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Unit 6 lesson 6 distance and the Pythagorean theorem 1-5

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If you're looking for an overview or summary of a lesson on distance and the Pythagorean Theorem, here's a brief guide that might help.

Distance and the Pythagorean Theorem

Objective: Understand how to calculate the distance between two points in a coordinate plane using the Pythagorean Theorem.

Key Concepts

  1. Pythagorean Theorem: The Pythagorean Theorem states that in a right 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 \(b\). This can be written as:

    \[ a^2 + b^2 = c^2 \]

  2. Distance Formula: The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a coordinate plane can be derived from the Pythagorean Theorem. The formula is:

    \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Examples

  1. Finding the Distance Between Two Points: For points \(A(3, 4)\) and \(B(7, 1)\):

    • Calculate the differences: \[ x_2 - x_1 = 7 - 3 = 4 \] \[ y_2 - y_1 = 1 - 4 = -3 \]
    • Apply the distance formula: \[ d = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

  2. Determining if a Triangle is a Right Triangle: Given points \(A(1, 1)\), \(B(4, 1)\), and \(C(1, 5)\):

    • Calculate the lengths of the sides using the distance formula: \[ AB = \sqrt{(4-1)^2 + (1-1)^2} = \sqrt{3^2} = 3 \] \[ BC = \sqrt{(1-4)^2 + (5-1)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5 \] \[ AC = \sqrt{(1-1)^2 + (5-1)^2} = \sqrt{0 + 16} = 4 \]
    • Check if \(AB^2 + AC^2 = BC^2\): \[ 3^2 + 4^2 = 9 + 16 = 25 \] \[ 5^2 = 25 \]
    • Since they are equal, triangle \(ABC\) is a right triangle.

Practice Problems

  1. Find the distance between the points \(P(2, 3)\) and \(Q(5, 7)\).
  2. Determine if the points \(D(0, 0)\), \(E(6, 8)\), and \(F(6, 0)\) form a right triangle.

Solutions to Practice Problems

  1. For points \(P(2, 3)\) and \(Q(5, 7)\): \[ d = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

  2. Checking if \(D(0, 0)\), \(E(6, 8)\), \(F(6, 0)\) form a right triangle: Lengths calculated:

    • \(DE = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = 10\)
    • \(EF = \sqrt{(6-6)^2 + (0-8)^2} = 8\)
    • \(DF = \sqrt{(0-6)^2 + (0-0)^2} = 6\) Check: \[ 6^2 + 8^2 = 36 + 64 = 100 \quad (10^2 = 100) \] Thus, \(DEF\) is a right triangle.

This framework should help you with understanding distance and the Pythagorean theorem. If you have specific exercises or problems you need assistance with, feel free to ask!