Distance Between Two Points

Deriving the distance formula

Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$. Their horizontal separation has length $|x_2-x_1|$ and their vertical separation has length $|y_2-y_1|$. By the Pythagorean theorem,

Distance is non-negative, so take the principal square root:

The absolute-value signs are unnecessary inside the squares because a number and its negative have the same square. The derivation, rather than the appearance of the diagram, is what justifies the formula. See the OpenStax distance-formula treatment.

Worked example: a 6–8–10 triangle

Find the distance from $A=(-2,1)$ to $B=(4,9)$.

Calculate both differences

Apply the formula

The distance is 10 coordinate units. If each unit represents a metre, it is 10 m. It is not $6+8=14$ units: that is the length of a route following the horizontal and vertical legs.

The answer is reasonable because the diagonal must be longer than either leg, but shorter than travelling both legs consecutively when both legs are non-zero.

An exact answer need not be an integer

For $P=(3,2)$ and $Q=(5,-1)$,

The exact distance is $\sqrt{13}$ units. The decimal is an approximation to three decimal places. Keep the radical during later algebra if an exact answer is needed; early rounding can change subsequent results.

The negative vertical change does not make the distance negative. It records direction before squaring, whereas distance records length.

Horizontal, vertical and identical points

If the points have the same $y$-coordinate, the vertical difference is zero, so the formula reduces to $d=|x_2-x_1|$. The distance from $(-3,4)$ to $(5,4)$ is therefore 8 units.

If the points have the same $x$-coordinate, use $d=|y_2-y_1|$. For identical points, both differences are zero and the distance is zero. No special exception to the formula is needed.

Changing the endpoint order leaves the result unchanged:

Both coordinate differences reverse sign together, but their squares do not change. This also provides a useful check on a calculation involving negative coordinates.

Units and scale

Suppose a rectangular site plan uses 2 m per horizontal coordinate unit and 3 m per vertical coordinate unit. A change of $(3,4)$ in the drawing represents physical changes of 6 m and 12 m. The physical separation is

Calculating $\sqrt{3^2+4^2}=5$ first and multiplying by a single scale factor would be wrong because no single scale factor applies to both directions.

Likewise, coordinates such as “seconds” and “degrees Celsius” should not be combined under this formula to claim a physical length. The coordinate axes must describe compatible spatial quantities.

Use squared distance when only a comparison is needed

Square roots preserve order for non-negative inputs. To decide which of two points is closer to the origin, compare their squared distances.

For $P=(2,3)$ and $Q=(4,0)$,

Since $13<16$, $P$ is closer. There is no need to approximate either square root. This same idea makes circle membership tests efficient: compare the squared distance from the centre with $r^2$.

Avoid mixing directions and lengths

Subtract corresponding coordinates: an $x$-coordinate from an $x$-coordinate, and a $y$-coordinate from a $y$-coordinate. Write parentheses around negative values. In particular, $4-(-2)=6$, not 2.

Square each complete difference before adding. The expression $(x_2-x_1)^2$ is not generally $x_2^2-x_1^2$. Finally, take one square root of the sum; adding the two separate square roots computes a different length.

Related topics

Midpoint Formula locates the centre of a segment. Circles in Coordinate Geometry uses a fixed distance to define an entire curve.