Lecture Notes: Distance, Midpoint, and Circles

Key Concepts Covered

• Distance Formula
• Midpoint
• Equation of a Circle

Distance Formula

• Objective: Find the distance between two points on a coordinate plane.
• Points: Consider points ((x_1, y_1)) and ((x_2, y_2)).
• Steps:
  1. Label the points: first point ((x_1, y_1)), second point ((x_2, y_2)).
  2. Connect the points with a line segment, label its length as (d).
  3. Use the Pythagorean theorem:
     • Derive horizontal distance: (x_2 - x_1).
     • Derive vertical distance: (y_2 - y_1).
  4. Express as: [ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

Examples

Example 1

• Points: ((-2, -5)) and ((-4, 3))
  • Substitute into the formula:
    • (x_2 - x_1 = -2 + 4 = 2)
    • (y_2 - y_1 = -5 - 3 = -8)
  • (d = \sqrt{2^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68})
  • Simplify: (\sqrt{68} = 2\sqrt{17})

Example 2

• Points: ((-\sqrt{3}, 5\sqrt{6})) and ((2\sqrt{3}, \sqrt{6}))
  • Simplification:
    • (x_2 - x_1 = -3\sqrt{3})
    • (y_2 - y_1 = 4\sqrt{6})
  • (d = \sqrt{(-3\sqrt{3})^2 + (4\sqrt{6})^2} = \sqrt{27 + 96} = \sqrt{123})
  • Already simplified.

Example 3

• Points: ((\frac{1}{3}, \frac{6}{5})) and ((\frac{7}{3}, \frac{1}{5}))
  • Simplification:
    • (x_2 - x_1 = -2)
    • (y_2 - y_1 = 1)
  • (d = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5})

Conclusion

• Understanding and applying the distance formula involves substituting the differences of x and y coordinates into the formula derived from the Pythagorean theorem.
• Simplification can often be required to find the exact value.
• The distance is expressed in units (e.g., meters, feet) depending on the context.
• Practice with different types of points (including those containing square roots and fractions) to become proficient in application.