Distance Formula

The distance formula uses horizontal and vertical change with the Pythagorean theorem to measure straight-line separation between coordinates.

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A segment’s horizontal and vertical changes form a right triangle for distance and midpoint reasoning. Two coordinate points are joined by a segment with dashed rise and run legs. The midpoint is marked halfway along the segment, connecting coordinate averages with the Pythagorean distance calculation.

The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

Built from a right triangle

Horizontal change $\Delta x=x_2-x_1$ and vertical change $\Delta y=y_2-y_1$ form perpendicular legs of a right triangle. The segment joining the points is its hypotenuse, so

d^2=(\Delta x)^2+(\Delta y)^2.

Taking the positive square root gives distance because length cannot be negative.

A reliable method

Label the two points consistently.
Calculate horizontal and vertical differences.
Square each difference.
Add.
Take the principal square root.
State exact or approximate form with units.

Because differences are squared, reversing both point labels gives the same distance.

Worked example

This is a scaled $3$–$4$–$5$ right triangle.

Exact and approximate form

For points $(1,2)$ and $(5,7)$, distance is

\sqrt{4^2+5^2}=\sqrt{41}.

$\sqrt{41}$ is exact; $6.40$ is an approximation to two decimal places. Keep the radical during work and round only when the context requests a decimal.

Common mistakes

Forgetting brackets around negative coordinates. Write $4-(-2)$ explicitly.

Adding differences before squaring. Square horizontal and vertical changes separately.

Stopping at $d^2$. Take the square root for distance.

Rounding too early. Preserve exact values until the final step.

Quick self-check

Are my coordinate differences paired correctly?
Did I square each complete difference?
Is my answer nonnegative and at least as large as either leg?
Are units and required precision included?