How to Calculate Distances Between Points in a Coordinate System?

Understanding how to calculate distances between points in a coordinate system is essential in mathematics, especially in geometry and algebra. It’s a fundamental concept that has practical applications in various fields, such as computer graphics, navigation, and even in everyday problem-solving.

The Distance Formula

The distance between two points
$(x_1, y_1)$ and
$(x_2, y_2)$ in a 2D coordinate system can be calculated using the distance formula derived from the Pythagorean theorem. The formula is:

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

Example

Let’s consider two points A(1, 2) and B(4, 6). To find the distance between these points, we substitute the coordinates into the distance formula:

$d = sqrt{(4 – 1)^2 + (6 – 2)^2}$

$d = sqrt{3^2 + 4^2}$

$d = sqrt{9 + 16}$

$d = sqrt{25}$

$d = 5$

So, the distance between points A and B is 5 units.

The 3D Distance Formula

When dealing with three-dimensional space, the distance formula extends to include the z-coordinates. For points
$(x_1, y_1, z_1)$ and
$(x_2, y_2, z_2)$, the formula is:

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

Example

Consider points C(1, 2, 3) and D(4, 6, 8). To find the distance between these points:

$d = sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2}$

$d = sqrt{3^2 + 4^2 + 5^2}$

$d = sqrt{9 + 16 + 25}$

$d = sqrt{50}$

$d = 5sqrt{2}$

So, the distance between points C and D is
$5sqrt{2}$ units.

Special Cases

Horizontal and Vertical Distances

If the points lie on a horizontal line, the distance is simply the difference in the x-coordinates. For points
$(x_1, y)$ and
$(x_2, y)$:

$d = |x_2 – x_1|$

Similarly, if the points lie on a vertical line, the distance is the difference in the y-coordinates. For points
$(x, y_1)$ and
$(x, y_2)$:

$d = |y_2 – y_1|$

Example

For horizontal distance, consider points E(2, 3) and F(5, 3):

$d = |5 – 2|$

$d = 3$

For vertical distance, consider points G(4, 1) and H(4, 5):

$d = |5 – 1|$

$d = 4$

Real-World Applications

Navigation

Calculating distances between points is crucial in navigation systems like GPS. For example, finding the shortest path between two locations on a map relies on understanding these distance calculations.

Computer Graphics

In computer graphics, distances between points help in rendering objects accurately. For instance, determining how far a camera is from an object can influence how the object is displayed on the screen.

Engineering and Architecture

Engineers and architects use these calculations to design structures, ensuring that measurements are precise and that components fit together correctly.

Conclusion

Calculating distances between points in a coordinate system is a fundamental skill in mathematics with numerous practical applications. Whether in two-dimensional or three-dimensional space, the distance formula provides a straightforward method to find the distance between any two points. By understanding and applying this formula, you can solve a wide range of real-world problems more effectively.

Citations

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