Skip to main content
K12 LibreTexts

4.40: Applications of the Distance Formula

  • Page ID
    4978
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Length between two points based on a right triangle.

    Distance Formula in the Coordinate Plane

    The distance between two points \((x_1, y_1)\) and \((x_2,y_2)\) can be defined as \(d=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}\). This is called the distance formula. Remember that distances are always positive!

    What if you were given the coordinates of two points? How could you find how far apart these two points are?

    Example \(\PageIndex{1}\)

    Find the distance between \((-2, -3)\) and \((3, 9)\).

    Solution

    Use the distance formula, plug in the points, and simplify.

    \(\begin{align*}d&=\sqrt{(3−(−2))^2+(9−(−3))^2} \\ &=\sqrt{(5)^2+(12)^2} \\ &= \sqrt{25+144} \\ &= \sqrt{169}=13\text{ units }\end{align*}\)

    Example \(\PageIndex{2}\)

    Find the distance between \((12, 26)\) and \((8, 7)\).

    Solution

    Use the distance formula, plug in the points, and simplify.

    \(\begin{align*}d&=\sqrt{(8−12)^2+(7−26)^2} \\ &= \sqrt{(−4)^2+(−19)^2} \\ &= \sqrt{16+361} \\ &= \sqrt{377}\approx 19.42\text{ units }\end{align*}\)

    Example \(\PageIndex{3}\)

    Find the distance between \((4, -2)\) and \((-10, 3)\).

    Solution

    Plug in \((4, -2)\) for \((x_1, y_1)\) and \((-10, 3)\) for \((x_2,y_2)\) and simplify.

    \(\begin{align*}d&=\sqrt{(−10−4)^2+(3+2)^2} \\ &= \sqrt{(−14)^2+(5)^2} \\ &= \sqrt{196+25} \\ &= \sqrt{221}\approx 14.87\text{ units }\end{align*}\)

    Example \(\PageIndex{4}\)

    Find the distance between \((3, 4)\) and \((-1, 3)\).

    Solution

    Plug in (3, 4)\) for \((x_1, y_1)\) and \((-1, 3)\) for \((x_2,y_2)\) and simplify.

    \(\begin{align*}d &=\sqrt{(−1−3)^2+(3−4)^2} \\ &= \sqrt{(−4)^2+(−1)^2} \\ &= \sqrt{16+1} \\ &= \sqrt{17} \approx 4.12\text{ units }\end{align*}\)

    Example \(\PageIndex{5}\)

    Find the distance between \((4, 23)\) and \((8, 14)\).

    Solution

    Plug in \((4, 23)\) for \((x_1, y_1)\) and \((8, 14)\) for \((x_2,y_2)\) and simplify.

    \(\begin{align*} d&=\sqrt{(8−4)^2+(14−23)^2} \\ &=\sqrt{(4)^2+(−9)^2} \\ &=\sqrt{16+81} \\ & =\sqrt{97} \approx 9.85\text{ units }\end{align*} \)

    Review

    Find the distance between each pair of points. Round your answer to the nearest hundredth.

    1. \((4, 15)\) and \((-2, -1)\)
    2. \((-6, 1)\) and \((9, -11)\)
    3. \((0, 12)\) and \((-3, 8)\)
    4. \((-8, 19)\) and \((3, 5)\)
    5. \((3, -25)\) and \((-10, -7)\)
    6. \((-1, 2)\) and\((8, -9)\)
    7. \((5, -2)\) and \((1, 3)\)
    8. \((-30, 6)\) and \((-23, 0)\)
    9. \((2, -2)\) and \((2, 5)\)
    10. \((-9, -4)\) and \((1, -1) \)

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 3.10.

    Resource

    Vocabulary

    Term Definition
    Distance Formula The distance between two points \((x_1, y_1)\) and \((x_2,y_2)\) can be defined as \(d=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}\).
    Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by \(a^2+b^2=c^2\), where \(a\) and \(b\) are legs of the triangle and c is the hypotenuse of the triangle.

    Additional Resources

    Interactive Element

    Video: The Distance Formula

    Activities: Distance Formula in the Coordinate Plane Discussion Questions

    Study Aids: Segments Study Guide

    Practice: Applications of the Distance Formula

    Real World: Distance Formula in the Coordinate Plane


    This page titled 4.40: Applications of the Distance Formula is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    CK-12 Foundation
    LICENSED UNDER
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License
    • Was this article helpful?