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# Distance Formula in the Coordinate Plane

## Length between two points using a right triangle.

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Practice Distance Formula in the Coordinate Plane
Progress
Estimated17 minsto complete
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Distance Formula in the Coordinate Plane

What if you were given the coordinates of two points? How could you find how far apart these two points are? After completing this Concept, you'll be able to find the distance between two points in the coordinate plane using the Distance Formula.

### Guidance

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 is called the distance formula . Remember that distances are always positive!

#### Example A

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

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

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

#### Example B

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

Plug in (3, 4) for $(x_1, y_1)$ and (-1, 3) for $(x_2, y_2)$ and simplify.

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

#### Example C

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

Plug in (4, 23) for $(x_1, y_1)$ and (8, 14) for $(x_2, y_2)$ and simplify.

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

### Guided Practice

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

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

3. Find the distance between (5, 2) and (6, 1)

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

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

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

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

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

$d & = \sqrt{(6-5)^2 + (1-2)^2}\\& = \sqrt{(1)^2 + (-1)^2}\\& = \sqrt{1+1}\\& = \sqrt{2} = 1.41 \ units$

### Practice

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)

### Vocabulary Language: English Spanish

Distance Formula

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

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.