<meta http-equiv="refresh" content="1; url=/nojavascript/"> Distance Formula and the Pythagorean Theorem ( Read ) | Geometry | CK-12 Foundation

# Distance Formula and the Pythagorean Theorem

%
Progress
Practice Distance Formula and the Pythagorean Theorem
Progress
%
Identify and Use the Distance Formula

Have you ever designed a vegetable garden? Take a look at this dilemma.

The eighth grade is doing a community service project, and all the homerooms 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 Concept 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.

### Guidance

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. You can apply the distance formula and understand the relationship between points on a coordinate grid.

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

Let's look at how we can do this.

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.

Here is the distance formula.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Now let's apply the distance formula.

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.

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.

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 A

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

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

#### Example B

What is the distance between (6, -1) and (6, 3)?

Solution:  The distance between the two points is 4 units.

#### Example C

What is the distance between (1, 5) and (6, 4)?

Solution:  The distance between the two points is 5.1 units.

Now let's go back to the dilemma from the beginning of the Concept.

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

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, $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

### Guided Practice

Here is one for you to try on your own.

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?

Solution

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.

### Explore More

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)?

Directions: Answer each of the following questions.

1. The map below shows Bryan’s town. What is the distance between the pet store and town hall?

1. The map below shows Bryan’s town. What is the distance between the pet store and the courthouse?
2. The map below shows Bryan’s town. What is the distance between the courthouse and the library?
3. The map below shows Bryan’s town. What is the distance between the library and the town hall?
4. The map below shows Bryan’s town. What is the distance between the pet store and the library?