The distance between two points

A right triangle hides in every pair of points

Two points $A(x_1, y_1)$ and $B(x_2, y_2)$ are on a slant. How far apart are they?

Draw a horizontal line from $A$ and a vertical line down from $B$. They meet at a right angle, and we have a right triangle with $AB$ as the hypotenuse.

• The horizontal side has length $|x_2 - x_1|$.
• The vertical side has length $|y_2 - y_1|$.

By the Pythagoras theorem,

The distance formula

You may choose either point as the first one. The squares make every number positive, so the answer is the same.

The distance from the origin

When one point is the origin $(0, 0)$, the formula is shorter:

Example: the point $(3, 4)$ is $\sqrt{9 + 16} = \sqrt{25} = 5$ units from the origin.

Worked example

Find the distance between $A(1, 2)$ and $B(5, 5)$.

1. The horizontal step: $5 - 1 = 4$.
2. The vertical step: $5 - 2 = 3$.
3. $d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ units.

This is the well-known $3, 4, 5$ right triangle.

With negative coordinates. Find the distance between $(-2, 3)$ and $(3, -9)$. The steps are $3 - (-2) = 5$ and $-9 - 3 = -12$. The sign does not matter, because we square it. $d = \sqrt{25 + 144} = \sqrt{169} = 13$ units. This is the triple $5, 12, 13$.

Make the two points share a $y$. The vertical step vanishes, and the distance is simply the horizontal step.

Using the formula

• Equal distances. To show that a point is the same distance from two others, find the two distances and compare them. Compare the squares to avoid square roots.
• Finding an unknown. If $d$ is given, square both sides, solve for the unknown, and check every answer. A square can give two values.

Example. Finding an unknown. The distance between $(2, -3)$ and $(k, 1)$ is $5$. Find $k$.

So $k = 5$ or $k = -1$. There are two such points, one on each side.

Example. A point on the x-axis, equally far from two points. Find the point $(x, 0)$ that is equidistant from $(2, -5)$ and $(-2, 9)$.

Expanding: $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$. The $x^2$ and $4$ cancel, and $-8x = 56$, so $x = -7$. The point is $(-7, 0)$.

Where is the formula used?

On a map with a grid, the distance between two towns follows from their coordinates. The satellite of a phone uses the same idea in three dimensions, to find how far you are from each satellite. A game uses it to find whether two objects are touching.

A common mistake

Do not add the coordinates. The distance between $(1, 2)$ and $(4, 6)$ is not $\sqrt{(4 + 1)^2 + (6 + 2)^2}$. We take the difference of the coordinates: $\sqrt{3^2 + 4^2} = 5$ units.

Practise with new numbers

Try it in Python

This program finds the distance between two points.

Remember these

• $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, from the Pythagoras theorem.
• The distance from the origin is $\sqrt{x^2 + y^2}$.
• Take the difference of the coordinates, never the sum.
• Compare squares to avoid square roots, and check the answers.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the distance of $(6, 8)$ from the origin?

2. What is the distance between $(2, 3)$ and $(2, 9)$?

3. For which numbers $y$ is the point $(0, y)$ at a distance of $5$ units from the origin?

4. Is $(1, 5)$ closer to $(0, 0)$ or to $(4, 4)$?