# Distance Formula in the Coordinate Plane

## Length between two points using a right triangle.

%
Progress

MEMORY METER
This indicates how strong in your memory this concept is
Progress
%
Distance Formula in the Coordinate Plane

### Distance Formula in the Coordinate Plane

The distance between two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*} can be defined as \begin{align*}d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\end{align*}. This is called the distance formula. Remember that distances are always positive!

What if you were given the coordinates of two points? How could you find how far apart these two points are?

### Examples

#### Example 1

Find the distance between (-2, -3) and (3, 9).

Use the distance formula, plug in the points, and simplify.

\begin{align*}d & = \sqrt{(3-(-2))^2 + (9-(-3))^2}\\ & = \sqrt{(5)^2 + (12)^2}\\ & = \sqrt{25+144}\\ & = \sqrt{169} = 13 \ units\end{align*}

#### Example 2

Find the distance between (12, 26) and (8, 7).

Use the distance formula, plug in the points, and simplify.

\begin{align*}d & = \sqrt{(8-12)^2 + (7-26)^2}\\ & = \sqrt{(-4)^2 + (-19)^2}\\ & = \sqrt{16+361}\\ & = \sqrt{377} \approx 19.42 \ units\end{align*}

#### Example 3

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

Plug in (4, -2) for \begin{align*}(x_1, y_1)\end{align*} and (-10, 3) for \begin{align*}(x_2, y_2)\end{align*} and simplify.

\begin{align*}d& = \sqrt{(-10-4)^2+(3+2)^2}\\ & = \sqrt{(-14)^2 + (5)^2}\\ & = \sqrt{196+25}\\ & = \sqrt{221} \approx 14.87 \ units\end{align*}

#### Example 4

Find the distance between (3, 4) and (-1, 3).

Plug in (3, 4) for \begin{align*}(x_1, y_1)\end{align*} and (-1, 3) for \begin{align*}(x_2, y_2)\end{align*} and simplify.

\begin{align*}d& = \sqrt{(-1-3)^2+(3-4)^2}\\ & = \sqrt{(-4)^2 + (-1)^2}\\ & = \sqrt{16+1}\\ & = \sqrt{17} \approx 4.12 \ units\end{align*}

#### Example 5

Find the distance between (4, 23) and (8, 14).

Plug in (4, 23) for \begin{align*}(x_1, y_1)\end{align*} and (8, 14) for \begin{align*}(x_2, y_2)\end{align*} and simplify.

\begin{align*}d& = \sqrt{(8-4)^2+(14-23)^2}\\ & = \sqrt{(4)^2 + (-9)^2}\\ & = \sqrt{16+81}\\ & = \sqrt{97} \approx 9.85 \ units\end{align*}

### Review

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)
9. (2, -2) and (2, 5)
10. (-9, -4) and (1, -1)

To see the Review answers, open this PDF file and look for section 3.10.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English Spanish

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.