Do I always need to use the distance formula, or can I just use the Pythagorean theorem?

The distance formula is really just the Pythagorean theorem written in a specific way for coordinate points. If you understand the Pythagorean theorem, you can always draw a right triangle on the coordinate plane and solve it step by step. The formula is faster once you're comfortable with it, but either method works.

What does 'Delta' (Δ) mean in Δx and Δy?

Delta is a symbol that means 'change in'. So Δx means the change in x-coordinate (how far apart the points are horizontally) and Δy means the change in y-coordinate (how far apart they are vertically). It's just a shorthand way of saying 'the difference between the coordinates'.

In a 3-D problem, how do I know which dimensions to use first?

You can pick any two dimensions to find the first diagonal. Once you have that diagonal, use it with the third dimension to find the space diagonal. The order doesn't matter—you'll get the same answer. A helpful tip: pick two dimensions that form a face you can visualize easily (like the bottom or front of the box).

Can the distance between two points be negative?

No, distance is always positive. Even if you subtract the coordinates 'backwards' (like x_1 - x_2 instead of x_2 - x_1), when you square the result the answer is still positive. Then when you take the square root, you get a positive distance. This is why the distance formula works no matter which point you call 'point 1' and which you call 'point 2'.

M8MATH-9.4

Pythagorean Applications & Distance Between Points

Apply the Pythagorean theorem to solve real-world problems in 2-D and 3-D, and find distances between points on a coordinate plane.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Pythagorean Applications & Distance Between Points, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Pythagorean Applications & Distance Between Points free

What this lesson covers

The Pythagorean theorem is one of the most useful tools in mathematics because it shows up everywhere—from hanging a ladder safely against a wall, to figuring out how far apart two cities are on a map, to designing the inside of a box. In this lesson, you'll learn how to use the Pythagorean theorem to solve practical problems in two dimensions (like finding a TV screen diagonal or the distance between two points), and even in three dimensions (like finding how far a corner of a box is from the opposite corner). By the end, you'll see that once you can spot the right triangle hidden in a problem, the Pythagorean theorem does the heavy lifting.

Applying the Pythagorean Theorem to 2-D Real-World Problems

Many everyday situations hide right triangles inside them. When a ladder leans against a wall, the ladder forms the hypotenuse, and the wall and ground form the two legs. If you know two of these measurements, you can always find the third using

a^2 + b^2 = c^2

.

Example: A ladder is 10 feet long and rests 6 feet away from the base of a wall. How high up the wall does it reach? Here, the ladder is the hypotenuse (

c = 10

) and the ground distance is one leg (

a = 6

). Solve for

b

6^2 + b^2 = 10^2

36 + b^2 = 100

b^2 = 64

b = 8 \text{ feet}

TV screen diagonals work the same way. If a TV is 24 inches wide and 18 inches tall, the diagonal is found by treating the width and height as the two legs of a right triangle. Always identify which measurement is the hypotenuse (the longest side, opposite the right angle) and which two are the legs.

Finding Distance Between Two Points on the Coordinate Plane

To find the distance between two points on a coordinate plane, imagine drawing a right triangle where the two points form the endpoints of the hypotenuse. The two legs of this triangle are the horizontal and vertical distances between the points.

Let's say you have points

A(2, 3)

and

B(5, 7)

. The horizontal leg is

\Delta x = 5 - 2 = 3

. The vertical leg is

\Delta y = 7 - 3 = 4

. Now use the Pythagorean theorem:

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

d^2 = 3^2 + 4^2 = 9 + 16 = 25

d = 5

This gives you the distance formula:

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

. A common error is forgetting to take the square root at the end—always finish by finding the square root of the sum of the squared differences.

Applying the Pythagorean Theorem to 3-D Problems

Three-dimensional problems often require you to use the Pythagorean theorem twice. Picture a rectangular box (or prism) where you want to find the space diagonal—the straight-line distance from one corner to the opposite corner through the inside.

Start by finding the diagonal of the bottom face using two dimensions. Then use that diagonal as one leg of a new right triangle, where the height of the box is the other leg. Example: A box is 3 units long, 4 units wide, and 5 units tall. First, find the diagonal of the bottom:

d_{\text{bottom}}^2 = 3^2 + 4^2 = 9 + 16 = 25

d_{\text{bottom}} = 5

Now the space diagonal uses that 5 as one leg and the height (5) as the other:

d_{\text{space}}^2 = 5^2 + 5^2 = 25 + 25 = 50

d_{\text{space}} = \sqrt{50} = 5\sqrt{2} \approx 7.07 \text{ units}

The key is breaking the 3-D problem into two 2-D problems, each using the Pythagorean theorem once.

Common Mistakes and How to Avoid Them

A frequent error is confusing which side is the hypotenuse. Remember: the hypotenuse is always the longest side and is opposite the right angle. When you see a problem, sketch it and mark the right angle clearly.

Another mistake is arithmetic errors when squaring or taking square roots. Double-check your algebra: if

c^2 = 25

, then

c = 5

(not 2.5). Always work in exact form (like

5\sqrt{2}

) before approximating with a decimal.

With the distance formula on the coordinate plane, students sometimes subtract in the wrong order. The good news:

(x_2 - x_1)^2

and

(x_1 - x_2)^2

give the same result because you're squaring, so order doesn't matter. However, always subtract the coordinates in the same order for both

x

and

y

to stay organized.

Finally, in 3-D problems, don't try to jump straight to a three-dimensional formula—break it into two right triangles instead. This step-by-step approach is clearer and less error-prone.

Key terms

Pythagorean Theorem.: In a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: $a^2 + b^2 = c^2$ .
Hypotenuse.: The longest side of a right triangle, located opposite the right angle.
Legs.: The two sides of a right triangle that form the right angle.
Distance Formula.: The formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ gives the distance between two points on a coordinate plane.
Space Diagonal.: A line segment connecting two opposite corners of a three-dimensional figure through its interior.
Change in x (Δx).: The horizontal distance between two points, found by subtracting the x-coordinates.
Change in y (Δy).: The vertical distance between two points, found by subtracting the y-coordinates.

Worked example

A 13-meter ladder leans against a building. The top of the ladder reaches 12 meters up the wall. How far is the base of the ladder from the building? (Round to one decimal place if needed.)

First, identify the right triangle. The ladder is the hypotenuse (

c = 13

meters), the height up the wall is one leg (

b = 12

meters), and the distance from the building to the base is the other leg (

a = ?

).

Set up the Pythagorean theorem:

a^2 + b^2 = c^2

a^2 + 12^2 = 13^2

a^2 + 144 = 169

Subtract 144 from both sides:

a^2 = 169 - 144 = 25

Take the square root:

a = \sqrt{25} = 5

The base of the ladder is 5 meters from the building. Notice that this is a 5-12-13 right triangle, which is a Pythagorean triple you may have seen before.

Practice questions

Two points on a coordinate plane are

P(1, 2)

and

Q(4, 6)

. What is the distance between them?

Answer: 5

Use the distance formula:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

. A common mistake is forgetting to take the square root at the end, which would give you 25—but 25 is the square of the distance, not the distance itself.

A rectangular box has dimensions 6 centimeters by 8 centimeters by 10 centimeters. Find the length of the space diagonal (the distance from one corner to the opposite corner through the inside of the box).

Answer: $10\sqrt{2}$ cm or approximately 14.14 cm

First, find the diagonal of the base using the two horizontal dimensions:

d_{base}^2 = 6^2 + 8^2 = 36 + 64 = 100

, so

d_{base} = 10

cm. Now this diagonal becomes one leg of a right triangle with height 10 as the other leg:

d_{space}^2 = 10^2 + 10^2 = 100 + 100 = 200

. Therefore

d_{space} = \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2} \approx 14.14

cm. Breaking the 3-D problem into two 2-D problems makes it much simpler.

A television screen is 40 inches wide and 30 inches tall. What is the diagonal of the TV in inches?

50 inches
45 inches
55 inches
35 inches

Answer: 50 inches

The width and height of the screen form the two legs of a right triangle, and the diagonal is the hypotenuse. Use the Pythagorean theorem:

d^2 = 40^2 + 30^2 = 1600 + 900 = 2500

. Taking the square root:

d = \sqrt{2500} = 50

inches. This is another Pythagorean triple: 30-40-50.

FAQ

Do I always need to use the distance formula, or can I just use the Pythagorean theorem?: The distance formula is really just the Pythagorean theorem written in a specific way for coordinate points. If you understand the Pythagorean theorem, you can always draw a right triangle on the coordinate plane and solve it step by step. The formula is faster once you're comfortable with it, but either method works.
What does 'Delta' (Δ) mean in Δx and Δy?: Delta is a symbol that means 'change in'. So Δx means the change in x-coordinate (how far apart the points are horizontally) and Δy means the change in y-coordinate (how far apart they are vertically). It's just a shorthand way of saying 'the difference between the coordinates'.
In a 3-D problem, how do I know which dimensions to use first?: You can pick any two dimensions to find the first diagonal. Once you have that diagonal, use it with the third dimension to find the space diagonal. The order doesn't matter—you'll get the same answer. A helpful tip: pick two dimensions that form a face you can visualize easily (like the bottom or front of the box).
Can the distance between two points be negative?: No, distance is always positive. Even if you subtract the coordinates 'backwards' (like $x_1 - x_2$ instead of $x_2 - x_1$ ), when you square the result the answer is still positive. Then when you take the square root, you get a positive distance. This is why the distance formula works no matter which point you call 'point 1' and which you call 'point 2'.

Learn this with a teacher, not a page

The Crimsora tutor teaches Pythagorean Applications & Distance Between Points live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free