Distance Between Two Points on a Number Line

Imagine a number line, a straight line with numbers marked at equal intervals. Each point on this line represents a number. Finding the distance between two points on this line is a fundamental concept in mathematics. It’s the length of the line segment connecting those two points.

Understanding the Concept

The distance between two points on a number line is always a positive value. This is because distance is a measure of how far apart two points are, regardless of direction. We use the concept of absolute value to ensure the distance is always positive.

Absolute Value

The absolute value of a number is its distance from zero, regardless of its sign. It is represented by two vertical bars surrounding the number. For example, the absolute value of 5 is 5, and the absolute value of -5 is also 5.

Calculating the Distance

To find the distance between two points on a number line, follow these steps:

  1. Identify the coordinates: Determine the numbers represented by the two points on the number line. Let’s call these numbers a and b.
  2. Subtract the coordinates: Subtract the smaller coordinate from the larger coordinate. This gives you the difference between the two points.
  3. Take the absolute value: Since distance is always positive, take the absolute value of the difference you calculated in step 2. This will ensure the distance is a positive number.

Formula

The distance between two points a and b on a number line can be represented by the following formula:

$Distance = |a – b|$

Examples

Let’s illustrate this with some examples:

Example 1:

Find the distance between the points representing 3 and 7 on the number line.

  1. Coordinates: a = 3, b = 7
  2. Subtract: 7 – 3 = 4
  3. Absolute Value: |4| = 4

Therefore, the distance between the points representing 3 and 7 on the number line is 4 units.

Example 2:

Find the distance between the points representing -2 and 5 on the number line.

  1. Coordinates: a = -2, b = 5
  2. Subtract: 5 – (-2) = 7
  3. Absolute Value: |7| = 7

Therefore, the distance between the points representing -2 and 5 on the number line is 7 units.

Example 3:

Find the distance between the points representing -6 and -1 on the number line.

  1. Coordinates: a = -6, b = -1
  2. Subtract: -1 – (-6) = 5
  3. Absolute Value: |5| = 5

Therefore, the distance between the points representing -6 and -1 on the number line is 5 units.

Applications

The concept of distance on a number line is fundamental in various mathematical applications, including:

  • Graphing: Representing and analyzing data on a number line requires understanding the distance between points.
  • Algebra: Solving equations and inequalities often involves finding the distance between points on a number line.
  • Calculus: Derivatives and integrals are built upon the concept of distance and change in distance.
  • Real-world applications: Distance on a number line can be applied to various real-world scenarios, such as measuring the distance between two cities on a map or calculating the difference in temperature between two locations.

Conclusion

Understanding the distance between two points on a number line is crucial for comprehending various mathematical concepts and applying them to real-world situations. The concept of absolute value ensures that distance is always a positive value, reflecting the length of the line segment connecting the two points.

Citations

  1. 1. Math is Fun – Number Line
  2. 2. Khan Academy – Absolute Value
  3. 3. Purplemath – Absolute Value

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