The Coordinate Plane and the Distance Formula

11 min read

Plot two dots on graph paper and ask how far apart they are. You could walk across, then up, counting squares the whole way. But the straight line between the dots is shorter than that walk, and you can find its length without ever stretching a ruler along it. The trick is a right triangle hiding between the two points, and the rule that measures it is one you already know.

By the end of this page you can:

• Find the straight-line distance between two points from their coordinates.
• Find the midpoint of a segment by averaging the coordinates.
• Explain why this distance is the Pythagorean theorem in disguise.

The floor. At minimum, compute $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for two given points. Every later idea, including the unit circle, leans on this one calculation.

Quick Reference

Stewart 7e Section 1.9 establishes the rectangular coordinate plane, the distance formula, and the midpoint formula, all derived from the geometry of the coordinate grid. The OpenStax alternate is free and openly licensed.

Before You Start

You are ready for this page once you can do the following.

• State the Pythagorean theorem $a^2 + b^2 = c^2$ and identify the hypotenuse.
• Plot an ordered pair $(x, y)$ and name its quadrant.
• Square a number, take a square root, and subtract negatives.

A quick self-check on the one strong prerequisite:

The Idea

To measure the straight-line gap between two points, build a right triangle. The horizontal gap between them is one leg, the vertical gap is the other, and the distance you want is the hypotenuse. The horizontal gap is the difference of the $x$-coordinates, and the vertical gap is the difference of the $y$-coordinates. Feed those two legs into the Pythagorean theorem and out comes the distance.

There is no new idea here, only a new way to write the legs. If you can use $a^2 + b^2 = c^2$, you can already find the distance between any two points. Run the same picture in reverse and you get the midpoint: instead of measuring the gap, stand exactly halfway across and halfway up, which is the average of the two positions.

Hold one image as you read: a right triangle whose slanted side connects the two points. Distance reads that slanted side. Midpoint marks its center.

Prerequisite Hub

Coordinate-plane distance rests on one prerequisite and supports one later skill.

Builds On

Unlocks

graph LR
    A["Pythagorean<br/>Theorem"]
    B["Coordinate Plane<br/>and Distance Formula"]
    C["Unit Circle:<br/>Sine and Cosine"]

    A --> B
    B --> C

    style B fill:#d1fae5,stroke:#a565f0,stroke-width:3px

    click A "pythagorean-theorem.html"
    click B "coordinate-plane-distance.html"
    click C "unit-circle-sine-cosine.html"

No cross-course prerequisites are required for this node.

Quick Reference

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

The midpoint formula. The midpoint of the segment between $(x_1, y_1)$ and $(x_2, y_2)$ is $$M = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right).$$

The reading. Distance subtracts the coordinates to measure a gap. Midpoint adds the coordinates to average two positions.

Official Definitions

Distance Formula

Given endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the distance between them is $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ It is derived from the Pythagorean Theorem.

Source: OpenStax Algebra and Trigonometry 2e, Section 2.1 The Rectangular Coordinate Systems and Graphs.

Midpoint Formula

The midpoint of the segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $$\left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right).$$

Source: OpenStax Algebra and Trigonometry 2e, Section 2.1 The Rectangular Coordinate Systems and Graphs.

Both formulas come from the same right triangle. The distance formula reads its hypotenuse. The midpoint formula marks the center of its slanted side.

Where the formula comes from

Draw the right triangle whose slanted side joins $(x_1, y_1)$ and $(x_2, y_2)$. The horizontal leg has length $|x_2 - x_1|$, and the vertical leg has length $|y_2 - y_1|$. The Pythagorean theorem gives $d^2 = (\text{horizontal leg})^2 + (\text{vertical leg})^2$, so $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ The squares remove any need for absolute value signs, since squaring a negative gives a positive. The squares are also why the order of the two points never changes the answer: $(x_2 - x_1)^2$ and $(x_1 - x_2)^2$ are the same number.

Worked Examples

Each example uses predict-then-check: estimate the size or sign of the answer first, then compute, then compare.

Example 1: A clean distance

Find the distance between $(1, 2)$ and $(4, 6)$.

Predict. The horizontal gap is $3$ and the vertical gap is $4$. The distance must beat each single leg but fall short of their sum $3 + 4 = 7$. Estimate around $5$.

Compute. $$d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$

Check. The result $5$ sits between $4$ and $7$, matching the prediction. This is the $3$-$4$-$5$ right triangle.

Square each difference before adding. A tempting wrong move is $\sqrt{(4 - 1) + (6 - 2)} = \sqrt{7}$, dropping the squares. The squares are what make this the Pythagorean theorem. Without them, the result is not a distance.

Example 2: Order of the points does not matter

Find the distance between $(5, 7)$ and $(2, 3)$, taking the points in the opposite order from how they are listed.

Predict. The gaps are $3$ across and $4$ up again, so the answer is $5$ regardless of which point comes first.

Compute. Take $(2, 3)$ first and $(5, 7)$ second: $$d = \sqrt{(2 - 5)^2 + (3 - 7)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$

Check. The negative differences squared to the same positive values $9$ and $16$, so the distance is $5$ either way. Order is irrelevant because each difference is squared.

Example 3: The midpoint averages the coordinates

Find the midpoint of the segment joining $(2, 3)$ and $(8, 7)$.

Predict. The midpoint sits halfway between the points, so its $x$-coordinate is halfway between $2$ and $8$ (around $5$), and its $y$-coordinate is halfway between $3$ and $7$ (around $5$).

Compute. $$M = \left( \frac{2 + 8}{2}, \; \frac{3 + 7}{2} \right) = \left( \frac{10}{2}, \; \frac{10}{2} \right) = (5, 5).$$

Check. The point $(5, 5)$ lies between the endpoints in both coordinates, matching the prediction.

Midpoint adds; distance subtracts. Do not carry the subtraction from the distance formula into the midpoint. The midpoint is an average, so it uses sums divided by two.

Example 4: Distance with negative coordinates

Find the distance between $(-3, 2)$ and $(1, -1)$.

Predict. Moving from $x = -3$ to $x = 1$ is a horizontal gap of $4$. Moving from $y = 2$ to $y = -1$ is a vertical gap of $3$. The answer is near $5$ again.

Compute. $$d = \sqrt{(1 - (-3))^2 + (-1 - 2)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.$$

Check. Subtracting a negative widened the horizontal gap correctly to $4$, and the distance came out to $5$. The negative inside the second difference squared away.

Example 5: Building a distance condition (toward the unit circle)

Verify that the point $\left( \dfrac{3}{5}, \dfrac{4}{5} \right)$ is at distance $1$ from the origin.

Predict. Both coordinates are fractions less than $1$, yet a point can still sit a full unit away if it leans into both directions. The $3$-$4$-$5$ triangle scaled down by $5$ should land exactly on distance $1$.

Compute. $$d = \sqrt{\left( \frac{3}{5} - 0 \right)^2 + \left( \frac{4}{5} - 0 \right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1.$$

Check. The distance is exactly $1$, so this point sits one unit from the origin. The set of all such points is the unit circle, where sine and cosine come from in the next skill.

Two valid paths. The distance computation above is one way. Another is to check whether the coordinates satisfy $x^2 + y^2 = 1$: $\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1$. The two checks are the same statement, since squaring “distance equals $1$” gives $x^2 + y^2 = 1$.

Common Misconceptions

Leveled Practice

Level 1: Direct Application

Problem 1. Find the distance between $(0, 0)$ and $(3, 4)$.

Problem 2. Find the midpoint of the segment joining $(2, 6)$ and $(8, 2)$.

Level 2: Multiple Steps

Problem 3. Find the distance between $(-3, 2)$ and $(1, -1)$.

Problem 4. Find both the distance between $(-1, -2)$ and $(2, 2)$ and the midpoint of that segment.

Level 3: Deeper Problems

Problem 5. Show that the point $\left( \dfrac{5}{13}, \dfrac{12}{13} \right)$ is at distance $1$ from the origin.

Problem 6. Two points are $(1, 2)$ and $(4, 6)$. Find the midpoint of the segment joining them, then find the distance from the midpoint to $(1, 2)$.

Level 4: Reasoning and Justification

Problem 7. A classmate claims that the points $(0, 0)$, $(4, 0)$, and $(0, 3)$ form a right triangle whose longest side has length $5$. Decide whether the claim is correct, and justify your decision using distances.

Level 5: Extension and Generalization

Problem 8. A point $P = (x, y)$ is equally distant from $(0, 0)$ and from $(6, 0)$. Find an equation that all such points $P$ satisfy, and describe the set of points in words.

Check Yourself

Go Deeper (optional)

Mastery Checklist

Novice (Level 1-2):

• I can find the distance between two points, squaring each difference before adding.
• I can find the midpoint of a segment by averaging the coordinates.
• I can explain why the order of the two points does not change the distance.

Competent (Level 3-4):

• I can recognize the distance formula as the Pythagorean theorem written in coordinates.
• I can verify that a point is at a given distance from another point, including distance $1$ from the origin.
• I can justify whether three points form a right triangle using distances.

Proficient (Level 5):

• I can set two distances equal and simplify the result into the equation of a set of points.
• I can connect the distance condition “stay one unit from the origin” to the unit circle $x^2 + y^2 = 1$.

Connections

Looking back:

• The Pythagorean Theorem and Right Triangles supplies the $a^2 + b^2 = c^2$ relation that becomes the distance formula once the legs are written as coordinate differences.

Looking ahead:

• The Unit Circle: Sine and Cosine: trigonometry is built on the set of points at distance $1$ from the origin, where a point has coordinates $(\cos\theta, \sin\theta)$.

Last updated: 2026-06-24