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# 3.6: The Distance Formula

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the distance between two points.
• Find the shortest distance between a point and a line and two parallel lines.
• Determine the equation of a perpendicular bisector of a line segment in the coordinate plane.

## Review Queue

1. What is the equation of the line between (-1, 3) and (2, -9)?
2. Find the equation of the line that is perpendicular to $y=-2x+5$ through the point (-4, -5).
3. Find the equation of the line that is parallel to $y=\frac{2}{3}x-7$ through the point (3, 8).

Know What? The shortest distance between two points is a straight line. To the right is an example of how far apart cities are in the greater Los Angeles area. There are always several ways to get somewhere in Los Angeles. Here, we have the distances between Los Angeles and Orange. Which distance is the shortest? Which is the longest?

## The Distance Formula

The distance between two points $(x_1, \ y_1)$ and $(x_2, \ y_2)$ can be defined as $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. This formula will be derived in Chapter 9.

Example 1: Find the distance between (4, -2) and (-10, 3).

Solution: Plug in (4, -2) for $(x_1, \ y_1)$ and (-10, 3) for $(x_2, \ y_2)$ and simplify.

$d& = \sqrt{(-10-4)^2+(3+2)^2}\\& = \sqrt{(-14)^2+(5^2)} \qquad \qquad \quad Distances \ are \ always \ positive!\\& = \sqrt{196+25}\\& = \sqrt{221} \approx 14.87 \ units$

Example 2: The distance between two points is 4 units. One point is (1, -6). What is the second point? You may assume that the second point is made up of integers.

Solution: We will still use the distance formula for this problem, however, we know $d$ and need to solve for $(x_2, \ y_2)$.

$4 & = \sqrt{\left (1-x_2 \right )^2+(-6-y_2)^2}\\16 & = (1-x_2)^2 + (-6-y_2)^2$

At this point, we need to figure out two square numbers that add up to 16. The only two square numbers that add up to 16 are $16 + 0$.

$&16 = \underbrace{ (1-x_2)^2 }_{4^2} + \underbrace{ (-6-y_2)^2 }_{0^2} && \text{or} && 16 = \underbrace{ (1-x_2)^2 }_{0^2} + \underbrace{ (-6-y_2)^2 }_{4^2}\\&1-x_2 = \pm 4 \qquad \qquad \quad -6-y_2=0 &&&& 1-x_2=0 \qquad \qquad \quad -6-y_2=\pm 4\\& \ \ -x_2 = -5 \ \text{or} \ 3 \quad \ \text{and} \quad \ \ -y_2=6 && \text{or} && \ \ -x_2=-1 \quad \ \text{and} \ \qquad \qquad \-y_2=10 \ \text{or} \ 2\\&\qquad x_2 = 5 \ \text{or} \ -3 \qquad \qquad \quad \ y_2=-6 &&&& \qquad x_2=1 \qquad \qquad \qquad \qquad y_2=-10 \ \text{or} \ -2$

Therefore, the second point could have 4 possibilities: (5, -6), (-3, -6), (1, -10), and (1, -2).

## Shortest Distance between a Point and a Line

We know that the shortest distance between two points is a straight line. This distance can be calculated by using the distance formula. Let’s extend this concept to the shortest distance between a point and a line.

Just by looking at a few line segments from $A$ to line $l$, we can tell that the shortest distance between a point and a line is the perpendicular line between them. Therefore, $AD$ is the shortest distance between $A$ and line $l$.

Putting this onto a graph can be a little tougher.

Example 3: Determine the shortest distance between the point (1, 5) and the line $y=\frac{1}{3}x-2$.

Solution: First, graph the line and point. Second determine the equation of the perpendicular line. The opposite sign and reciprocal of $\frac{1}{3}$ is -3, so that is the slope. We know the line must go through the given point, (1, 5), so use that to find the $y-$intercept.

$y&=-3x+b\\5&=-3(1)+b \qquad \quad \text{The equation of the line is} \ y=-3x+8.\\8&=b$

Next, we need to find the point of intersection of these two lines. By graphing them on the same axes, we can see that the point of intersection is (3, -1), the green point.

Finally, plug (1, 5) and (3,-1) into the distance formula to find the shortest distance.

$d& =\sqrt{(3-1)^2+(-1-5)^2}\\&=\sqrt{(2)^2+(-6)^2}\\& = \sqrt{2+36}\\& = \sqrt{38} \approx 6.16 \ units$

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

Solution: Any line with $x = a$ number is a vertical line. In this case, we can just count the squares between the two lines. The two lines are $3-(-5)$ units apart, or 8 units.

You can use this same method with horizontal lines as well. For example, $y = -1$ and $y = 3$ are $3-(-1)$ units, or 4 units apart.

Example 4: What is the shortest distance between $y=2x+4$ and $y=2x-1$?

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

$d & = \sqrt{(1+1)^2+(1-2)^2}\\& = \sqrt{2^2+(-1)^2}\\& = \sqrt{4+1}\\& = \sqrt{5} \approx 2.24 \ units$

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.

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

Example 5: Find the distance between the two parallel lines below.

Solution: 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 $y-$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 $-\frac{1}{3}$ and the perpendicular slope would be 3.

Next, find two points on the lines. Let’s use the $y-$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.

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

## Perpendicular Bisectors in the Coordinate Plane

Recall that the definition of a perpendicular bisector is a perpendicular line that goes through the midpoint of a line segment. Using what we have learned in this chapter and the formula for a midpoint, we can find the equation of a perpendicular bisector.

Example 6: Find the equation of the perpendicular bisector of the line segment between (-1, 8) and (5, 2).

Solution: First, find the midpoint of the line segment.

$\left ( \frac{-1+5}{2}, \ \frac{8+2}{2} \right ) = \left ( \frac{4}{2}, \ \frac{10}{2} \right )= (2, \ 5)$

Second, find the slope between the two endpoints. This will help us figure out the perpendicular slope for the perpendicular bisector.

$m=\frac{2-8}{5+1}=\frac{-6}{6}=-1$

If the slope of the segment is -1, then the slope of the perpendicular bisector will be 1. The last thing to do is to find the $y-$intercept of the perpendicular bisector. We know it goes through the midpoint, (2, 5), of the segment, so substitute that in for $x$ and $y$ in the slope-intercept equation.

$y & = mx+b\\5 & = 1(2)+b\\5 & = 2+b\\3 & = b$

The equation of the perpendicular bisector is $y= x+3$.

Example 7: The perpendicular bisector of $\overline{AB}$ has the equation $y=-\frac{1}{3}x+1$. If $A$ is (-1, 8) what are the coordinates of $B$?

Solution: The easiest way to approach this problem is to graph it. Graph the perpendicular line and plot the point. See the graph to the left.

Second, determine the slope of $\overline{AB}$. If the slope of the perpendicular bisector is $-\frac{1}{3}$, then the slope of $\overline{AB}$ is 3.

Using the slope, count down 3 and over to the right 1 until you hit the perpendicular bisector. Counting down 6 and over 2, you land on the line at (-3, 2). This is the midpoint of $\overline{AB}$. If you count down another 6 and over to the right 2 more, you will find the coordinates of $B$, which are (-5, -4).

Know What? Revisited Draw two intersecting lines. Make sure they are not perpendicular. Label the 26.3 miles along hwy 5. The longest distance is found by adding the distances along the 110 and 405, or 41.8 miles.

## Review Questions

Find the distance between each pair of points. Round your answer to the nearest hundredth.

1. (4, 15) and (-2, -1)
2. (-6, 1) and (9, -11)
3. (0, 12) and (-3, 8)
4. (-8, 19) and (3, 5)
5. (3, -25) and (-10, -7)
6. (-1, 2) and (8, -9)
7. (5, -2) and (1, 3)
8. (-30, 6) and (-23, 0)

Determine the shortest distance between the given line and point. Round your answers to the nearest hundredth.

1. $y=\frac{1}{3}x+4; \ (5, \ -1)$
2. $y = 2x-4; \ (-7, \ -3)$
3. $y=-4x+1; \ (4, \ 2)$
4. $y=-\frac{2}{3}x-8; \ (7, \ 9)$

Use each graph below to determine how far apart each theparallel 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. $x = 5, \ x = 1$
2. $y = -6, \ y = 4$
3. $y=x+5, \ y=x-3$
4. $y=-\frac{1}{3}x+2, \ y=-\frac{1}{3}x-8$
5. $y=4x+9, \ y=4x-8$
6. $y=\frac{1}{2}x, \ y=\frac{1}{2}x-5$

Find the equation of the perpendicular bisector for pair of points.

1. (1, 5) and (7, -7)
2. (1, -8) and (7, -6)
3. (9, 2) and (-9, -10)
4. (-7, 11) and (-3, 1)
5. The perpendicular bisector of $\overline{CD}$ has the equation $y=3x-11$. If $D$ is (-3, 0) what are the coordinates of $C$?
6. The perpendicular bisector of $\overline{LM}$ has the equation $y=-x+5$. If $L$ is (6, -3) what are the coordinates of $M$?
7. Construction Plot the points (5, -3) and (-5, -9). Draw the line segment between the points. Construct the perpendicular bisector for these two points. (Construction was in Chapter 1). Determine the equation of the perpendicular bisector and the midpoint.
8. Construction Graph the line $y=-\frac{1}{2}x-5$ and the point (2, 5). Construct the perpendicular line, through (2, 5) and determine the equation of this line.
9. Challenge The distance between two points is 25 units. One point is (-2, 9). What is the second point? You may assume that the second point is made up of integers.
10. Writing List the steps you would take to find the distance between two parallel lines, like the two in #24.

1. $y = -4x -1$
2. $y = \frac{1}{2}x - 3$
3. $y = \frac{2}{3}x + 6$

Feb 23, 2012

Feb 12, 2015