Lesson Objectives

- Demonstrate an understanding of how to plot ordered pairs
- Learn about the Pythagorean Formula
- Learn how to use the distance formula to find the distance between two points

## How to Find the Distance between two Points

In our last lesson, we learned about square roots.
We can use our knowledge of square roots to find the distance between two points on the coordinate plane.

Leg a and Leg b » represent the two shorter sides

c » represents the longest side

c has a special name, it is referred to as the hypotenuse. Our Pythagorean formula states:

a

Suppose we have a right triangle where a is 3 and b is 4. What is the value of c? To find c, we will start by plugging in for a and b: a

3

9 + 16 = c

25 = c

How do we undo an exponent operation? Since c is squared, we can take the square root of each side: $$\sqrt{25} = \sqrt{c^2}$$ $$c = 5$$ (-5) can also be squared to produce 25. In this case, we are looking for a length, so (-5) is not included. Our value for c is 5. We can check with our formula:

3

9 + 16 = 25

25 = 25

Difference in x - values:

(7 - 2) = 5

Difference in y - values:

(6 - (-6)) = 12 If we relate this back to our original formula:

a

a and b are the two shorter sides, we can let a = 5 and b = 12 or vice versa. Let's plug into the formula and find c:

5

25 + 144 = c

169 = c

Let's take the square root of each side: $$\sqrt{c^2} = \sqrt{169}$$ c = 13

Here c, our hypotenuse represents the distance between the two points. In our case, the answer is 13.

We certainly don't have to whip out graph paper and draw points each time we want to find the distance between two points. We have a formula known as the distance formula which allows us to quickly find the distance between any two points on a coordinate plane. This formula is derived from the same thought process we used in our example.

Distance Formula: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ To use this formula, we simply label our points as:

(x

Just like with the slope formula, it does not matter which point is labeled as which. Once our points are labeled, we simply plug into the formula and simplify. Let's use our earlier example with the points:

(2,-6) and (7,6)

Let (2,-6) be point 1

Then (7,6) is point 2 $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ $$d = \sqrt{(7 - 2)^2 + (6 - (-6))^2}$$ $$d = \sqrt{5^2 + 12^2}$$ $$d = \sqrt{25 + 144}$$ $$d = \sqrt{169}$$ $$d = 13$$ Once again, we get the answer of 13.

We can easily see where this formula comes from. If we say leg a is our difference in x-values, and leg b is our difference in y-values:

a

(x

To solve for c, we can just take the square root of each side: $$c = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ We will replace c with d to stand for distance: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ Let's look at a few examples.

Example 1: Find the distance between the two points given

(1,9) and (9, -6)

Let's label (1,9) as point 1

Then (9,-6) is point 2

Plug into the distance formula: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ $$d = \sqrt{(9 - 1)^2 + (-6 - 9)^2}$$ $$d = \sqrt{(8)^2 + (-15)^2}$$ $$d = \sqrt{64 + 225}$$ $$d = \sqrt{289}$$ $$d = 17$$ The distance between the two points is 17.

Example 2: Find the distance between the two points given

(7,4) and (3, -1)

Let's label (7,4) as point 1

Then (3,-1) is point 2

Plug into the distance formula: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ $$d = \sqrt{(3 - 7)^2 + (-1 - 4)^2}$$ $$d = \sqrt{(-4)^2 + (-5)^2}$$ $$d = \sqrt{16 + 25}$$ $$d = \sqrt{41}$$ Since the square root of 41 is an irrational number, we can leave it in this form or approximate its value using a calculator and rounding. We will state our answer by saying the distance between the two points is the square root of 41.

### Pythagorean Formula

The Pythagorean formula tells us about the relationship between the sides in a right triangle: A right triangle is a triangle with one 90 degree angle. There are three sides:Leg a and Leg b » represent the two shorter sides

c » represents the longest side

c has a special name, it is referred to as the hypotenuse. Our Pythagorean formula states:

a

^{2}+ b^{2}= c^{2}Suppose we have a right triangle where a is 3 and b is 4. What is the value of c? To find c, we will start by plugging in for a and b: a

^{2}+ b^{2}= c^{2}3

^{2}+ 4^{2}= c^{2}9 + 16 = c

^{2}25 = c

^{2}How do we undo an exponent operation? Since c is squared, we can take the square root of each side: $$\sqrt{25} = \sqrt{c^2}$$ $$c = 5$$ (-5) can also be squared to produce 25. In this case, we are looking for a length, so (-5) is not included. Our value for c is 5. We can check with our formula:

3

^{2}+ 4^{2}= 5^{2}9 + 16 = 25

25 = 25

### Distance Formula

We have previously used something known as the distance formula to solve motion word problems (d = r x t). The distance formula in this section is different and allows us to find the distance between two points on the coordinate plane. It does this using the Pythagorean formula. We can find the vertical distance between the two points and the horizontal distance between the two points. These will represent our two shorter sides, leg a and leg b. We can then find the distance between the two points c using the Pythagorean formula. Let's suppose we wanted to find the distance between (2,-6) and (7,6): We plotted the two given points (2, -6) and (7, 6). We also plotted an additional point for illustration purposes (7,-6). If we want to find the length of the two shorter sides, we can use our points:Difference in x - values:

(7 - 2) = 5

Difference in y - values:

(6 - (-6)) = 12 If we relate this back to our original formula:

a

^{2}+ b^{2}= c^{2}a and b are the two shorter sides, we can let a = 5 and b = 12 or vice versa. Let's plug into the formula and find c:

5

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

^{2}169 = c

^{2}Let's take the square root of each side: $$\sqrt{c^2} = \sqrt{169}$$ c = 13

Here c, our hypotenuse represents the distance between the two points. In our case, the answer is 13.

We certainly don't have to whip out graph paper and draw points each time we want to find the distance between two points. We have a formula known as the distance formula which allows us to quickly find the distance between any two points on a coordinate plane. This formula is derived from the same thought process we used in our example.

Distance Formula: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ To use this formula, we simply label our points as:

(x

_{1}, y_{1}), (x_{2}, y_{2})Just like with the slope formula, it does not matter which point is labeled as which. Once our points are labeled, we simply plug into the formula and simplify. Let's use our earlier example with the points:

(2,-6) and (7,6)

Let (2,-6) be point 1

Then (7,6) is point 2 $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ $$d = \sqrt{(7 - 2)^2 + (6 - (-6))^2}$$ $$d = \sqrt{5^2 + 12^2}$$ $$d = \sqrt{25 + 144}$$ $$d = \sqrt{169}$$ $$d = 13$$ Once again, we get the answer of 13.

We can easily see where this formula comes from. If we say leg a is our difference in x-values, and leg b is our difference in y-values:

a

^{2}+ b^{2}= c^{2}(x

_{2}- x_{1})^{2}+ (y_{2}- y_{1})^{2}= c^{2}To solve for c, we can just take the square root of each side: $$c = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ We will replace c with d to stand for distance: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ Let's look at a few examples.

Example 1: Find the distance between the two points given

(1,9) and (9, -6)

Let's label (1,9) as point 1

Then (9,-6) is point 2

Plug into the distance formula: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ $$d = \sqrt{(9 - 1)^2 + (-6 - 9)^2}$$ $$d = \sqrt{(8)^2 + (-15)^2}$$ $$d = \sqrt{64 + 225}$$ $$d = \sqrt{289}$$ $$d = 17$$ The distance between the two points is 17.

Example 2: Find the distance between the two points given

(7,4) and (3, -1)

Let's label (7,4) as point 1

Then (3,-1) is point 2

Plug into the distance formula: $$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$ $$d = \sqrt{(3 - 7)^2 + (-1 - 4)^2}$$ $$d = \sqrt{(-4)^2 + (-5)^2}$$ $$d = \sqrt{16 + 25}$$ $$d = \sqrt{41}$$ Since the square root of 41 is an irrational number, we can leave it in this form or approximate its value using a calculator and rounding. We will state our answer by saying the distance between the two points is the square root of 41.

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