17.1: Appendix A- A review of Straight Lines
- Page ID
- 121181
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A.1 Geometric ideas: lines, slopes, equations
Straight lines have some important geometric properties, namely:
The slope of a straight line is the same everywhere along its length.
Slope of a straight line. We define the slope of a straight line as follows:
\[\text { Slope }=\frac{\Delta y}{\Delta x} \nonumber \]
where \(\Delta y\) means "change in the \(y\) value" and \(\Delta x\) means "change in the \(x\) value" between two points. See Figure A.1 for what this notation represents.
Equation of a straight line. Using this basic geometric property, we can find the equation of a straight line given any of the following information about the line:
- The \(y\)-intercept, \(b\), and the slope, \(m\) :
\[y=m x+b . \nonumber \]
- A point \(\left(x_{0}, y_{0}\right)\) on the line, and the slope, \(m\), of the line:
\[\frac{y-y_{0}}{x-x_{0}}=m \nonumber \]
- Two points on the line, say \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) :
\[\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \nonumber \]
Note: any of these can be rearranged or simplified to produce the standard form \(y=m x+b\), as discussed in the problem set.
The following examples serve as a refresher on finding the equation of the line that satisfies each of the given conditions.
In each case write the equation of the straight line that satisfies the given statements.
Note: you should also be able to easily sketch the line in each case.
(a) The line has slope 2 and y-intercept 4.
(b) The line goes through the points \((1,1)\) and \((3,-2)\).
(c) The line has \(y\)-intercept \(-1\) and \(x\)-intercept 3 .
(d) The line has slope \(-1\) and goes through the point \((-2,-5)\).
Solution
(a) We can use the standard form of the equation of a straight line, \(y=m x+b\) where \(m\) is the slope and \(b\) is the \(y\)-intercept to obtain the equation: \(y=\) \(2 x+4\).
(b) The line goes through the points \((1,1)\) and \((3,-2)\). We use the fact that the slope is the same all along the line. Thus,
\[\frac{\left(y-y_{0}\right)}{\left(x-x_{0}\right)}=\frac{\left(y_{1}-y_{0}\right)}{\left(x_{1}-x_{0}\right)}=m . \nonumber \]
Substituting in the values \(\left(x_{0}, y_{0}\right)=(1,1)\) and \(\left(x_{1}, y_{1}\right)=(3,-2)\),
\[\frac{(y-1)}{(x-1)}=\frac{(1+2)}{(1-3)}=-\frac{3}{2} \text {. } \nonumber \]
This tells us that the slope is \(m=-3 / 2\). We find that
\[y-1=-\frac{3}{2}(x-1)=-\frac{3}{2} x+\frac{3}{2}, \quad \Rightarrow \quad y=-\frac{3}{2} x+\frac{5}{2} . \nonumber \]
(c) The line has \(y\)-intercept \(-1\) and \(x\)-intercept 3 , i.e. goes through the points \((0,-1)\) and \((3,0)\). We can use the method in (b) to get
\[y=\frac{1}{3} x-1 \nonumber \]
Alternately, as a shortcut, we could find the slope,
\[m=\frac{\Delta y}{\Delta x}=\frac{1}{3} . \nonumber \]
Note: \(\Delta\) means "change in the value", i.e. \(\Delta y=y_{1}-y_{0}\).
Thus \(m=1 / 3\) and \(b=-1\) ( \(y\)-intercept), leading to the same result.
(d) The line has slope \(-1\) and goes through the point \((-2,-5)\). Then,
\[\frac{(y+5)}{(x+2)}=-1, \quad \Rightarrow \quad y+5=-1(x+2)=-x-2, \quad \Rightarrow \quad y=-x-7 . \nonumber \]