The Distance Between Two Points

The distance is the shortest route from one point to another. To measure the Euclidean distance between two points in a plane, we use the Pythagorean theorem: $$ \overline{AB} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $$

Here, (x₁, y₁) and (x₂, y₂) are the Cartesian coordinates of points A and B in the plane.

A Practical Example

Consider the Cartesian coordinates of two points in space:

$$ A = \begin{pmatrix} 2 \\ 4 \end{pmatrix} = (2, 4) $$

$$ B = \begin{pmatrix} 6 \\ 1 \end{pmatrix} = (6, 1) $$

The distance between these two points is calculated as:

$$ \overline{AB} = \sqrt{(6-2)^2+(1-4)^2} $$

$$ \overline{AB} = \sqrt{4^2+(-3)^2} = \sqrt{16+9} = \sqrt{25} = 5 $$

Verify with Geogebra. The calculated distance is also confirmed by Geogebra.

The Proof

Let's consider two points in the plane:

The segment AB represents the distance between points A and B.

We project the two points onto the Cartesian axes.

The projections form a right triangle ABC on the plane.

The segment AB is the hypotenuse of the right triangle ABC.

Therefore, to measure segment AB, we can use the Pythagorean theorem.

$$ \overline{AB} = \sqrt{\overline{AC}^2 + \overline{BC}^2} $$

Segment AC is the projection of segment AB onto the x-axis.

$$ \overline{AB} = \sqrt{(x_2-x_1)^2 + \overline{BC}^2} $$

Segment BC is the projection of segment AB onto the y-axis.

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

This is the formula we aimed to demonstrate.

Distance Measured with Polar Coordinates

The distance between two points in the plane can also be measured using the polar coordinates of the points: $$ \overline{AB} = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos (\alpha_2 - \alpha_1)} $$

Here, (r₁, α₁) and (r₂, α₂) are the polar coordinates of the points.

Example

Two points in the plane have the following polar coordinates:

$$ A [ \ 4.47 \ , \ 63.43° \ ] $$

$$ B [ \ 6.08 \ , \ 9.46° \ ] $$

Graphically, this looks like:

The distance between points A and B is:

$$ \overline{AB} = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos (\alpha_2 - \alpha_1)} $$

$$ \overline{AB} = \sqrt{(4.47)^2 + (6.08)^2 - 2 \cdot (4.47) \cdot (6.08) \cos (9.46° - 63.43°)} $$

$$ \overline{AB} = \sqrt{19.98 + 36.97 - 54.36 \cdot \cos (-53.97°)} $$

$$ \overline{AB} = \sqrt{25} = 5 $$

Proof. The distance formula using polar coordinates is derived from the cosine rule in a triangle.

And so on.