Enter the coordinates of point 1 and point 2 to find the distance between them. The slope and the equation of the line through the two points are shown too.
Enter the coordinates as numbers (decimals and negative numbers are OK). All four fields are required.
Result and graph
Enter the coordinates of two points in the fields on the left and press "Calculate". The result will appear here.
Table of Contents
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What you can do on this page
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What is this calculation used for?
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How to Use
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Formulas and figures
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Symbols and terms
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Good to know before you start
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How to calculate it in Excel
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How to calculate it in Google Sheets
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How to calculate it in Python
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How to write it in LaTeX and other math languages (copy and paste)
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How to have ChatGPT do the calculation
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DataChef Features
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Related Features
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NumberChef Calculators List
What you can do on this page
- Enter the coordinates of two points, and you get the distance \(d\) between them on the spot
- Along with the distance, it also finds the change in x \(\Delta x\), the change in y \(\Delta y\), the slope \(m\) and the equation of the line through the two points, \(y = mx + b\)
- The result is also shown on a graph, so you can see that the distance is the length of the line segment joining the two points
- A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page
This page finds the straight-line distance between two points in the plane (two dimensions). It also works for two points with the same x-coordinate (on a vertical line), but then the slope is undefined, so "Undefined" is shown, and the line is written as x = c (a constant) instead of y = mx + b.
What is this calculation used for?
Collision detection and distance to enemies in games (programming)
In a 2D game, the positions of your character, enemies and items are kept as coordinates, and the game counts a "hit" when the distance becomes less than a set amount. For example, if your character is at (1, 1) and an enemy is at (4, 5), the distance is \(\sqrt{3^2 + 4^2} = 5\). Repeating this calculation is how distance-based collision detection works.
The distance formula is one of the first practical formulas you write when you start learning game development.
Finding the distance between two locations in surveying and mapping (surveying and civil engineering)
Surveyors record property corners and construction reference points as coordinates, for example in the State Plane Coordinate System used across the United States. Once you know the coordinates of two points, the horizontal distance between them can be found with the distance formula on this page.
This calculation supports the building of roads, buildings and other infrastructure. (To find a distance from latitude and longitude, taking the curve of the Earth into account, a different formula is used.)
Measuring how similar data are in data analysis and AI (Euclidean distance)
In data analysis, a pair of numbers such as height and weight is treated as a point on the coordinate plane, and the closer two points are, the more similar the data are considered to be. This distance (the Euclidean distance) is used to group similar customers (clustering) and to make predictions from nearby data (the k-nearest neighbors method).
The formula on this page is the basic way of measuring distance that appears in every AI and machine learning textbook.
Reading diagonal lengths from drawings and CAD (design and manufacturing)
In technical drawings and CAD, positions such as the centers of holes in a part are given as coordinates. The diagonal distance between the centers of two holes, which is often not written on the drawing, can be found from the differences in coordinates with the distance formula.
For example, two holes 1.5 in apart horizontally and 2 in apart vertically have centers \(\sqrt{1.5^2 + 2^2} = \sqrt{6.25} = 2.5\) in apart. It is an everyday calculation in machining and inspection.
Formulas and figures
Distance between two points \(d\) (the distance formula)
Figure
Standard notation
(the usual math form)
\(d\)
\(=\)
\(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
In words
(symbols replaced with words)
②
\(d\): distance between the points
\(=\)
①
square root of the sum of the squared differences
The formula in words
①
Take the
square root of the sum of the squared differences, \((x_2 - x_1)^2 + (y_2 - y_1)^2\)
(square the difference of the x-coordinates and the difference of the y-coordinates, add them, and take the square root)
②
and you get the
\(d\): distance between the points
Quick example
The distance between the points (1, 1) and (4, 5) is
distance \(d\)
\(=\)
square root of the sum of the squared differences ((4−1)² + (5−1)² = 25)
\(d = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
Key idea
A common mistake is to take the square roots separately, as in \(\sqrt{(x_2-x_1)^2} + \sqrt{(y_2-y_1)^2}\). Add up everything under the square root sign first, and take the square root only once, at the end.
Even if a difference of coordinates is negative, its square is positive, so the distance is the same whichever point you enter first. The answer is often a number that does not come out even, such as \(\sqrt{13}\); in that case, this calculator shows a decimal (an approximate value).
Why this formula gives the distance (the Pythagorean theorem)
Figure
Standard notation
(the usual math form)
\(\Delta x^2\)
\(+\)
\(\Delta y^2\)
\(=\)
\(d^2\)
In words
(symbols replaced with words)
①
square of the change in x, \(\Delta x = x_2 - x_1\)
\(+\)
②
square of the change in y, \(\Delta y = y_2 - y_1\)
\(=\)
③
square of the distance \(d\) (the hypotenuse)
The formula in words
①
Add the
square of the change in x, \(\Delta x\) (how far apart the points are horizontally)
②
and the
square of the change in y, \(\Delta y\) (how far apart the points are vertically)
③
and you get exactly the
square of the distance \(d\) between the points (the hypotenuse)
(the Pythagorean theorem)
Quick example
Checking with the points (1, 1) and (4, 5) (Δx = 3, Δy = 4)
Δx squared (3² = 9)
\(+\)
Δy squared (4² = 16)
\(=\)
distance squared (25 = 5²)
\(3^2 + 4^2 = 9 + 16 = 25 = 5^2\)
Key idea
Picture a right triangle whose hypotenuse is the line segment joining the two points, and whose two legs are the horizontal change \(\Delta x\) and the vertical change \(\Delta y\). Then the Pythagorean theorem applies directly. The distance formula is just this equation solved for \(d\).
Once you see that "the distance formula" and "the Pythagorean theorem" are not two different formulas but one and the same theorem, you have one less thing to memorize.
The slope \(m\) and the equation \(y = mx + b\) of the line through two points
Graph
Standard notation
(the usual math form)
\(m\)
\(=\)
\(y_2 - y_1\)
\(\div\)
\(x_2 - x_1\)
\(y\)
\(=\)
\(m\)
\(\times\)
\(x\)
\(+\)
\(b\)
In words
(symbols replaced with words)
③
\(m\): slope of the line
\(=\)
①
change in y, \(\Delta y = y_2 - y_1\)
\(\div\)
②
change in x, \(\Delta x = x_2 - x_1\)
⑦
\(y\): y-coordinate of a point on the line
\(=\)
④
\(m\): slope
\(\times\)
⑤
\(x\): x-coordinate
\(+\)
⑥
\(b\): y-intercept
The formula in words
①
Take the
change in y, \(\Delta y\)
②
divide it by the
change in x, \(\Delta x\)
③
and you get the
\(m\): slope of the line
④
Next, take the
\(m\): slope
⑤
multiply it by the
\(x\): x-coordinate
⑥
add the
\(b\): y-intercept
(found as \(b = y_1 - m x_1\))
⑦
and you get the
\(y\): y-coordinate of a point on the line
(this is the equation of the line through the two points)
Quick example
For the line through the points (1, 5) and (3, 2)
slope \(m\)
\(=\)
change in y (2 − 5 = −3)
\(\div\)
change in x (3 − 1 = 2)
\(m = \frac{2 - 5}{3 - 1} = \frac{-3}{2} = -1.5\)
\(b = y_1 - m x_1 = 5 - (-1.5) \times 1 = 6.5\)
\(y = -1.5x + 6.5\)
Key idea
As a bonus to the distance, this calculator also shows information about the line through the two points. The slope \(m\) tells how much the line goes up or down for each 1 step to the right, and the y-intercept \(b\) is the height where the line crosses the y-axis.
However, when the two points have the same x-coordinate (\(x_1 = x_2\), one directly above the other), you would have to divide by \(\Delta x = 0\), so the slope is undefined. This line cannot be written as \(y = mx + b\); it takes the form \(x = c\), where \(c\) is a constant (a vertical line). The distance can still be found as \(|\Delta y|\).
For a more detailed explanation of slope and the angle of inclination, see the related page "Slope Calculator".
To find the distance between two points, square the difference of the x-coordinates and the difference of the y-coordinates, add them, and take the square root. This is the Pythagorean theorem itself. Picture it as finding the length of the hypotenuse from the horizontal and vertical differences, and you will not forget it.
Symbols and terms
Symbols
| \(d\) | dee | The distance between the two points, from the first letter of "distance". |
| \((x_1, y_1)\), \((x_2, y_2)\) | x sub one, y sub one | The coordinates of the two points. \(x\) is the horizontal position and \(y\) is the vertical position. The small number at the lower right (the subscript) tells the first point from the second. |
| \(\Delta x\), \(\Delta y\) | delta x, delta y | The change (the difference of the coordinates). \(\Delta\) (delta) is a Greek letter that stands for "change" or "difference": \(\Delta x = x_2 - x_1\) and \(\Delta y = y_2 - y_1\). |
| \(\sqrt{\phantom{a}}\) | square root (radical sign) | The sign for the square root (the number that gives this number when squared). \(\sqrt{25} = 5\) (because 5² = 25). In the distance formula, you take the square root only once, at the end. |
| \(m\) | em | The slope of the line through the two points. It tells how much the line goes up or down for each 1 step to the right. If it is positive, the line rises to the right; if negative, it falls to the right. |
| \(b\) | bee | The y-intercept: the height where the line crosses the y-axis (the value of \(y\) when \(x = 0\)). |
Terms
| coordinates | A pair of numbers (horizontal position, vertical position) that gives the position of a point. The point (4, 5) is 4 to the right of the origin and 5 up. |
| coordinate plane | A plane where positions are given by a horizontal x-axis and a vertical y-axis. The distance on this page is the straight-line length between two points on the coordinate plane. |
| line segment | A straight line that joins two points, with those points as its ends. The distance between two points is the length of this line segment. |
| Pythagorean theorem | The theorem that in a right triangle, the sum of the squares of the two legs equals the square of the hypotenuse (\(a^2 + b^2 = c^2\)). The distance formula is this theorem applied to coordinates. It is taught in Grade 8. |
| hypotenuse | The longest side of a right triangle, opposite the right angle. In the distance formula, the line segment joining the two points is the hypotenuse. |
| Euclidean distance | The formal name for the straight-line distance between two points that this page calculates. In data analysis and AI, this name is often used for the basic way to measure how close points are. |
| slope | A number that shows how steep a line is: how much y increases when x increases by 1 ("rise over run"). It is the same as the rate of change of a linear function. |
| vertical line | A line perpendicular to the x-axis (going straight up and down). The line through two points with the same x-coordinate is vertical. The change in x is 0, so its slope is undefined. Its equation has the form \(x = c\), where \(c\) is a constant. |
Good to know before you start
Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back over these topics is the fastest way forward.
| The coordinate plane (Grade 6) |
|
| Operations with positive and negative numbers (Grade 7) |
|
| Square roots (Grade 8) |
|
| The Pythagorean theorem (Grade 8) |
|
| Linear functions (Grade 8) |
|
How to calculate it in Excel
Copy the whole table below and paste it into cell A1 in Excel. It works as is.
Table to find the distance d between two points
| Point 1 x-coordinate x₁ | 1 |
| Point 1 y-coordinate y₁ | 1 |
| Point 2 x-coordinate x₂ | 4 |
| Point 2 y-coordinate y₂ | 5 |
| Distance between the points d | =SQRT((B3-B1)^2+(B4-B2)^2) |
Table to check with the Pythagorean theorem
| Point 1 x-coordinate x₁ | 1 |
| Point 1 y-coordinate y₁ | 1 |
| Point 2 x-coordinate x₂ | 4 |
| Point 2 y-coordinate y₂ | 5 |
| Change in x Δx | =B3-B1 |
| Change in y Δy | =B4-B2 |
| Sum of squares Δx²+Δy² | =B5^2+B6^2 |
| Distance between the points d | =SQRT(B7) |
Table to find the slope m and the equation of the line
| Point 1 x-coordinate x₁ | 1 |
| Point 1 y-coordinate y₁ | 5 |
| Point 2 x-coordinate x₂ | 3 |
| Point 2 y-coordinate y₂ | 2 |
| Slope m | =(B4-B2)/(B3-B1) |
| y-intercept b | =B2-B5*B1 |
After pasting, the upper rows are your inputs, and the formulas in the lower rows are calculated automatically.
"SQRT(…)" is the square root, and "^2" is squared.
The first table gives the distance 5. The second table finds the same distance 5 from Δx = 3, Δy = 4 and the sum of squares 25, so you can also see the values along the way.
The third table gives the slope −1.5 and the y-intercept 6.5 (the line y = −1.5x + 6.5). If the two points have the same x-coordinate, you would be dividing by 0 and an error (#DIV/0!) appears; this tells you that the slope is undefined (a vertical line).
"SQRT(…)" is the square root, and "^2" is squared.
The first table gives the distance 5. The second table finds the same distance 5 from Δx = 3, Δy = 4 and the sum of squares 25, so you can also see the values along the way.
The third table gives the slope −1.5 and the y-intercept 6.5 (the line y = −1.5x + 6.5). If the two points have the same x-coordinate, you would be dividing by 0 and an error (#DIV/0!) appears; this tells you that the slope is undefined (a vertical line).
How to calculate it in Google Sheets
Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.
Table to find the distance d between two points
| Point 1 x-coordinate x₁ | 1 |
| Point 1 y-coordinate y₁ | 1 |
| Point 2 x-coordinate x₂ | 4 |
| Point 2 y-coordinate y₂ | 5 |
| Distance between the points d | =SQRT((B3-B1)^2+(B4-B2)^2) |
Table to check with the Pythagorean theorem
| Point 1 x-coordinate x₁ | 1 |
| Point 1 y-coordinate y₁ | 1 |
| Point 2 x-coordinate x₂ | 4 |
| Point 2 y-coordinate y₂ | 5 |
| Change in x Δx | =B3-B1 |
| Change in y Δy | =B4-B2 |
| Sum of squares Δx²+Δy² | =B5^2+B6^2 |
| Distance between the points d | =SQRT(B7) |
Table to find the slope m and the equation of the line
| Point 1 x-coordinate x₁ | 1 |
| Point 1 y-coordinate y₁ | 5 |
| Point 2 x-coordinate x₂ | 3 |
| Point 2 y-coordinate y₂ | 2 |
| Slope m | =(B4-B2)/(B3-B1) |
| y-intercept b | =B2-B5*B1 |
The same formulas as in Excel (SQRT and ^2) work as is in Google Sheets. Copy the whole table, paste it into cell A1, and replace the input numbers with your own coordinates.
How to calculate it in Python
import math
x1, y1 = 1.0, 1.0 # coordinates of point 1
x2, y2 = 4.0, 5.0 # coordinates of point 2
delta_x = x2 - x1 # change in x
delta_y = y2 - y1 # change in y
distance = math.hypot(delta_x, delta_y) # distance between the points (same as √(Δx²+Δy²))
print(f"Distance between the points d: {distance}")
# Bonus: slope and equation of the line through the two points (undefined when Δx is 0, a vertical line)
if delta_x != 0:
slope = delta_y / delta_x # slope m
y_intercept = y1 - slope * x1 # y-intercept b
sign = "+" if y_intercept >= 0 else "-" # split off the sign so a negative intercept is not shown as "+ -0.3"
print(f"Slope m: {slope}")
print(f"y-intercept b: {y_intercept}")
print(f"Equation of the line: y = {slope}x {sign} {abs(y_intercept)}")
else:
print(f"Vertical line (x = {x1}), so the slope is undefined")
Runs with the standard library only. math.hypot calculates "the square root of the sum of the squares" in one step, so it is the distance formula itself. Change the coordinates at the top and run it.
How to write it in LaTeX and other math languages (copy and paste)
Distance between two points \(d\) (the distance formula)
d = √((x₂ − x₁)² + (y₂ − y₁)²)
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>d</mi>
<mo>=</mo>
<msqrt>
<mrow>
<msup>
<mrow><mo>(</mo><msub><mi>x</mi><mn>2</mn></msub><mo>−</mo><msub><mi>x</mi><mn>1</mn></msub><mo>)</mo></mrow>
<mn>2</mn>
</msup>
<mo>+</mo>
<msup>
<mrow><mo>(</mo><msub><mi>y</mi><mn>2</mn></msub><mo>−</mo><msub><mi>y</mi><mn>1</mn></msub><mo>)</mo></mrow>
<mn>2</mn>
</msup>
</mrow>
</msqrt>
</mrow>
</math>
d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
Sqrt[(x2 - x1)^2 + (y2 - y1)^2]
d := sqrt((x2 - x1)^2 + (y2 - y1)^2);
d = sqrt((x2 - x1)^2 + (y2 - y1)^2);
d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)
Why this formula gives the distance (the Pythagorean theorem)
Δx² + Δy² = d²
\Delta x^2 + \Delta y^2 = d^2
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<msup><mrow><mi mathvariant="normal">Δ</mi><mi>x</mi></mrow><mn>2</mn></msup>
<mo>+</mo>
<msup><mrow><mi mathvariant="normal">Δ</mi><mi>y</mi></mrow><mn>2</mn></msup>
<mo>=</mo>
<msup><mi>d</mi><mn>2</mn></msup>
</mrow>
</math>
(Delta x)^2 + (Delta y)^2 = d^2
dx^2 + dy^2 == d^2
dx^2 + dy^2 = d^2;
dx^2 + dy^2 == d^2
Δx^2 + Δy^2 = d^2
The slope \(m\) and the equation \(y = mx + b\) of the line through two points
m = (y₂ − y₁) ÷ (x₂ − x₁), y = mx + b (b = y₁ − m·x₁)
m = \frac{y_2 - y_1}{x_2 - x_1}, \quad y = mx + b, \quad b = y_1 - m x_1
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>m</mi>
<mo>=</mo>
<mfrac>
<mrow><msub><mi>y</mi><mn>2</mn></msub><mo>−</mo><msub><mi>y</mi><mn>1</mn></msub></mrow>
<mrow><msub><mi>x</mi><mn>2</mn></msub><mo>−</mo><msub><mi>x</mi><mn>1</mn></msub></mrow>
</mfrac>
<mo>,</mo>
<mi>y</mi>
<mo>=</mo>
<mi>m</mi><mi>x</mi>
<mo>+</mo>
<mi>b</mi>
<mo>,</mo>
<mi>b</mi>
<mo>=</mo>
<msub><mi>y</mi><mn>1</mn></msub>
<mo>−</mo>
<mi>m</mi><msub><mi>x</mi><mn>1</mn></msub>
</mrow>
</math>
m = (y_2 - y_1)/(x_2 - x_1), y = m x + b, b = y_1 - m x_1
m = (y2 - y1)/(x2 - x1); b = y1 - m*x1; y = m*x + b
m := (y2 - y1)/(x2 - x1); b := y1 - m*x1; y := m*x + b;
m = (y2 - y1)/(x2 - x1); b = y1 - m*x1; y = m*x + b;
m = (y_2 - y_1)/(x_2 - x_1), y = mx + b, b = y_1 - mx_1
How to have ChatGPT do the calculation
You are a math calculation assistant. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing). For the two points (1, 1) and (4, 5) on the coordinate plane, find each of the following: 1. The distance d between the two points (use the distance formula d = √((x2−x1)² + (y2−y1)²)) 2. The change in x, Δx, and the change in y, Δy 3. The slope m of the line through these two points, and the equation of the line y = mx + b (also show the value of the y-intercept b) Show the formulas you used and the numbers from the execution result.
How to Use
-
1Enter your numbersType the numbers you want to calculate with into the input fields
-
2CalculatePress the "Calculate" button
-
3Check the resultThe result appears on the spot. The same page also explains the idea behind the calculation and the formula
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