Distance Between Two Points Calculator

Published on: September 15, 2024
Final Answer: Free • Full Steps: Plus

This Distance Between Two Points Calculator finds the straight-line distance between two points on a coordinate plane, and shows the working line by line. The horizontal gap and the vertical gap between the points are the two shorter sides of a right triangle, so each difference is squared, the squares are added, and the square root of that total is the distance. Following the same pattern every time is what makes the distance formula easy to check and repeat.

Line between two points

Enter the coordinates for \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\).

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two pairs of coordinates into the formula.
  4. Subtract to find the horizontal and vertical differences.
  5. Square each difference.
  6. Add the two squares under the root.
  7. Take the square root to get the distance.

Formula:

\(d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\)
See Example 1 Hide Example 1

Problem

\((x_{1}, y_{1}) = (2, 3),\; (x_{2}, y_{2}) = (14, 11)\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two pairs of coordinates into the formula.
  4. Subtract to find the horizontal and vertical differences.
  5. Square each difference.
  6. Add the two squares under the root.
  7. Take the square root to get the distance.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the distance between the points \((2, 3)\) and \((14, 11)\).

\((x_{1}, y_{1}) = (2, 3),\; (x_{2}, y_{2}) = (14, 11)\)

Step 2

Step 2 - Write the formula.

In this problem: The horizontal gap and the vertical gap between the points are the two shorter sides of a right triangle, and the straight line joining the points is its hypotenuse.

\(d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\)

Step 3

Step 3 - Substitute the two pairs of coordinates into the formula.

In this problem: Replacing each coordinate in \(d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\) gives \(d = \sqrt{\left(14 - 2\right)^{2} + \left(11 - 3\right)^{2}}\).

\(d = \sqrt{\left(14 - 2\right)^{2} + \left(11 - 3\right)^{2}}\)

Step 4

Step 4 - Subtract to find the horizontal and vertical differences.

In this problem: Under the root both brackets are still written as subtractions, so each one is worked out in turn.

\(d = \sqrt{\left(14 - 2\right)^{2} + \left(11 - 3\right)^{2}}\)

Step 4a

Step 4a - Subtract to find the horizontal and vertical differences.

In this problem: Subtracting the x-coordinates gives the horizontal difference: \(14 - 2 = 12\).

\(d = \sqrt{12^{2} + \left(11 - 3\right)^{2}}\)

Step 4b

Step 4b - Subtract to find the horizontal and vertical differences.

In this problem: Subtracting the y-coordinates gives the vertical difference: \(11 - 3 = 8\).

\(d = \sqrt{12^{2} + 8^{2}}\)

Step 5

Step 5 - Square each difference.

In this problem: Both differences are still written as squares, so each one is worked out in turn.

\(d = \sqrt{12^{2} + 8^{2}}\)

Step 5a

Step 5a - Square each difference.

In this problem: Squaring the horizontal difference gives \(12^{2} = 144\).

\(d = \sqrt{144 + 8^{2}}\)

Step 5b

Step 5b - Square each difference.

In this problem: Squaring the vertical difference gives \(8^{2} = 64\).

\(d = \sqrt{144 + 64}\)

Step 6

Step 6 - Add the two squares under the root.

In this problem: Adding the two squares gives \(144 + 64 = 208\).

\(d = \sqrt{208}\)

Step 7

Step 7 - Take the square root to get the distance.

In this problem: Taking the square root undoes the squaring and gives \(\sqrt{208} \approx 14.42\).

\(d = \sqrt{208} \approx 14.42\)

Final Answer

\(d \approx 14.42\)
See Example 2 Hide Example 2

Problem

\((x_{1}, y_{1}) = (-5, 2),\; (x_{2}, y_{2}) = (15, 9)\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two pairs of coordinates into the formula.
  4. Subtract to find the horizontal and vertical differences.
  5. Square each difference.
  6. Add the two squares under the root.
  7. Take the square root to get the distance.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the distance between the points \((-5, 2)\) and \((15, 9)\).

\((x_{1}, y_{1}) = (-5, 2),\; (x_{2}, y_{2}) = (15, 9)\)

Step 2

Step 2 - Write the formula.

In this problem: The horizontal gap and the vertical gap between the points are the two shorter sides of a right triangle, and the straight line joining the points is its hypotenuse.

\(d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\)

Step 3

Step 3 - Substitute the two pairs of coordinates into the formula.

In this problem: Replacing each coordinate in \(d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\) gives \(d = \sqrt{\left(15 - \left(-5\right)\right)^{2} + \left(9 - 2\right)^{2}}\).

\(d = \sqrt{\left(15 - \left(-5\right)\right)^{2} + \left(9 - 2\right)^{2}}\)

Step 4

Step 4 - Subtract to find the horizontal and vertical differences.

In this problem: Under the root both brackets are still written as subtractions, so each one is worked out in turn.

\(d = \sqrt{\left(15 - \left(-5\right)\right)^{2} + \left(9 - 2\right)^{2}}\)

Step 4a

Step 4a - Subtract to find the horizontal and vertical differences.

In this problem: Subtracting the x-coordinates gives the horizontal difference: \(15 - \left(-5\right) = 20\).

\(d = \sqrt{20^{2} + \left(9 - 2\right)^{2}}\)

Step 4b

Step 4b - Subtract to find the horizontal and vertical differences.

In this problem: Subtracting the y-coordinates gives the vertical difference: \(9 - 2 = 7\).

\(d = \sqrt{20^{2} + 7^{2}}\)

Step 5

Step 5 - Square each difference.

In this problem: Both differences are still written as squares, so each one is worked out in turn.

\(d = \sqrt{20^{2} + 7^{2}}\)

Step 5a

Step 5a - Square each difference.

In this problem: Squaring the horizontal difference gives \(20^{2} = 400\).

\(d = \sqrt{400 + 7^{2}}\)

Step 5b

Step 5b - Square each difference.

In this problem: Squaring the vertical difference gives \(7^{2} = 49\).

\(d = \sqrt{400 + 49}\)

Step 6

Step 6 - Add the two squares under the root.

In this problem: Adding the two squares gives \(400 + 49 = 449\).

\(d = \sqrt{449}\)

Step 7

Step 7 - Take the square root to get the distance.

In this problem: Taking the square root undoes the squaring and gives \(\sqrt{449} \approx 21.19\).

\(d = \sqrt{449} \approx 21.19\)

Final Answer

\(d \approx 21.19\)

Example Problems

Click one to load it into the box and solve it.