Finding the Distance Between Two Points Using the Pythagorean Theorem - 8th Grade Math

Section 1

Where Did This Idea Come From?

Have you ever tried to figure out how far apart two places are on a map? If the places are on the same street, you can just count blocks. But what if one place is north and east of another? You need a shortcut — and that shortcut was discovered thousands of years ago.

The Pythagorean Theorem is one of the most famous ideas in all of mathematics. It connects the sides of a right triangle in a beautifully simple way, and people have been using it for everything from building pyramids to programming video games.

~1800 BCE

Ancient Babylonian clay tablets show lists of numbers that follow the Pythagorean relationship. This means people understood the pattern more than a thousand years before Pythagoras was born!

~570–495 BCE

Pythagoras of Samos, a Greek mathematician, is credited with proving why the relationship always works for every right triangle — not just specific examples.

~300 BCE

Euclid included a careful proof of the theorem in his book Elements, which became the most influential math textbook in history.

1637 CE

René Descartes invented the coordinate plane (the x-y grid you use in class). This made it possible to combine the Pythagorean Theorem with coordinates to find the distance between any two points.

Here's the big question Descartes helped us answer: If you know the coordinates of two points, how do you find the exact distance between them? That's what this lesson is all about.

Section 2

Core Principles & Definitions

Before we jump into finding distances, let's make sure you're comfortable with a few key ideas. Each one is a building block you'll need.

Right Triangle

A triangle with exactly one 90° angle (a perfect corner). The longest side — across from the right angle — is called the hypotenuse.

Pythagorean Theorem

In a right triangle with legs a and b and hypotenuse c: a² + b² = c². The squares of the two short sides add up to the square of the long side.

Coordinate Plane

A flat grid made by a horizontal number line (x-axis) and a vertical number line (y-axis). Every point is described by an ordered pair like (3, 5).

Distance

The length of the straight line segment connecting two points. It's always positive and measured in the same units as the axes.

★

✦ Key Takeaway

Think of the coordinate plane like a city grid. If you want to walk from one corner to another, you could go along the streets (horizontal then vertical). But the shortest path is a diagonal — a straight line cutting across. The Pythagorean Theorem tells you exactly how long that diagonal is, using just the horizontal and vertical distances.

Section 3

Seeing It on the Coordinate Plane

Let's place two points on a coordinate plane and see how a right triangle appears. Look at points A (1, 2) and B (5, 5). If you draw a horizontal line from A and a vertical line from B, they meet at a corner — and that corner forms a right angle!

The right triangle formed between two coordinate points

Notice what happened. The horizontal leg measures how far apart the two points are left-to-right (the difference in their x-coordinates: 5 − 1 = 4). The vertical leg measures how far apart they are up-and-down (the difference in their y-coordinates: 5 − 2 = 3). The hypotenuse is the actual straight-line distance between the points — and that's what we want to find.

Since we have a right triangle, we can use the Pythagorean Theorem: 4² + 3² = d². That gives us 16 + 9 = 25, so d = 5. The distance between A and B is 5 units.

Section 4

The Formulas You Need

Let's turn what we just saw into a formula that works for any two points. Say you have Point 1 at (x₁, y₁) and Point 2 at (x₂, y₂).

The Pythagorean Theorem

a² + b² = c²

Where a and b are the legs, and c is the hypotenuse.

On the coordinate plane, the two legs of the right triangle are:

Horizontal Distance (Leg a)

a = x₂ − x₁

The difference in the x-coordinates (how far apart they are left-to-right).

Vertical Distance (Leg b)

b = y₂ − y₁

The difference in the y-coordinates (how far apart they are up-and-down).

Now plug those into the Pythagorean Theorem and solve for the distance d (which is the hypotenuse):

The Distance Formula

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Take the square root of the sum of the squared differences. This is the Pythagorean Theorem in coordinate form!

Don't let the square root scare you. All this formula does is follow three steps: (1) find the horizontal and vertical distances, (2) square each one and add them together, (3) take the square root of that sum. You already did this in Section 3 — we're just writing it in a general way now.

★

✦ Key Takeaway

The distance formula isn't something new to memorize — it is the Pythagorean Theorem, just wearing a coordinate-plane outfit. Every time you use it, you're really building a right triangle and finding the hypotenuse. If you ever forget the formula, just sketch the triangle!

Section 5

Step-by-Step Breakdown

Let's look at the process you'll follow every time. The diagram below shows the steps visually with a new pair of points.

Distance between (−2, 1) and (4, 4) with negative coordinates

Here's a quick reference for the three steps you'll repeat every time:

Step	What You Do	Example: (−2, 1) to (4, 4)
1	Subtract the x-coordinates and subtract the y-coordinates	`4 − (−2) = 6 \| 4 − 1 = 3`
2	Square each result and add them together	`6² + 3² = 36 + 9 = 45`
3	Take the square root	`√45 = 3√5 ≈ 6.71`

Notice that even though one x-coordinate was negative, the method still works perfectly. When you subtract a negative number, it becomes addition: 4 − (−2) = 4 + 2 = 6. The squaring step also helps because it makes everything positive anyway!

Section 6

Worked Example

Let's work through a full problem together, nice and slow.

ℹ

Problem

Find the distance between the points (−3, 2) and (5, −4). Give an exact answer and a decimal rounded to the nearest hundredth.

Worked Example

1
Step 1 — Identify your coordinates

Label the two points. It doesn't matter which one is "Point 1" and which is "Point 2" — the answer comes out the same.

(x₁, y₁) = (−3, 2) (x₂, y₂) = (5, −4)
2
Step 2 — Find the horizontal distance

Subtract the x-coordinates: x₂ − x₁.

5 − (−3) = 5 + 3 = 8
3
Step 3 — Find the vertical distance

Subtract the y-coordinates: y₂ − y₁. It's negative, but that's fine — we'll square it next, which makes it positive.

−4 − 2 = −6
4
Step 4 — Square each distance and add

Square each difference and add them together.

8² + (−6)² = 64 + 36 = 100
5
Step 5 — Take the square root

Take the square root to find the distance.

d = √100 = 10
6
Final Answer

The distance between (−3, 2) and (5, −4) is exactly 10 units. That's already a whole number, so the decimal answer is also 10.00. Here's a nice check: we got a perfect answer because 8, 6, 10 form a Pythagorean triple (it's just the 3-4-5 triple multiplied by 2). Not every problem gives a whole number answer — many will give you a square root like √45 — and that's totally normal.

Section 7

Strengths, Limitations & Common Mistakes

The distance formula is powerful, but there are a few things to watch out for. Let's compare situations where it shines and where students often slip up.

Situation	What Works	Watch Out For
Both coordinates are positive	Straightforward subtraction	Don't accidentally add the coordinates instead of subtracting
One or both coordinates are negative	Formula still works perfectly	Subtracting a negative: 3 − (−5) = 8, not −2
Points on the same horizontal or vertical line	One leg is zero, so the distance is just the other leg	You don't really need the full formula here — just subtract!
Answer is not a whole number	Leave it as a simplified square root (exact) or use a calculator (decimal)	Don't round too early — simplify the square root first
Forgetting to square before adding	—	The most common mistake! You must square each difference before adding

★

✦ Key Takeaway

Think of the squaring step like packing each distance into a box before combining them. You can't just throw two distances together — you have to "square-pack" them first, add the packed values, then "unpack" by taking the square root. Skipping the packing step gives you the wrong answer every time.

Section 8

Where Does This Lead?

The distance formula you just learned is your first step into a much bigger world. Here's a peek at how this idea grows as you move through math.

What You Know Now	What Comes Next
Distance between two points on a flat (2D) coordinate plane	Distance in 3D space (add a z-coordinate): d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
Building a right triangle from two points	The midpoint formula finds the exact center of the hypotenuse
Finding one distance at a time	Using distance to write the equation of a circle (all points at the same distance from the center)
Working with specific number pairs	Using distance in proofs to show that shapes are congruent, isosceles, or right-angled

In high school geometry and algebra, you'll use the distance formula constantly — to prove triangles are congruent, to find the radius of a circle, and even to analyze real-world data. Every time you use it, remember: it's just the Pythagorean Theorem in disguise.

Section 9

Practice Problems

Try these on your own! Work through each one step by step, then click "Show Answer" to check. Remember: subtract, square, add, square root.

Question 1: Problem 1 — Conceptual

When you use the distance formula, you're really finding the length of which part of a right triangle?

Show answer

You're finding the length of the hypotenuse — the longest side, which is across from the right angle. The hypotenuse is the straight-line distance between the two points. The legs of the triangle are the horizontal and vertical distances. Great start if you got this — it means you understand the big idea behind the formula!

Question 2: Problem 2 — Basic Calculation

Find the distance between the points (2, 3) and (6, 6).

Show answer

Horizontal distance: 6 − 2 = 4 Vertical distance: 6 − 3 = 3 Apply the formula: d = √(4² + 3²) = √(16 + 9) = √25 = 5 The distance is 5 units. This is the classic 3-4-5 right triangle! Nice work if you recognized that.

Question 3: Problem 3 — Intermediate

Find the distance between (−1, 4) and (3, −2). Give your answer as a simplified square root.

Show answer

Horizontal distance: 3 − (−1) = 3 + 1 = 4 Vertical distance: −2 − 4 = −6 Apply the formula: d = √(4² + (−6)²) = √(16 + 36) = √52 Simplify: √52 = √(4 × 13) = 2√13, which is about 7.21 units. Remember, when you simplify a square root, look for perfect square factors (like 4) hiding inside.

Question 4: Problem 4 — Applied

On a treasure map, the "Start" is at coordinates (1, 1) and the treasure is buried at (7, 9). Each unit on the map represents 10 meters. How far is the treasure from the starting point in meters?

Show answer

Horizontal distance: 7 − 1 = 6 Vertical distance: 9 − 1 = 8 Apply the formula: d = √(6² + 8²) = √(36 + 64) = √100 = 10 units Convert to meters: 10 units × 10 meters per unit = 100 meters The treasure is 100 meters from the start. Notice that 6-8-10 is another Pythagorean triple (it's the 3-4-5 triple doubled). Always finish by answering the question in the right units!

Question 5: Problem 5 — Challenge

Triangle PQR has vertices at P(0, 0), Q(6, 0), and R(3, 4). Find the length of all three sides. Then use your results to decide: is this triangle equilateral (all sides equal), isosceles (two sides equal), or scalene (no sides equal)?

Show answer

Side PQ: Both points have y = 0, so PQ = 6 − 0 = 6 Side PR: d = √((3−0)² + (4−0)²) = √(9 + 16) = √25 = 5 Side QR: d = √((3−6)² + (4−0)²) = √(9 + 16) = √25 = 5 The three sides are 6, 5, and 5. Two sides are equal (PR = QR = 5), so this is an isosceles triangle. Awesome job if you figured this out! You used the distance formula three times and then compared the results. That's exactly how mathematicians classify triangles using coordinates.

Summary

Lesson Summary

Summary

The Pythagorean Theorem — the rule that a² + b² = c² in a right triangle — can be applied directly on a coordinate plane to find the distance between any two points. You do this by treating the horizontal gap (x₂ − x₁) and the vertical gap (y₂ − y₁) as the two legs of a right triangle. The distance formula — d = √[(x₂ − x₁)² + (y₂ − y₁)²] — is simply the Pythagorean Theorem rewritten in coordinate form.

The three-step process is the same every time: subtract the coordinates to get each leg, square and add those values, then take the square root. This works whether your coordinates are positive, negative, or zero. If the answer isn't a perfect square, simplify the radical by factoring out perfect squares. This foundational skill connects to circles, triangle classification, 3D distance, and many topics you'll see in high school geometry.