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# Distance Between Parallel Lines

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Practice Distance Between Parallel Lines
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Distance Between Parallel Lines

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 two vertical lines, two horizontal lines, and two non-vertical, non-horizontal parallel lines using the perpendicular slope.

### Guidance

All vertical lines are in the form $x = a$ , where a is the $x-$ 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 $y = b$ , where $b$ is the $y-$ intercept.

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

Remember that distances are always positive!

#### Example A

Find the distance between $x = 3$ and $x = -5$ .

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

#### Example B

Find the distance between $y = 5$ and $y = -8$ .

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

#### Example C

Find the distance between $y = x+6$ and $y=x-2$ .

Step 1: Find the perpendicular slope.

$m = 1$ , so $m_\perp=-1$

Step 2: Find the $y-$ intercept of the top line, $y=x+6$ . (0, 6)

Step 3: Use the slope and count down 1 and to the right 1 until you hit $y=x-2$ .

Always rise/run the same amount for $m = 1$ or -1.

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

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

### Guided Practice

1. Find the distance between $y=-x-1$ and $y=-x-3$ .

2. Find the distance between $y = 2$ and $y = -4$ .

3. Find the distance between $x = -5$ and $x = -10$ .

1. Step 1: Find the perpendicular slope.

$m = -1$ , so $m_\perp = 1$

Step 2: Find the $y-$ intercept of the top line, $y=-x-1$ . (0, -1)

Step 3: Use the slope and count down 1 and to the left 1 until you hit $y=x-3$ .

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

$d & =\sqrt{(0-(-1))^2 + (-1-(-2))^2}\\& = \sqrt{(1)^2+(1)^2}\\& = \sqrt{1+1}\\& = \sqrt{2} = 1.41 \ units$

2. The two lines are 2 – (-4) units apart, or 6 units apart.

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

### Practice

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.

1. $x = 5, x = 1$
2. $y = -6, y = 4$
3. $y = 3, y = 15$
4. $x = -10, x = -1$
5. $x = 8, x = 0$
6. $y = 7, y = -12$

Find the distance between the given parallel lines.

1. $y=x-3, \ y=x+11$
2. $y=-x+4, \ y=-x$
3. $y=-x-5, \ y = -x+1$
4. $y = x+12 , \ y=x-6$