unit 6 lesson 8 Distance and the Pythagorean theorem

Unit 6, Lesson 8 on Distance and the Pythagorean Theorem typically focuses on understanding how to find the distance between two points in a coordinate system using the Pythagorean Theorem. Here's a brief overview of the key concepts:

### Key Concepts

1. **The Pythagorean Theorem**:
- The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b).
- Formula: \( a^2 + b^2 = c^2 \)

2. **Distance Formula**:
- The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a Cartesian coordinate system can be derived from the Pythagorean theorem.
- The distance \(d\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

3. **Finding the Distance**:
- To find the distance between two points:
- Identify the coordinates of the points.
- Substitute these coordinates into the distance formula.
- Simplify to find the distance.

4. **Applications**:
- This concept is useful in various fields including mathematics, engineering, physics, and computer science for calculating distances in 2D and 3D spaces.

### Example Problem

**Example**: Find the distance between the points \(A(2, 3)\) and \(B(5, 7)\).

**Solution**:
1. Identify the coordinates:
- \(x_1 = 2\), \(y_1 = 3\)
- \(x_2 = 5\), \(y_2 = 7\)

2. Use the distance formula:
\[
d = \sqrt{(5 - 2)^2 + (7 - 3)^2}
\]
\[
d = \sqrt{(3)^2 + (4)^2}
\]
\[
d = \sqrt{9 + 16}
\]
\[
d = \sqrt{25}
\]
\[
d = 5
\]

Therefore, the distance between points \(A\) and \(B\) is \(5\) units.

This lesson may also include exercises where students practice finding the distance between various pairs of points and solving real-world problems using the distance formula and the Pythagorean theorem. If you have any specific questions or need further examples, feel free to ask!