Distance and Midpoint Quiz Geometry - Coordinate Plane Study Guide

Fundamental Coordinate Plane Formulas

• Geometry study topics for the coordinate plane include calculating straight-line segment length and locating segment bisecting points.

• Standard variables for coordinate pairs:

  • Point 1: $(x_1, y_1)$

  • Point 2: $(x_2, y_2)$

• Distance Formula:

  • Standard Equation: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

  • Purpose: Computes the geometric distance $d$ between any two points on a Cartesian coordinate plane based on the Pythagorean theorem.

• Midpoint Formula:

  • Standard Equation: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

  • Purpose: Finds the exact central coordinate point $M$ equidistant from both endpoints of a segment.

Part 1: Distance Formula Exercises

• General instructions: Calculate the distance between each pair of coordinate points. Round final values to the nearest hundredth when necessary.

• Item 1: Points $(2, 3)$ and $(5, 7)$

  • Step-by-step substitution: $d = \sqrt{(5 - 2)^2 + (7 - 3)^2}$

  • Simplification: $d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}$

  • Evaluated Answer: $5$

• Item 2: Points $(-1, 4)$ and $(3, 1)$

  • Step-by-step substitution: $d = \sqrt{(3 - (-1))^2 + (1 - 4)^2}$

  • Simplification: $d = \sqrt{(3 + 1)^2 + (-3)^2} = \sqrt{4^2 + 9} = \sqrt{16 + 9} = \sqrt{25}$

  • Evaluated Answer: $5$

• Item 3: Points $(0, 0)$ and $(6, 8)$

  • Step-by-step substitution: $d = \sqrt{(6 - 0)^2 + (8 - 0)^2}$

  • Simplification: $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$

  • Evaluated Answer: $10$

• Item 4: Points $(-3, -2)$ and $(2, 10)$

  • Step-by-step substitution: $d = \sqrt{(2 - (-3))^2 + (10 - (-2))^2}$

  • Simplification: $d = \sqrt{(2 + 3)^2 + (10 + 2)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169}$

  • Evaluated Answer: $13$

• Item 5: Points $(5, -1)$ and $(5, 6)$

  • Step-by-step substitution: $d = \sqrt{(5 - 5)^2 + (6 - (-1))^2}$

  • Simplification: $d = \sqrt{0^2 + (6 + 1)^2} = \sqrt{0^2 + 7^2} = \sqrt{49}$

  • Evaluated Answer: $7$

Part 2: Midpoint Formula Exercises

• General instructions: Calculate the midpoint coordinates $M$ for the line segment connecting each pair of coordinate points.

• Item 1: Points $(2, 3)$ and $(8, 7)$

  • Step-by-step substitution: $M = \left( \frac{2 + 8}{2}, \frac{3 + 7}{2} \right)$

  • Simplification: $M = \left( \frac{10}{2}, \frac{10}{2} \right)$

  • Evaluated Answer: $(5, 5)$

• Item 2: Points $(-4, 5)$ and $(2, -1)$

  • Step-by-step substitution: $M = \left( \frac{-4 + 2}{2}, \frac{5 + (-1)}{2} \right)$

  • Simplification: $M = \left( \frac{-2}{2}, \frac{4}{2} \right)$

  • Evaluated Answer: $(-1, 2)$

• Item 3: Points $(0, 0)$ and $(10, 6)$

  • Step-by-step substitution: $M = \left( \frac{0 + 10}{2}, \frac{0 + 6}{2} \right)$

  • Simplification: $M = \left( \frac{10}{2}, \frac{6}{2} \right)$

  • Evaluated Answer: $(5, 3)$

• Item 4: Points $(-6, -3)$ and $(4, 9)$

  • Step-by-step substitution: $M = \left( \frac{-6 + 4}{2}, \frac{-3 + 9}{2} \right)$

  • Simplification: $M = \left( \frac{-2}{2}, \frac{6}{2} \right)$

  • Evaluated Answer: $(-1, 3)$

• Item 5: Points $(7, 2)$ and $(-1, -2)$

  • Step-by-step substitution: $M = \left( \frac{7 + (-1)}{2}, \frac{2 + (-2)}{2} \right)$

  • Simplification: $M = \left( \frac{6}{2}, \frac{0}{2} \right)$

  • Evaluated Answer: $(3, 0)$

Part 3: Mixed Practice Exercises

• Item 1: Given Point $A$ at $(1, 2)$ and Point $B$ at $(7, 10)$.

  • Segment Identification: Line segment $\overline{AB}$

  • Distance Evaluation between Point $A$ and Point $B$:

  • Substitution: $d = \sqrt{(7 - 1)^2 + (10 - 2)^2}$

  • Simplification: $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$

  • Distance Result: $10$

  • Midpoint Evaluation of Segment $\overline{AB}$:

  • Substitution: $M = \left( \frac{1 + 7}{2}, \frac{2 + 10}{2} \right)$

  • Simplification: $M = \left( \frac{8}{2}, \frac{12}{2} \right)$

  • Midpoint Result: $(4, 6)$

• Item 2: Given line segment with endpoints $(-5, 3)$ and $(3, -3)$.

  • Length (Distance) Evaluation:

  • Substitution: $d = \sqrt{(3 - (-5))^2 + (-3 - 3)^2}$

  • Simplification: $d = \sqrt{(3 + 5)^2 + (-6)^2} = \sqrt{8^2 + 36} = \sqrt{64 + 36} = \sqrt{100}$

  • Length Result: $10$

  • Midpoint Evaluation:

  • Substitution: $M = \left( \frac{-5 + 3}{2}, \frac{3 + (-3)}{2} \right)$

  • Simplification: $M = \left( \frac{-2}{2}, \frac{0}{2} \right)$

  • Midpoint Result: $(-1, 0)$

Practice Quiz: Fundamental Coordinate Plane Formulas 1. Distance Formula Questions a. Calculate the distance between points $(1, 1)$ and $(4, 5)$. b. What is the distance between points $(2, -3)$ and $(3, 3)$? c. Find the distance between points $(0, 0)$ and $(3, 4)$. d. Determine the distance between points $(5, 5)$ and $(5, -3)$. e. What is the distance between points $(-2, -2)$ and $(2, 2)$? 2. Midpoint Formula Questions a. What is the midpoint between points $(2, 6)$ and $(8, 2)$? b. Calculate the midpoint between points $(1, 1)$ and $(4, 4)$. c. Find the midpoint between points $(-1, 2)$ and $(3, -2)$. d. Determine the midpoint between points $(0, 0)$ and $(8, 10)$. e. What is the midpoint between points $(-4, -4)$ and $(0, 0)$? 3. Mixed Practice Questions a. Given point $A$ at $(2, 3)$ and point $B$ at $(8, -1)$, find the distance between points $A$ and $B$ and their midpoint. b. Given points $(-2, 4)$ and $(2, 0)$, evaluate the distance and midpoint. 4. Bonus - Theoretical Questions a. Why is the Pythagorean theorem fundamental in deriving the distance formula? b. In what situations would you need to use the midpoint formula?