- 53 Views
- Uploaded on
- Presentation posted in: General

Introduction

Introduction

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Introduction

Finding the distance between two points on a coordinate system is similar to finding the distance between two points on a number line. It is different in that finding the distance between two points on a coordinate system makes use of two dimensions, but the distance along a number line is only one dimension. Also, we can’t always easily measure or count the number of units between two points on a coordinate system. Instead, we can use the Pythagorean Theoremto find the number.

6.1.1: Applying the Pythagorean Theorem

Introduction, continued

The Pythagorean Theorem relates the length of the hypotenuse of a right triangle (c) to the lengths of its legs (a and b), a2 + b2 = c2, to calculate the distance between the two points. Remember that distance is always positive; therefore, we must take the absolute value, or the distance from zero, of distances calculated.

6.1.1: Applying the Pythagorean Theorem

Key Concepts

Reviewing Distance on a Number Line

To find the distance between two points, a and b, on a number line, find the absolute value of the difference of a and b. This can be expressed algebraically as

|a – b| or |b – a|.

For example, to find the distance between –4 and 5, take the absolute value of the difference of –4 and 5.

|–4 – 5| = |–9| = 9 or |5 – –4| = |5 + 4| = |9| = 9

6.1.1: Applying the Pythagorean Theorem

Key Concepts, continued

The distance between the numbers –4 and 5 is 9 units.

It may not be that easy to find the distance on a coordinate system.

To find the distance between two points on a coordinate system, we must use the Pythagorean Theorem.

6.1.1: Applying the Pythagorean Theorem

Key Concepts, continued

Reviewing the Pythagorean Theorem

Right triangles are triangles with one right (90˚) angle.

The side that is the longest and is always across from the right angle is called the hypotenuse. The two shorter sides are referred to as the legs of the right triangle.

We can use the Pythagorean Theorem to calculate the length of any one of the three sides.

6.1.1: Applying the Pythagorean Theorem

Key Concepts, continued

For example, to find the length of the hypotenuse of a triangle with legs of 5 and 7 units, we use the Pythagorean Theorem.

a2 + b2= c2Pythagorean Theorem

52+ 72= c2Substitute known values.

25 + 49 = c2Simplify.

74 = c2Simplify.

Take the square root of both sides of the equation.

6.1.1: Applying the Pythagorean Theorem

Key Concepts, continued

The length of the hypotenuse of the right triangle with side lengths 5 and 7 is , or approximately 8.6 units.

The Pythagorean Theorem can help us find the distance between two points on a coordinate system.

6.1.1: Applying the Pythagorean Theorem

Key Concepts, continued

6.1.1: Applying the Pythagorean Theorem

Common Errors/Misconceptions

incorrectly identifying the x- and y-values

substituting the x- and y-values incorrectly to find the lengths of the triangle’s sides

forgetting to take the square root of c in order to find the distance between the points

6.1.1: Applying the Pythagorean Theorem

Guided Practice

Example 2

Tyler and Arsha have mapped out locations for a game of manhunt. Tyler’s position is represented by the point (–2, 1). Arsha’s position is represented by the point (–7, 9). Each unit is equivalent to 100 feet. What is the approximate distance between Tyler and Arsha?

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 2, continued

Plot the points on a coordinate system.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 2, continued

Draw lines to form a right triangle, using each point as the end of the hypotenuse.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 2, continued

Calculate the length of the vertical side, a, of the right triangle.

Let (x1, y1) = (–2, 1) and (x2, y2) = (–7, 9).

|y2 – y1| = |9 – 1| = |8| = 8

The length of side a is 8 units.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 2, continued

Calculate the length of the horizontal side, b, of the right triangle.

|x2 – x1| = |–7 – –2| = |–5| = 5

The length of side b is 5 units.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 2, continued

Use the Pythagorean Theorem to calculate the length of the hypotenuse, c.

a2b2= c2Pythagorean Theorem

82+ 52= c2Substitute values for a and b.64 + 25 = c2Simplify each term.

89 = c2Simplify.

Take the square root of both sides of the equation.

The distance between Tyler and Arsha is approximately 9.4 units or 940 feet.

✔

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 2, continued

6.1.1: Applying the Pythagorean Theorem

Guided Practice

Example 3

Kevin is standing 2 miles due north of the school. James is standing 4 miles due west of the school. What is the distance between Kevin and James?

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

Plot the points on a coordinate system.

Let (0, 0) represent the location of the school.

Kevin is standing 2 miles due north, so his location is 2 units above the origin, or at the point (0, 2).

James is standing 4 units due west, so his location is 4 units to the left of the origin, or at the point (–4, 0).

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

Draw lines to form a right triangle, using each point as the end of the hypotenuse.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

Calculate the length of the vertical side, a, of the right triangle.

Let (x1, y1) = (0, 2) and (x2, y2) = (–4, 0).

|y2 – y1| = |0 – 2| = |–2| = 2

The length of side a is 2 units.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

Calculate the length of the horizontal side, b, of the right triangle.

|x2 – x1| = |–4 – 0| = |–4| = 4

The length of side b is 4 units.

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

Use the Pythagorean Theorem to calculate the length of the hypotenuse, c.

a2b2 = c2Pythagorean Theorem

22+ 42= c2Substitute values for a and b.4 + 16 = c2Simplify each term.

20 = c2Simplify.

Take the square root of both sides of the equation.

The distance between Kevin and James is approximately 4.5 miles.

✔

6.1.1: Applying the Pythagorean Theorem

Guided Practice: Example 3, continued

6.1.1: Applying the Pythagorean Theorem