x_2, y_2) can be defined as d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.
Perpendicular Perpendicular lines are lines that intersect at a 90^{\circ} angle. The product of the slopes of two perpendicular lines is -1.

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Difficulty Level:
At Grade
Date Created:
Jul 17, 2012
Last Modified:
Oct 05, 2016
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