AB = \sqrt(1-0)^2 + (0-1)^2 + (0-0)^2 = \sqrt1 + 1 = \sqrt2

Understanding the Distance Formula: Deriving AB = √2 Using the Coordinate Plane

In mathematics, especially in geometry and coordinate systems, calculating distances between points is a fundamental skill. One elegant example involves finding the distance between two points, A and B, using the 2D coordinate plane. This article explores how the distance formula works, with a clear step-by-step derivation of the distance formula AB = √[(1−0)² + (0−1)² + (0−0)²] = √2.

Understanding the Context

What is the Distance Between Two Points?

When two points are defined on a coordinate plane by their ordered pairs — for example, A = (1, 0) and B = (0, 1) — the distance formula allows us to compute how far apart they are. This formula comes directly from the Pythagorean theorem.

Given two points A = (x₁, y₁) and B = (x₂, y₂), the distance AB is calculated as:

$$
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

[Solved]: A Prove that the sequence sqrt(2), sqrt(2 sqrt(2)

Image Gallery

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$Solved a) \\( \\int_{0}^{1} x \\sqrt{2-\\sqrt{1-x^{2}}} d x | Chegg.com$

$inequality - Mathematical Induction Proof: $1/\sqrt1 + 1/\sqrt2+...+1 ...$

$algebra precalculus - If $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2 ...$

Key Insights

In our case,

A = (1, 0) → $x_1 = 1$, $y_1 = 0$
- B = (0, 1) → $x_2 = 0$, $y_2 = 1$

Applying the Formula to Points A(1, 0) and B(0, 1)

Substitute these coordinates into the distance formula:

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Final Thoughts

$$
AB = \sqrt{(0 - 1)^2 + (1 - 0)^2}
$$

Simplify the differences inside the parentheses:

$$
AB = \sqrt{(-1)^2 + (1)^2}
$$

Now calculate the squares:

$$
AB = \sqrt{1 + 1} = \sqrt{2}
$$

Why This Formula Works: The Pythagorean Theorem in 2D

The distance formula is nothing more than an application of the Pythagorean theorem in a coordinate system. If we visualize the points A(1, 0) and B(0, 1), connecting them forms a right triangle with legs along the x-axis and y-axis.

The horizontal leg has length $ |1 - 0| = 1 $
- The vertical leg has length $ |0 - 1| = 1 $

Then, the distance AB becomes the hypotenuse: