Lesson Objectives

- Demonstrate an understanding of the coordinate plane
- Learn how to derive the distance formula from the Pythagorean Theorem
- Learn how to find the distance between two points using the distance formula

## How to Find the Distance Between Two Points

The Pythagorean Theorem relates the lengths of the sides of a right triangle (a triangle with a 90-degree angle). The side opposite of the 90-degree angle is called the hypotenuse. This is the longer side and is usually labeled as "c". The two shorter sides are labeled as "a" and "b".

The Pythagorean Theorem states the following:

a

This means if we square the length of leg a and add this to the square of the length of leg b, we will get a result that is equal to the length of the hypotenuse or c squared.

Suppose we had the following right triangle: From our image above, we can see that leg a is 5, and leg b is 12. What is the length of c or the hypotenuse? Let's plug into the Pythagorean Theorem.

a

(5)

25 + 144 = c

169 = c

To solve this, we can just take the square root of each side.

c

We can immediately use the Pythagorean Theorem to find the distance between two points on the coordinate plane. Suppose we wanted to find the distance between the point (3,4) and (7,7). Let's plot each point on the coordinate plane and draw a right triangle. Looking at the graph, we can see the horizontal distance between the two points and the vertical distance between the two points. We have added an additional point, (7,4) to make everything easier to see. What is the horizontal distance between the two points? This would be the length of leg b:

3 - 7 = -4

What is the vertical distance between the two points? This would be the length of leg a:

4 - 7 = -3

The negatives here don't matter since we will be squaring our numbers. It would have been just as valid to say:

leg b: 7 - 3 = 4

leg a: 7 - 4 = 3

We will get the same answer either way.

Let's plug into the Pythagorean Formula and find the length of the hypotenuse, which is the distance between the two points.

a

(-3)

9 + 16 = c

25 = c

If we take the square root of each side: $$\sqrt{c^2}=\sqrt{25}$$ $$c=5$$ This tells us the distance between the point (3,4) and (7,7) is 5.

(x

We can find the distance between the two points using the distance formula. $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ We can think of x

Example 1: Find the distance between the points (6, -3) and (13, 21)

We label one of the points as point 1 and the other as point 2. Let's let (6, -3) be point 1 and then (13, 21) will be point 2. Let's plug into our formula. $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ $$d=\sqrt{(13 - 6)^2 + (21 - (-3))^2}$$ $$d=\sqrt{(7)^2 + (24)^2}$$ $$d=\sqrt{49 + 576}$$ $$d=\sqrt{625}$$ $$d=25$$ Example 2: Find the distance between the points (-1, 9) and (-3, 6)

We label one of the points as point 1 and the other as point 2. Let's let (-1, 9) be point 1 and then (-3, 6) will be point 2. Let's plug into our formula. $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ $$d=\sqrt{(-3 - (-1))^2 + (6 - 9)^2}$$ $$d=\sqrt{(-2)^2 + (3)^2}$$ $$d=\sqrt{4 + 9}$$ $$d=\sqrt{13}$$

The Pythagorean Theorem states the following:

a

^{2}+ b^{2}= c^{2}This means if we square the length of leg a and add this to the square of the length of leg b, we will get a result that is equal to the length of the hypotenuse or c squared.

Suppose we had the following right triangle: From our image above, we can see that leg a is 5, and leg b is 12. What is the length of c or the hypotenuse? Let's plug into the Pythagorean Theorem.

a

^{2}+ b^{2}= c^{2}(5)

^{2}+ (12)^{2}= c^{2}25 + 144 = c

^{2}169 = c

^{2}To solve this, we can just take the square root of each side.

c

^{2}= 169 $$\sqrt{c^2}=\sqrt{169}$$ $$c=13$$ In most cases, we would take the positive and negative square root of 169. An answer of (-13) does not do any good here. The length of our hypotenuse is a distance, which can't be negative. Therefore we only need to take the principal or positive square root of 169 to obtain c.We can immediately use the Pythagorean Theorem to find the distance between two points on the coordinate plane. Suppose we wanted to find the distance between the point (3,4) and (7,7). Let's plot each point on the coordinate plane and draw a right triangle. Looking at the graph, we can see the horizontal distance between the two points and the vertical distance between the two points. We have added an additional point, (7,4) to make everything easier to see. What is the horizontal distance between the two points? This would be the length of leg b:

3 - 7 = -4

What is the vertical distance between the two points? This would be the length of leg a:

4 - 7 = -3

The negatives here don't matter since we will be squaring our numbers. It would have been just as valid to say:

leg b: 7 - 3 = 4

leg a: 7 - 4 = 3

We will get the same answer either way.

Let's plug into the Pythagorean Formula and find the length of the hypotenuse, which is the distance between the two points.

a

^{2}+ b^{2}= c^{2}(-3)

^{2}+ (-4)^{2}= c^{2}9 + 16 = c

^{2}25 = c

^{2}If we take the square root of each side: $$\sqrt{c^2}=\sqrt{25}$$ $$c=5$$ This tells us the distance between the point (3,4) and (7,7) is 5.

### The Distance Formula

Instead of having to graph things each time, we can plug into the distance formula. If we have two points:(x

_{1}, y_{1}), (x_{2}, y_{2})We can find the distance between the two points using the distance formula. $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ We can think of x

_{2}- x_{1}as the horizontal distance between the two points on the coordinate plane. Similarly, y_{2}- y_{1}is the vertical distance. Each represents the length of one of the shorter sides of the right triangle. If these are squared and added together, the result will be the hypotenuse or distance between the two points squared. To get distance by itself, we simply take the square root of each side. Let's look at a few examples.Example 1: Find the distance between the points (6, -3) and (13, 21)

We label one of the points as point 1 and the other as point 2. Let's let (6, -3) be point 1 and then (13, 21) will be point 2. Let's plug into our formula. $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ $$d=\sqrt{(13 - 6)^2 + (21 - (-3))^2}$$ $$d=\sqrt{(7)^2 + (24)^2}$$ $$d=\sqrt{49 + 576}$$ $$d=\sqrt{625}$$ $$d=25$$ Example 2: Find the distance between the points (-1, 9) and (-3, 6)

We label one of the points as point 1 and the other as point 2. Let's let (-1, 9) be point 1 and then (-3, 6) will be point 2. Let's plug into our formula. $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ $$d=\sqrt{(-3 - (-1))^2 + (6 - 9)^2}$$ $$d=\sqrt{(-2)^2 + (3)^2}$$ $$d=\sqrt{4 + 9}$$ $$d=\sqrt{13}$$

#### Skills Check:

Example #1

Find the distance between each pair of points. $$(7, 1), (15, 16)$$

Please choose the best answer.

A

$$\sqrt{5}$$

B

$$\sqrt{17}$$

C

$$13$$

D

$$17$$

E

$$9$$

Example #2

Find the distance between each pair of points. $$(2, 8), (5, 12)$$

Please choose the best answer.

A

$$\sqrt{11}$$

B

$$5$$

C

$$\sqrt{5}$$

D

$$11$$

E

$$2$$

Example #3

Find the distance between each pair of points. $$(-1, 3), (4, 5)$$

Please choose the best answer.

A

$$\sqrt{13}$$

B

$$\sqrt{29}$$

C

$$\sqrt{21}$$

D

$$7$$

E

$$12$$

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