11.2: Equations of Lines - Slope-Intercept Form
- Page ID
- 173513
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Media
Definitions and Theorems
The slope-intercept form of the equation of a nonvertical line is\[y = mx + b,\nonumber\]where \(m\) is the slope of the line and \(b\) is the \(y\)-coordinate of the point where the line crosses the \(y\)-axis.
The graph of \(y = mx + b\) is the nonvertical line whose slope is \(m\) and whose \(y\)-intercept is the point \((0,b)\). Conversely, every nonvertical line is the graph of exactly one equation of this form.
Any linear equation in which the \(y\)-term has a nonzero coefficient can be rewritten in slope-intercept form by solving the equation for \(y\). Once \(y\) is isolated, the coefficient of \(x\) is the slope \(m\), and the constant term is the \(y\)-intercept value \(b\).
A vertical line has the equation \(x = a\) for a constant \(a\); it has no slope and cannot be written in slope-intercept form. A horizontal line has slope \(m = 0\), so its equation reduces to \(y = b\).
The equation of the horizontal line containing the point \( \left( k,c \right) \), where \( k \) is any real number, is\[ y = c. \nonumber \]The equation of the vertical line containing the point \( \left( c, k \right) \), where \( k \) is any real number, is\[ x = c. \nonumber \]
Examples
Identify the slope and the \(y\)-intercept of the line \(y = 3x - 7\).
- Solution
-
The equation is already in slope-intercept form \(y = mx + b\). Matching terms, the coefficient of \(x\) is the slope and the constant term is the \(y\)-coordinate of the \(y\)-intercept:\[m = 3, \qquad b = -7.\nonumber\]The slope is \(3\), and the \(y\)-intercept is the point \((0,-7)\).
Identify the slope and the \(y\)-intercept of the line \(y = 4 - x\).
- Solution
-
Reorder the right-hand side so the \(x\)-term is written first, matching the form \(y = mx + b\):\[y = -x + 4.\nonumber\]An unwritten coefficient of \(x\) is understood to be \(1\), so the coefficient here is \(-1\). Thus \(m = -1\) and \(b = 4\). The slope is \(-1\), and the \(y\)-intercept is \((0,4)\).
Write \(2x + y = 9\) in slope-intercept form, then state its slope and \(y\)-intercept.
- Solution
-
Solve for \(y\) by subtracting \(2x\) from both sides:\[y = -2x + 9.\nonumber\]The equation is now in slope-intercept form. The slope is \(m = -2\), and the \(y\)-intercept is \((0,9)\).
Write \(3x + 4y = 12\) in slope-intercept form, then state its slope and \(y\)-intercept.
- Solution
-
Isolate the \(y\)-term, then divide every term by \(4\):\[4y = -3x + 12 \quad\Longrightarrow\quad y = -\dfrac{3}{4}x + 3.\nonumber\]The slope is \(m = -\dfrac{3}{4}\), and the \(y\)-intercept is \((0,3)\).
Write \(5x - 2y = 8\) in slope-intercept form, then state its slope and \(y\)-intercept.
- Solution
-
Isolate the \(y\)-term, then divide every term by \(-2\). Dividing by a negative number changes the sign of each term:\[-2y = -5x + 8 \quad\Longrightarrow\quad y = \dfrac{5}{2}x - 4.\nonumber\]The slope is \(m = \dfrac{5}{2}\), and the \(y\)-intercept is \((0,-4)\).
Write \(3y - 12 = 0\) in slope-intercept form, then state its slope and \(y\)-intercept and describe the graph.
- Solution
-
The equation contains no \(x\)-term. Solve for \(y\):\[3y = 12 \quad\Longrightarrow\quad y = 4.\nonumber\]Written in the form \(y = mx + b\), this is \(y = 0x + 4\), so \(m = 0\) and \(b = 4\). The slope is \(0\), and the \(y\)-intercept is \((0,4)\). Because the slope is zero, the graph is a horizontal line through \((0,4)\).
Sources
Several parts of this text use modifications from the following source:
- Wikipedia article: "Linear equation"
This source is released under the Creative Commons Attribution-Share-Alike License 4.0.