How to Find the Length of AB?

Finding the length of a line segment AB in a coordinate plane is a fundamental concept in geometry. This involves understanding the distance formula, which is derived from the Pythagorean theorem. Let’s walk through the process step-by-step.

Understanding Coordinates

First, it’s essential to understand what coordinates are. In a 2D coordinate plane, any point can be represented by an ordered pair $(x, y)$, where x is the horizontal distance from the origin (0,0) and y is the vertical distance.

For example, if point A is at $(x_1, y_1)$ and point B is at $(x_2, y_2)$, these coordinates tell us exactly where each point is located on the plane.

The Distance Formula

The distance formula is a direct application of the Pythagorean theorem. If you have two points, A and B, with coordinates $(x_1, y_1)$ and $(x_2, y_2)$, the distance between these points, denoted as $d$, can be calculated using the formula:

$d = sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

Breaking Down the Formula

  1. Subtract the x-coordinates: $(x_2 – x_1)$
  2. Subtract the y-coordinates: $(y_2 – y_1)$
  3. Square both results: $(x_2 – x_1)^2$ and $(y_2 – y_1)^2$
  4. Add the squared results: $(x_2 – x_1)^2 + (y_2 – y_1)^2$
  5. Take the square root: $sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

Example Calculation

Let’s say we have two points: A(3, 4) and B(7, 1). To find the length of AB, follow these steps:

  1. Subtract the x-coordinates: $7 – 3 = 4$
  2. Subtract the y-coordinates: $1 – 4 = -3$
  3. Square both results: $4^2 = 16$ and $(-3)^2 = 9$
  4. Add the squared results: $16 + 9 = 25$
  5. Take the square root: $sqrt{25} = 5$

So, the length of AB is 5 units.

Applications of the Distance Formula

The distance formula is not just a theoretical concept; it has practical applications in various fields:

  1. Navigation: Calculating the shortest path between two locations.
  2. Physics: Determining the distance between two particles in space.
  3. Computer Graphics: Measuring distances between pixels on a screen.
  4. Engineering: Finding distances in design and construction projects.

Extending to 3D Space

The distance formula can also be extended to three-dimensional space. If you have points A($x_1, y_1, z_1$) and B($x_2, y_2, z_2$), the distance between them is calculated as:

$d = sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}$

Example in 3D

Suppose we have points A(1, 2, 3) and B(4, 6, 8). To find the distance between them:

  1. Subtract the x-coordinates: $4 – 1 = 3$
  2. Subtract the y-coordinates: $6 – 2 = 4$
  3. Subtract the z-coordinates: $8 – 3 = 5$
  4. Square all results: $3^2 = 9$, $4^2 = 16$, $5^2 = 25$
  5. Add the squared results: $9 + 16 + 25 = 50$
  6. Take the square root: $sqrt{50} approx 7.07$

So, the distance between points A and B in 3D space is approximately 7.07 units.

Common Mistakes to Avoid

  1. Sign Errors: Ensure you correctly subtract the coordinates.
  2. Squaring Mistakes: Remember that squaring a negative number results in a positive number.
  3. Order of Operations: Follow the correct order—subtract, square, add, then take the square root.

Practice Problems

  1. Find the distance between points (2, 3) and (5, 7).
  2. Calculate the distance between points (-1, -1) and (4, 2).
  3. Determine the length of the line segment connecting (0, 0, 0) and (3, 4, 5).

Conclusion

Understanding how to find the length of a line segment AB using the distance formula is a crucial skill in geometry and various real-world applications. By mastering this formula, you can easily calculate distances in both two-dimensional and three-dimensional spaces.

Citations

  1. 1. Khan Academy – Distance Formula
  2. 2. Math is Fun – Distance Between 2 Points
  3. 3. Purplemath – Distance Formula

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ