Distance on the Coordinate Plane

When the points line up

If two points share a coordinate, they sit on a grid line and no formula is needed.

$$(2, -3) \text{ to } (2, 5): \qquad |{-3} - 5| = 8$$

The $x$ matches, so the points lie on a vertical line, and the distance is the difference in $y$ with the sign stripped by absolute value.

When they do not

Two points sharing neither coordinate are the ends of a hypotenuse.

Draw a horizontal leg and a vertical leg. They meet at a right angle, so the Pythagorean theorem applies:

$$\text{across} = |7 - 1| = 6 \qquad \text{up} = |4 - 1| = 3$$ $$6^2 + 3^2 = c^2 \quad\Rightarrow\quad 36 + 9 = 45$$ $$c = \sqrt{45} \approx 6.7$$

The distance formula is that, written down

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

Compare it with $a^2 + b^2 = c^2$:

Theorem	Formula	Meaning
$a$	$x_2 - x_1$	the horizontal leg
$b$	$y_2 - y_1$	the vertical leg
$c$	$d$	the hypotenuse

The formula is not a new fact. It is the Pythagorean theorem with the legs expressed as coordinate differences, and the square root moved to the front because you want $c$ rather than $c^2$.

Working an example

Find the distance from $(-2, 3)$ to $(4, -5)$.

$$x_2 - x_1 = 4 - (-2) = 6 \qquad y_2 - y_1 = -5 - 3 = -8$$ $$d = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

The $-8$ squared to $+64$, so the negative never reached the answer.

Polygons on a grid

A polygon’s vertices are coordinates, so its perimeter is a sum of distances.

Two sides lie along grid lines, so they are subtractions:

$$\text{base} = |6 - 0| = 6 \qquad \text{right side} = |8 - 0| = 8$$

The third is slanted, so it needs the formula:

$$d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$$ $$\text{perimeter} = 6 + 8 + 10 = 24$$

The area is easier, because the right angle means the two grid-line sides are the base and height:

$$A = \tfrac{1}{2} \times 6 \times 8 = 24 \text{ square units}$$

That the perimeter and area both come to $24$ is a coincidence of this triangle, not a rule — one is a length and the other an area.

Why the coordinate plane makes distance computable

Before coordinates, finding a distance meant measuring it. With coordinates it becomes arithmetic on two pairs of numbers, and that changes what is possible.

A computer has no ruler. Every distance it calculates — how far a character is from a wall, whether two circles overlap, which of a thousand points is nearest — runs through this formula. The screen has coordinates, so the geometry becomes subtraction, squaring and a square root.

It also runs in three dimensions, with one more term under the root:

$$d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$$

The reason is the same. The third axis is perpendicular to the other two, so the theorem applies again.

Worked examples

Common mistakes

Practice problems