The Distance & Midpoint Formulas on the Coordinate Plane

Two points on a coordinate plane raise two natural questions: how far apart are they, and what point sits exactly halfway between them? The distance formula answers the first: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . The midpoint formula answers the second: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ .

Neither formula is new math. The distance formula is the Pythagorean theorem wearing coordinates, and the midpoint formula is nothing more than averaging. Once you see where each one comes from, there's very little left to memorize.

The distance formula is the Pythagorean theorem

Take the segment below from $(1, 2)$ to $(4, 6)$ . Slide right from one endpoint and up to the other, and the dashed legs form a right triangle: a horizontal leg of $4 - 1 = 3$ and a vertical leg of $6 - 2 = 4$ . The segment itself is the hypotenuse, so by the Pythagorean theorem its length is $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$ .

That's all the distance formula says: $x_2 - x_1$ is the horizontal leg, $y_2 - y_1$ is the vertical leg, and the square root puts the hypotenuse back together. It doesn't matter which point you call $(x_1, y_1)$ — the differences get squared, so any negative signs wash out.

The midpoint formula is an average

The point halfway between two numbers is their average. The midpoint of a segment does that twice — once for the $x$ -coordinates and once for the $y$ -coordinates: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ .

The most common midpoint mistake is subtracting instead of adding. Subtraction belongs to the distance formula; the midpoint adds the coordinates and halves them. If your midpoint doesn't land between the two endpoints, check that first.

In the figure below, $M$ sits exactly halfway along the segment — its coordinates are the averages of the two endpoints'.

Answers that aren't whole numbers

Distances often come out as square roots — that's normal. If the legs are $5$ and $5$ , the distance is $\sqrt{50}$ , which simplifies to $5\sqrt{2}$ . Leave it as a simplified radical unless the problem asks for a decimal.

Midpoints can have fraction coordinates for the same reason averages can: the midpoint of $(0, 0)$ and $(5, 8)$ is $\left(\dfrac{5}{2}, 4\right)$ . A fraction there is a correct answer, not a mistake.

Worked examples

Example 1: a whole-number distance

Find the distance between $(2, 3)$ and $(5, 7)$ .

Write the distance formula

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

Substitute the coordinates

d = \sqrt{(5 - 2)^2 + (7 - 3)^2}

Simplify inside the radical

d = \sqrt{9 + 16} = \sqrt{25}

Take the square root

d = 5

Answer: $d = 5$

Example 2: a radical answer

Find the exact distance between $(2, 1)$ and $(7, 6)$ .

Substitute into the distance formula

d = \sqrt{(7 - 2)^2 + (6 - 1)^2}

Simplify inside

d = \sqrt{25 + 25} = \sqrt{50}

Simplify the radical:

50 = 25 \cdot 2

d = 5\sqrt{2}

Answer: $d = 5\sqrt{2} \approx 7.07$

Example 3: finding a midpoint

Find the midpoint of the segment with endpoints $(-2, 5)$ and $(6, 1)$ .

Write the midpoint formula

M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)

Substitute — add, don't subtract

M = \left(\dfrac{-2 + 6}{2}, \dfrac{5 + 1}{2}\right)

Simplify each coordinate

M = (2, 3)

Answer: $M = (2, 3)$

Example 4: finding a missing endpoint

The midpoint of a segment is $(3, 2)$ , and one endpoint is $(1, 6)$ . Find the other endpoint.

Set up the

x

-coordinate equation

\dfrac{1 + x}{2} = 3

Multiply by

2

, then subtract

1

1 + x = 6, \quad x = 5

Set up the

y

-coordinate equation

\dfrac{6 + y}{2} = 2

Multiply by

2

, then subtract

6

6 + y = 4, \quad y = -2

Check: the midpoint of

(1, 6)

and

(5, -2)

\left(\dfrac{6}{2}, \dfrac{4}{2}\right) = (3, 2)

✓

Answer: $(5, -2)$

Try one yourself

Common questions

Does it matter which point I call

(x_1, y_1)

No. In the distance formula the differences get squared, so swapping the points flips the signs but not the answer. In the midpoint formula you're adding, and addition works in either order. Pick whichever assignment keeps the arithmetic comfortable.

How are the distance formula and the Pythagorean theorem related?

They're the same statement. The horizontal change $x_2 - x_1$ and the vertical change $y_2 - y_1$ are the legs of a right triangle, and the segment between the points is the hypotenuse. The distance formula is just $a^2 + b^2 = c^2$ solved for $c$ with coordinates plugged in.

How do I keep the two formulas from getting mixed up?

Match the operation to the job. Distance measures a gap, and gaps come from subtracting; the midpoint is an average, and averages come from adding and dividing by $2$ . If you catch yourself subtracting inside a midpoint, stop and re-set-up.

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