3.5: The Pythagorean Theorem, Distance Formula, and Midpoint Formula
- Page ID
- 190948
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definitions and Theorems
The Pythagorean Theorem describes the fundamental relationship among the three sides of a right triangle. Applied to points in the coordinate plane, it yields two of the most frequently used tools in coordinate geometry: the distance formula and the midpoint formula.
In a right triangle, if \(c\) is the length of the hypotenuse, and the lengths of the two legs are denoted by \(a\) and \(b\) (as shown in the figure below), then\[a^2 + b^2 = c^2. \nonumber \]
The Pythagorean Theorem applies only to right triangles, and \(c\) must always be the hypotenuse—the side opposite the right angle, which is also the longest side.
A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\).
A right triangle whose sides form a Pythagorean triple is called a Pythagorean triangle.
The distance \(d\) between two points \(P_1\left(x_1, y_1\right)\) and \(P_2\left(x_2, y_2\right)\) is\[d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}. \nonumber\]
The distance formula is the Pythagorean Theorem applied to the horizontal change \(x_2 - x_1\) and the vertical change \(y_2 - y_1\) between the two points, which serve as the legs of a right triangle whose hypotenuse is the segment connecting the points. Because each difference is squared, the order in which you subtract does not matter; you may label either point as \((x_1, y_1)\).
The midpoint of a line segment is the point on the segment that is equidistant from its two endpoints. Equivalently, it is the point that divides the segment into two equal parts.
The midpoint \(M\) of the line segment with endpoints \(A = (x_1, y_1)\) and \(B = (x_2, y_2)\) is\[M = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right).\nonumber\]That is, each coordinate of the midpoint is the average of the corresponding coordinates of the endpoints.
Examples
A right triangle has legs of length \(8\) and \(15\). Find the length of the hypotenuse.
- Solution
- The legs have lengths \(a = 8\) and \(b = 15\), and the hypotenuse has unknown length \(c\). By the Pythagorean Theorem,\[c^2 = a^2 + b^2 = 8^2 + 15^2 = 64 + 225 = 289.\nonumber\]Taking the positive square root, since a length cannot be negative, gives \(c = \sqrt{289} = 17\).
A right triangle has a hypotenuse of length \(13\) and one leg of length \(5\). Find the length of the other leg.
- Solution
- The hypotenuse has length \(c = 13\) and one leg has length \(a = 5\); the other leg has unknown length \(b\). Rearranging the Pythagorean Theorem to isolate the unknown leg,\[b^2 = c^2 - a^2 = 13^2 - 5^2 = 169 - 25 = 144.\nonumber\]Thus \(b = \sqrt{144} = 12\).
Find the distance between the points \(A = (1, 2)\) and \(B = (4, 6)\).
- Solution
- Apply the Distance Formula, substituting the coordinates of \(A\) and \(B\):\[d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.\nonumber\]The distance between the two points is \(5\).
Find the distance between the points \(A = (-3, 2)\) and \(B = (3, -4)\).
- Solution
- Substitute the coordinates into the Distance Formula, taking care with the signs:\[d = \sqrt{(3 - (-3))^2 + (-4 - 2)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72}.\nonumber\]Simplify the radical by factoring out the largest perfect square, \(36\):\[\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}.\nonumber\]The distance between the two points is \(6\sqrt{2}\).
Find the midpoint of the line segment with endpoints \(A = (-5, 7)\) and \(B = (3, -1)\).
- Solution
- Average the coordinates of the endpoints using the Midpoint Formula:\[M = \left( \dfrac{-5 + 3}{2}, \dfrac{7 + (-1)}{2} \right) = \left( \dfrac{-2}{2}, \dfrac{6}{2} \right) = (-1, 3).\nonumber\]The midpoint is \((-1, 3)\).
The midpoint of a line segment is \(M = (5, 1)\), and one endpoint is \(A = (2, -4)\). Find the other endpoint.
- Solution
- Let the unknown endpoint be \(B = (x, y)\). Because each coordinate of the midpoint is the average of the corresponding coordinates of the endpoints, the Midpoint Formula produces one equation per coordinate:\[\dfrac{2 + x}{2} = 5 \qquad \text{and} \qquad \dfrac{-4 + y}{2} = 1.\nonumber\]Solving the first equation, \(2 + x = 10\), so \(x = 8\). Solving the second, \(-4 + y = 2\), so \(y = 6\). The other endpoint is \(B = (8, 6)\).
Sources
Several parts of this text use modifications from the following sources:
- Wikipedia article: "Pythagorean theorem"
- Wikipedia article: "Euclidean distance"
- Wikipedia article: "Midpoint"
All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.