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# 7.5: The Distance Formula

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## Introduction

The Vegetable Garden

The community service project is in full swing and all the eighth grade homeroom classes have selected their projects. Mr. Henry’s class has decided to make a vegetable garden. They hope that if they are successful that the seventh graders will help them and that they can give a percentage of the food grown to charity.

The students have drawn a map of the garden plan. The biggest obstacle is where to put the garden. They know that it needs to be a level area free of obstructions, but it also needs to be accessible to a water source.

“This is my plan. We can figure out the distance from the water source to the center of the garden. Then if we can buy a hose the correct length and a sprinkler, we should be able to water the garden,” Belinda said to the class.

“It’s a good idea. Why did you put it on a grid?” Carmen asked.

“Because that way we can figure out the exact distance between the two points and each square on the grid represents one foot. I measured it out yesterday. But the exact distance from the water to the center was a little too tough to figure out using a tape measure. That is why I drew it on the grid. Now we can use the distance formula,” Belinda explained.

The class looked puzzled.

Are you puzzled? The distance formula is a great way to figure out exact distances using coordinates and a coordinate grid. Use this lesson to learn all about it and then you can figure out the distance from the water source to the center of the garden at the end.

What You Will Learn

By the end of this lesson you will be able to demonstrate the following skills.

• Recognize the Distance Formula as an application of the Pythagorean Theorem.
• Find the distance between two points on a coordinate grid.
• Find a missing coordinate value for a point, given the other coordinate value, another point and the distance.
• Solve real-world problems involving applications of the Distance Formula.

Teaching Time

I. Recognize the Distance Formula as an Application of the Pythagorean Theorem

When working with points and lines on coordinate grids, there are many different ways to solve problems. You can use your knowledge of algebra, rational numbers, and the Pythagorean Theorem to help you. In this lesson you will learn to use the distance formula to understand the relationship between points on a coordinate grid. In the first lesson, you will learn how to derive this formula mathematically. While it is helpful to memorize the distance formula, it is even better to understand how and why it works. That way, you can always check your work to make sure you have used it properly.

You learned in the last lesson that you can use the Pythagorean Theorem to understand different types of right triangles, find missing lengths, and identify Pythagorean triples. In this lesson, you will apply the Pythagorean Theorem to a coordinate grid and learn how to use it to find distances between points.

Example

Look at the points on the grid below. Then find the distance of the line represented.

The question asks you to identify the length of the line. How can we do this accurately? We can think of this line as the hypotenuse of a right triangle. Draw a vertical line at $x=1$ and a horizontal line at $y=2$ and find the point of intersection. This point represents the third vertex in the right triangle.

You can easily count the lengths of the legs of this triangle on the grid. The vertical leg extends from (1,2) to (1,5), so it is 3 units long. The horizontal leg extends from (1,2) to (5,2), so it is 4 units long. Use the Pythagorean Theorem with these values to identify the length of the hypotenuse.

$a^2+b^2 &= c^2\\3^2+4^2 &= c^2\\(3 \times 3)+(4 \times 4) &= c^2\\9+16 &= c^2\\25 &= c^2\\\sqrt{25} &= \sqrt{c^2}\\5 &= c$

The hypotenuse is 5 units long.

Mathematicians have simplified this process and created a formula which uses these steps to find the distance. This formula is called the distance formula. If you use the distance formula, you don’t have to draw the extra lines.

Example

Use the distance formula $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ to find the distance between the points (1,5) and (5,2) on a coordinate grid.

You already know from the first example that the distance will be 5 units, but you can practice using the Distance formula to make sure it works. In this formula, substitute 1 for $x_1$, 5 for $y_1$, 5 for $x_2$, and 2 for $y_2$ because (1,5) and (5,2) are the two points in question.

$D &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\D &= \sqrt{(5-1)^2+(2-5)^2}\\D &= \sqrt{(4)^2+(-3)^2}\\D &= \sqrt{16+9}\\D &= \sqrt{25}\\D &= 5$

So you see, no matter which way you solve this problem, you find that the distance between (1,5) and (5,2) on a coordinate grid is 5 units.

Notice that the distance formula helps you eliminate the need to graph the line and count all the units. We can use the formula to solve the problem mathematically.

II. Find the Distance between Two Points on a Coordinate Grid

Now let’s practice using the distance formula to solve problems. It is important to become comfortable applying the distance formula to many types of problems and situations. Remember that either points can be considered $(x_1, y_1)$ or $(x_2, y_2)$, but it crucial to keep your assignments consistent through the problem. The most common error students make when using the distance formula is in incorrect substitution. Keep your variables straight and your algebra careful and you’ll be fine.

Example

Use the distance formula to find the distance between the points (-3,2) and (4,-5) on a coordinate grid.

Because we know the distance formula, we don’t even have to draw this out on a coordinate grid. All you have to do is substitute the values in the problem into the distance formula and solve. In this formula, substitute -3 for $x_1$, 2 for $y_1$, 4 for $x_2$, and -5 for $y_2$ because (-3,.2) and (4,-5) are the two points in question.

$D &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\D &= \sqrt{(4-(-3))^2+((-5)-2)^2}\\D &= \sqrt{(7)^2+(-7)^2}\\D &= \sqrt{49+49}\\D &= \sqrt{98}$

You can leave the answer as the radical as shown, or use your calculator to find the approximate value of 9.899 units.

Notice that this answer is not a Pythagorean Triple, so finding a perfect square root is not possible. When this happens, you can either leave the answer in the radical form or find an approximate answer by using a calculator and rounding.

Example

Use the distance formula to find the distance between the points (2,3) and (7,15) on a coordinate grid.

All we have to do is substitute the values in the problem into the distance formula and solve. In this formula, substitute 2 for $x_1$, 3 for $y_1$, 7 for $x_2$, and 15 for $y_2$ because (2,3) and (7,15) are the two points in question.

$D &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\D &= \sqrt{(7-2)^2+(15-3)^2}\\D &= \sqrt{(5)^2+(12)^2}\\D &= \sqrt{25+144}\\D &= \sqrt{169}\\D &= 13$

The distance between the two points in the problem is 13 units.

Now let’s look at how we can use the distance formula to solve problems when we are missing some of the necessary information.

III. Find a Missing Coordinate Value for a Point, Given the Other Coordinate Value, another Point and the Distance

Now that you are comfortable using the distance formula to find the space between two points on a coordinate grid, there is a slightly different type of problem you can learn to solve. You may be given the distance between two points, the coordinates to one point, and one value in the coordinate pair of the second point. You can simply substitute into the distance formula all the values that are provided, and solve algebraically for the missing value.

Example

The distance between two points on a grid is 17 units. If one of the points has the coordinates (-2, 10) and the other point has an $x-$coordinate of 13, what is the $y-$coordinate of the second point?

Use the distance formula, substituting 17 for $D$, -2 for $x_1$, 10 for $y_1$, and 13 for $x_2$. You will have to leave $y_2$ as a variable and solve for its numerical value.

$D &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\17 &= \sqrt{(13-(-2))^2+(y_2-10)^2}\\17^2 &= \sqrt{\left((13-(-2))^2+(y_2-10)^2\right)}^2\\289 &= (15)^2+(y_2-10)^2\\289 &= 225+(y_2-10)^2\\289-225 &= 225+(y_2-10)^2-225\\64 &= (y_2-10)^2\\\sqrt{64} &= \sqrt{(y_2-10)^2}\\8 &= y_2-10\\8+10 &= y_2-10+10\\18 &= y_2$

The other missing point is 18.

There are a few steps to note as we solved this problem. First, notice that we used squaring to cancel out a radical. If we have a radical around values and we want to cancel the radical, then we can square both sides of the equation to cancel out the radical.

Similarly, when we had a value squared, we could use the inverse or square root of both sides of the equation to cancel out the exponent.

Go back through the last problem and make a note of the places where square roots and squaring was used to cancel and simplify.

Write the distance formula down in your notebook.

IV. Solve Real – World Problems Involving Applications of the Distance Formula

Just like any other mathematical concept, you may be tested on this material in a real-world context. Just translate the information into mathematical data and proceed as usual. No new tools or skills are required to solve these problems.

Example

The map below shows the location of various points in Helene’s town.

What is the distance, between the library and the school in Helene’s town?

All you have to do for this problem is identify the coordinates of the school (-1,9) and the library (5,1) on the map and substitute them into the distance formula. Then solve as usual.

$D &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\D &= \sqrt{(5-(-1))^2+(1-9)^2}\\D &= \sqrt{(6)^2+(-8)^2}\\D &= \sqrt{36+64}\\D &= \sqrt{100}\\D &= 10$

So, the distance between the two places is 10 units. You can see on the scale that one unit is equal to one mile, so the real distance is 10 miles.

Now let’s apply the distance formula to the problem from the introduction.

## Real-Life Example Completed

The Vegetable Garden

Here is the problem from the introduction. Reread it and then use the distance formula to figure out the distance from the water source to the center of the garden. Remember that each unit represents 1 foot.

The community service project is in full swing and all the eighth grade homeroom classes have selected their projects. Mr. Henry’s class has decided to make a vegetable garden. They hope that if they are successful that the seventh graders will help them and that they can give a percentage of the food grown to charity.

The students have drawn a map of the garden plan. The biggest obstacle is where to put the garden. They know that it needs to be a level area free of obstructions, but it also needs to be accessible to a water source.

“This is my plan. We can figure out the distance from the water source to the center of the garden. Then if we can buy a hose the correct length and a sprinkler, we should be able to water the garden,” Belinda said to the class.

“It’s a good idea. Why did you put it on a grid?” Carmen asked.

“Because that way we can figure out the exact distance between the two points and each square on the grid represents one foot. I measured it out yesterday. But the exact distance from the water to the center was a little too tough to figure out using a tape measure. That is why I drew it on the grid. Now we can use the distance formula,” Belinda explained.

The class looked puzzled.

Solution to Real – Life Example

To solve this problem, first you will need the coordinates of each point on the grid. This is the distance that you are measuring. In this problem, you will be measuring from point $A$ to point $B$.

Water Source $= A (8, 5)$

Center of Garden $= B (1, 0)$

Now substitute these values into the distance formula and solve.

$D &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\D &= \sqrt{(1-8)^2+(5-0)^2}\\D &= \sqrt{7^2+5^2}\\D &= \sqrt{74}\\D &= 8.6 \ feet$

The students will need a hose that is at least 9 feet long.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

The Pythagorean Theorem
$a^2+b^2=c^2$ - a way of solving for any leg of a right triangle given the lengths of the other two.
The Distance Formula
a formula designed to measure the distance between points on a coordinate grid without drawing all of the lines and counting units, $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

## Time to Practice

Directions: Use the distance formula to find the distance between the following pairs of points. You may round to the nearest tenth when necessary.

1. What is the distance between (3, 6) and (-1, 3)?
2. What is the distance between (-2,-2) and (10, 3)?
3. What is the distance between (1,9) and (9,1)?
4. What is the distance between (-5,-5) and (-2,-1)?
5. What is the distance between (2, 12) and (3,7)?
6. What is the distance between (2, 2) and (8, 2)?
7. What is the distance between (-3, 4) and (2, 0)?
8. What is the distance between (3, 4) and (3, -4)?
9. What is the distance between (-4, -3) and (1, -1)?
10. What is the distance between (-6, 2) and (-3, 1)?
11. What is the distance between (6, -1) and (6, 3)?
12. What is the distance between (1, 5) and (6, 4)?

1. The map below shows Bryan’s town. What is the distance between the pet store and town hall?
2. Malik knows that the distance between two points is 10. One point is (-4,8) and the $x-$coordinate for the other is 4. What is the missing $y-$coordinate? There are two possible answers for this problem.
3. Louise knows that the distance between two points is 13. One point is (-2,-1) and the $y-$coordinate for the other is 4. What is the missing $x-$coordinate? There are two possible answers for this problem.
4. The distance between two points is 17. If one point is (10,5) and the other has an $x-$coordinate of -5, what is the missing $y-$coordinate? There are two possible answers for this problem.
5. Tinsley plots a point at (3,9). She wants to find another point that is 20 units away and has an $x-$coordinate of 15. What will be the $y-$coordinate of Tinsley’s second point? There are two possible answers for this problem.

Jan 14, 2013

Sep 23, 2014