# Distance Formula and the Pythagorean Theorem

## Determine distance between ordered pairs

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Identify and Use the Distance Formula

As Ross was cruising on his boat toward the cottage, he was concentrating on finding a new mooring for his boat. He had previously sunk two buoy markers in the water near the cottage and both locations had ample water below for the boat. Ross looked at his chart plot which showed the coordinates for the cottage and the two buoys. He decided to choose the buoy closest to the cottage.

How can Ross use the coordinates from the chart plot to figure out which buoy is closest to the cottage?

In this concept, you will learn to identify and use the distance formula.

### Distance Formula

When plotting points or graphing lines on a Cartesian grid, the coordinates of the points, or of the endpoints, of the graphed line are very useful when solving real-world problems.

The following graph shows an orange line that represents the distance between the pet store (1, 6) and the drug store (8, 2). The green dot on the grid represents Timothy who is sitting on a bench in the Mall. To his right he can see the drug store and straight ahead he can see the pet store. He decides the pet store is closer so he goes there first. He exits the pet store and tries to decide whether to go back to the bench and then to the drug store or to walk diagonally to the drug store. Which way would be the shorter of the two distances if each side of the squares represents one yard?

On the grid, draw a horizontal line from Timothy on the bench to the drug store. The length of this line can be found by counting the tops of the squares that the line passes through. The length of the line is 7 yards. Now, draw a vertical line from Timothy on the bench to the pet store. The length of this line can be found by counting the sides of the squares that the line passes through. The length of this line is 4 yards. The three lines form a right triangle and the Pythagorean Theorem can be used to find the diagonal distance between the pet store and the drug store. This diagonal distance is in fact the hypotenuse of the right triangle.

First, write the Pythagorean Theorem.

\begin{align*}c^2 = a^2 + b^2\end{align*}

Next, determine the values of \begin{align*}(a, b, c)\end{align*} for the Pythagorean Theorem.

\begin{align*}\begin{array} {rcl} a &=& 4 \\ b &=& 7 \\ c &=& ? \end{array}\end{align*}

Next, fill these values into the Pythagorean Theorem.

\begin{align*}\begin{array}{rcl} c^2 &=& a^2 + b^2 \\ c^2 &=& (4)^2 + (7)^2 \end{array}\end{align*}

Next, perform the indicated squaring on the right side of the equation and simplify.

\begin{align*}\begin{array}{rcl} c^2 &=& (4)^2 + (7)^2 \\ c^2 &=& (4 \times 4) + (7 \times 7) \\ c^2 &=& 16 + 49 \\ c^2 &=& 65 \end{array}\end{align*}

Then, solve for the variable ‘\begin{align*}c\end{align*}’ by taking the square root of both sides of the equation.

\begin{align*}\begin{array}{rcl} c^2 &=& 65 \\ \sqrt{c^2} &=& \sqrt{65} \\ c &=& 8.06 \end{array}\end{align*}

The diagonal distance between the pet store and the drug store is 8.06 yards.

It would be a shorter distance for Timothy to walk than to return the 4 yards to the bench and then walk another 7 yards to the drug store.

This method of finding the distance between two points requires that the diagram be drawn on grid paper which may be neither available nor convenient. Another way of finding the distance between two points with known coordinates is to use the distance formula. The distance formula is the square root of the square of the difference between the \begin{align*}x\end{align*}-coordinates of the two points plus the square of the difference between the \begin{align*}y\end{align*}-coordinates of the two points. The distance formula is written as:

\begin{align*}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\end{align*}

such that \begin{align*}(x_1, y_1)\end{align*} are \begin{align*}(x_2, y_2)\end{align*} the coordinates of the point chosen as the first point and the point chosen as the second point respectively. Once the points have been designated, the corresponding values must remain consistent and be substituted into the formula as named.

Let’s apply this formula to find the distance between the pet store and the drug store. Remember the coordinates of the pet store were (1, 6) and those of the drug store were (8, 2).

First, choose the first and second points. The answer will be the same regardless of how the points are chosen. Naming the points as shown below will help match the correct value with its corresponding variable.

\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 1, & 6 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 8, & 2 \end{pmatrix}\end{align*}

Next, write the distance formula and fill in the values for the variables.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(8-1)^2 + (2-6)^2} \end{array}\end{align*}

Next, perform the operations shown under the square root sign.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(8-1)^2 + (2-6)^2} \\ d &=& \sqrt{(7)^2+ (-4)^2} \\ d &=& \sqrt{(7 \times 7) + (-4 \times -4)} \\ d &=& \sqrt{49 + 16} \\ d &=& \sqrt{65} \end{array}\end{align*}

Then, take the square root of 65 and round the answer to the nearest hundredth.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{65} \\ d &=& 8.06 \end{array}\end{align*}

The distance is 8.06 yards.

### Examples

#### Example 1

Earlier, you were given a problem about Ross and the two buoys. He needs to figure out which buoy is closer to the cottage, so he can moor his boat. Ross needs to use the distance formula to figure out which buoy is closer to the cottage.

First, write the coordinates for each distance to be calculated and use the distance formula to calculate each distance to the nearest hundredth.

\begin{align*}\begin{array}{rcl} \text{Cottage to Buoy One} & \quad & \text{Cottage to Buoy Two} \\ (160, 120) \ \text{and} \ (40, 100) & \quad & (160, 120) \ \text{and} \ (220, 40) \end{array}\end{align*}

Distance from cottage to Buoy One:

First, name the first and second points.

\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 160, & 120 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 40, & 100 \end{pmatrix}\end{align*}

Next, write the distance formula and fill in the values for the variables.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(40 - 160)^2 + (100 - 120)^2} \end{array}\end{align*}

Next, perform the operations shown under the square root sign.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(40 - 160)^2 + (100 - 120)^2} \\ d &=& \sqrt{(-120)^2 + (-20)^2} \\ d &=& \sqrt{(-120 \times -120) + (-20 \times -20)} \\ d &=& \sqrt{14400 + 400} \\ d &=& \sqrt{14800} \end{array}\end{align*}

Then, take the square root of 14800 and round the answer to the nearest hundredth.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{14800} \\ d &=& 121.66 \end{array}\end{align*}

The distance from the cottage to Buoy One is 121.66 meters.

Distance from cottage to Buoy Two:

First, name the first and second points.

\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 160, & 120 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 220, & 40 \end{pmatrix}\end{align*}

Next, write the distance formula and fill in the values for the variables.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(220 - 160)^2 + (40 - 120)^2} \end{array}\end{align*}

Next, perform the operations shown under the square root sign.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(220 - 160)^2 + (40 - 120)^2} \\ d &=& \sqrt{(60)^2 + (-80)^2} \\ d &=& \sqrt{(60 \times 60) + (-80 \times -80)} \\ d &=& \sqrt{3600 + 6400} \\ d &=& \sqrt{10000} \end{array}\end{align*}

Then, take the square root of 10000 and round the answer to the nearest hundredth.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{10000} \\ d &=& 100 \end{array}\end{align*}

The distance from the cottage to Buoy Two is 100 meters. Ross will moor the boat at Buoy Two.

#### Example 2

Use the distance formula to find the distance between these points to the nearest tenth

\begin{align*}A (54, 120)\end{align*} and \begin{align*}B (113, 215)\end{align*}

First, name the first and second points.

\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 54, & 120 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 113, & 215 \end{pmatrix}\end{align*}

Next, write the distance formula and fill in the values for the variables.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(113 - 54)^2 + (215 - 120)^2} \end{array}\end{align*}

Next, perform the operations shown under the square root sign.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(113 - 54)^2 + (215 - 120)^2} \\ d &=& \sqrt{(59)^2 + (95)^2} \\ d &=& \sqrt{(59 \times 59) + (95 \times 95)} \\ d &=& \sqrt{3481 + 9025} \\ d &=& \sqrt{12506} \end{array}\end{align*}

Then, take the square root of 12506 and round the answer to the nearest tenth.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{12506} \\ d &=& 111.8 \end{array}\end{align*}

#### Example 3

Use the distance formula to find the distance between the points \begin{align*}D (3, -4)\end{align*} and \begin{align*}E (-2, -10)\end{align*}.

First, name the first and second points.

\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 3, & -4 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2 ,& y_2 \\ -2, & -10 \end{pmatrix}\end{align*}

Next, write the distance formula and fill in the values for the variables.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(-2 -3)^2 + (-10 - (-4))^2} \end{array}\end{align*}

Next, perform the operations shown under the square root sign.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(-2 -3)^2 + (-10 - (-4))^2} \\ d &=& \sqrt{(-5)^2 + (-6)^2} \\ d &=& \sqrt{(-5 \times -5) + (-6 \times -6)} \\ d &=& \sqrt{25+36} \\ d &=& \sqrt{61} \\ \end{array}\end{align*}

Then, take the square root of 61 and round the answer to the nearest tenth.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{61} \\ d &=& 7.8 \end{array}\end{align*}

The distance from \begin{align*}D\end{align*} to \begin{align*}E\end{align*} is 7.8 units.

#### Example 4

The following circle has its center at the point (5, 3) and a radius of 3 inches. Use the distance formula to determine if the point (8, 5) is inside the circle, on the circle or outside the circle.

You must use the distance formula to find the distance between the center of the circle (5, 3) and the point (8, 5).

First, name the first and second points.

\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 5, & 3 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 8, & 5 \end{pmatrix}\end{align*}

Next, write the distance formula and fill in the values for the variables.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(8-5)^2 + (5-3)^2} \end{array}\end{align*}

Next, perform the operations shown under the square root sign.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{(8-5)^2 + (5-3)^2} \\ d &=& \sqrt{(3)^2 + (2)^2} \\ d &=& \sqrt{(3 \times 3) + (2 \times 2)} \\ d &=& \sqrt{9+4} \\ d &=& \sqrt{13} \end{array}\end{align*}

Then, take the square root of 13 and round the answer to the nearest tenth.

\begin{align*}\begin{array}{rcl} d &=& \sqrt{13} \\ d &=& 3.6 \end{array}\end{align*}

The distance from the center of the circle to the point (8, 5) is 3.6 inches. The point will be outside the circle since this distance is greater than the length of the radius.Follow Up

### Review

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)?

Answer each of the following questions using the map below of Bryan’s town.

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

12. The map above shows Bryan’s town. What is the distance between the pet store and the courthouse?

13. The map above shows Bryan’s town. What is the distance between the courthouse and the library?

14. The map above shows Bryan’s town. What is the distance between the library and the town hall?

15. The map above shows Bryan’s town. What is the distance between the pet store and the library?

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### Vocabulary Language: English

TermDefinition
Distance Formula The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.