I had to reinstall my system OS last week. (Windows 10)
I had been using Windows 7 and MikTex.
I created my file with TexMaker and before my OS install the preview compiled fine
Now with a brand new install of everything I'm seeing errors that don't make sense when it was working before.
The author of TexMaker suggests there is a problem with the MikTex install
I don't know how to figure out what's missing, would appreciate some help in learning how to track down the problem.
The file (with the beginning and end of the document)
\documentclass[10pt,letterpaper]{article}
\usepackage{xcolor}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\begin{document}
\textbf{Section 1.2 - Distance Formula}
\newline \textbf{Problems}
\newline \textbf{A}
\newline \textit{In problems 1-8, find the distance between the given points}
\\
\newline 1. $A=(1,-3), B=(2,5)$ $\color{blue} => x_1=1, x_2=2$, $\color{green}y_1=-3, y_2=5 $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(2-1)^2+(5-(-3))^2}=\sqrt{(1)^2+(5+3)^2}=\sqrt{1^2+8^2}=\sqrt{1+64}=\sqrt{65}$\\
\newline 2. $A=(4,13), B=(-1,5)$ $\color{blue} => x_1 =4, x_2=-1$, $\color{green}y_1=13, y_2=5 $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(-1-4)^2+(13-5)^2}=\sqrt{(-5)^2+(8)^2}=\sqrt{25+64}=\sqrt{89}$\\
\newline 3. $A=(3, -2), B=(3, -4)$ $\color{blue} => x_1 =3, x_2=-2$, $\color{green}y_1=3, y_2=-4 $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(-2-3)^2 + (-4-3)^2}=\sqrt{(-5)^2+(-7)^2}=\sqrt{25+49}=\sqrt{74}$\\
\newline 4. $A=(-5, 1), B=(0, -10)$ $\color{blue} => x_1=-5, x_2=0$, $\color{green}y_1=1, y_2=-10 $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(0-(-5))^2+(-10-1)^2}=\sqrt{(0+5)^2+(-11)^2}=\sqrt{5^2+(-11)^2}=\sqrt{25+121}=\sqrt{146}$\\
\newline 5. $A=(\frac{1}{2}, \frac{3}{2}), B=(\frac{-5}{2}, 2)$ $\color{blue} => x_1=\frac{1}{2}, x_2=\frac{-5}{2}$, $\color{green}y_1=\frac{3}{2}, y_2=2 $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(\frac{-5}{2}-\frac{1}{2})^2+(2-\frac{3}{2})^2}=\sqrt{(\frac{-6}{2})^2+(\frac{4}{2}-\frac{3}{2})^2}=\sqrt{(-3)^2+(\frac{1}{2})^2}=\sqrt{9+\frac{1}{4}}=\sqrt{\frac{36}{4}+\frac{1}{4}}=\sqrt{\frac{37}{4}}=\frac{\sqrt{37}}{2}$\\
\newline 6. $A=(\frac{2}{3}, \frac{1}{3}), B=(\frac{-4}{3}, \frac{4}{3})$ $\color{blue} => x_1=\frac{2}{3}, x_2=\frac{-4}{3}$, $\color{green}y_1=\frac{1}{3}, y_2=\frac{4}{3} $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(\frac{-4}{3}-\frac{2}{3})^2+(\frac{4}{3}-\frac{1}{3})^2}=\sqrt{(\frac{-6}{3})^2+(\frac{3}{3})^2}=\sqrt{(-2)^2+(1)^2}=\sqrt{4+1}=\sqrt{5}$\\
\newline 7. $A=(\sqrt{2}, 1), B=(2\sqrt{2}, 3)$ $\color{blue} => x_1=\sqrt{2}, x_2=2$, $\color{green}y_1=1, y_2=\sqrt{2} $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(2\sqrt{2}-\sqrt{2})^2+(3-1)^2}=\sqrt{(\sqrt{2})^2+(2)^2}=\sqrt{2+4}=\sqrt{6}$\\
\newline 8. $A=(\sqrt{3}, -\sqrt{2}), B=(-3\sqrt{3}, \sqrt{2})$ $\color{blue} => x_1=\sqrt{3}, x_2=-3\sqrt{3}$, $\color{green}y_1=13, y_2= $\\
\textbf{\underline {Solution}:}\\
\newline$\overline{AB}=\sqrt{(-3\sqrt{3}-\sqrt{3})^2+(\sqrt{2}-(-\sqrt{2}))^2}=\sqrt{(-4\sqrt{3})^2+(\sqrt{2}+\sqrt{2})^2}=\sqrt{(-4\sqrt{3})^2+(2\sqrt{2})}=\sqrt{(-4)^2(\sqrt{3})^2+(2)^2(\sqrt{2})^2}=\sqrt{(16)(3)+(4)(2)}=\sqrt{48+8}=\sqrt{56}$\\
\\
\newline \textit{In Problems 9-14, determine whether the three given points are collinear}
\begin{quote}
[Given 3 points A, B, C, these points are collinear if $\overline{AC}=\overline{AB}+\overline{BC}$ or $\overline{AC}=\overline{BC}-\overline{AB}$]
\end{quote}
9. $A=(2, 1), B=(4, 3), C=(-1, -2)$\\
\textbf{\underline {Solution}:}\\
\newline $\overline{AB} =\sqrt{(4-2)^2+(3-1)^2}=\sqrt{(2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
\newline $\overline{BC} =\sqrt{(-1-4)^2+(-2-3)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}=\sqrt{25}\sqrt{2}=5\sqrt{2}$
\newline $\overline{AC} =\sqrt{(-1-2))^2+(-2-1)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=\sqrt{9}\sqrt{2}=3\sqrt{2}$
\\
\newline 10. $A=(3, 2), B=(4, 6), C=(0, -8)$\\
\textbf{\underline {Solution}:}\\
\newline $\overline{AB} =\sqrt{(4-3)^2+(6-2)^2}=\sqrt{(1)^2+(4)^2}=\sqrt{1+16}=\sqrt{17}$
\newline $\overline{BC} =\sqrt{(0-4)^2+(-8-6)^2}=\sqrt{(-4)^2+(-14)^2}=\sqrt{16+196}=\sqrt{212}=\sqrt{4}\sqrt{53}=2\sqrt{53}$
\newline $\overline{AC} =\sqrt{(0-3)^2+(-8-2)^2}=\sqrt{(-3)^2+(-10)^2}=\sqrt{9+100}=\sqrt{109}$
\\
\newline 11. A=$(-2, 3), B=(7, -2), C=(2, 5)$\\
\textbf{\underline {Solution}:}\\
\newline $\overline{AB} =\sqrt{(7-(-2))^2+(-2-3)^2}=\sqrt{(7+2)^2+(-5)^2}=\sqrt{9^2+25}=\sqrt{81+25}=\sqrt{106}=\sqrt{4}\sqrt{26}=2\sqrt{26}$
\newline $\overline{BC} =\sqrt{(2-7)^2+(5-(-2))^2}=\sqrt{(-5)^2+(5+2)^2}=\sqrt{25+7^2}=\sqrt{25+49}=\sqrt{74}$
\newline $\overline{AC} =\sqrt{(2-(-2))^2+(5-3)^2}=\sqrt{(2+2)^2+(2)^2}=\sqrt{4^2+4}=\sqrt{16+4}=\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}$
\\
\newline 12. A=$(2, -1), B=(-1, 4), C=(5, -6)$\\
\textbf{\underline {Solution}:}\\
\newline $\overline{AB} =\sqrt{(-1-2)^2+(4-(-1))^2}=\sqrt{(-3)^2+(4+1)^2}=\sqrt{9+(5)^2}=\sqrt{9+25}=\sqrt{34}$
\newline $\overline{BC} =\sqrt{(5-(-1))^2+(-6-4)^2}=\sqrt{(5+1)^2+(-10)^2}=\sqrt{(6)^2+100}==\sqrt{36+100}=\sqrt{136}=4\sqrt{34}$
\newline $\overline{AC} =\sqrt{(5-2)^2+(-6-(-1))^2}=\sqrt{(3)^2+(-6+1)^2}=\sqrt{9+(-5)^2}=\sqrt{9+25}=\sqrt{34}$
\\
\newline 13. A=$(1, -1), B=(3, 3), C=(0, -3)$\\
\textbf{\underline {Solution}:}\\
\newline $\overline{AB} =\sqrt{(3-1)^2+(3-(-1))^2}=\sqrt{(2)^2+(3+1)^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}$
\newline $\overline{BC} =\sqrt{(0-3)^2+(-3-3)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45}=\sqrt{9}\sqrt{5}=3\sqrt{5}$
\newline $\overline{AC} =\sqrt{(0-1)^2+(-3-(-1))^2}=\sqrt{(-1)^2+(-3+1)^2}=\sqrt{1+(-2)^2}=\sqrt{1+4}=\sqrt{5}$
\\
\newline 14. A=$(1, \sqrt{2}), B=(4, 3\sqrt{2}), C=(10, 6\sqrt{2})$\\
\textbf{\underline {Solution}:}\\
\newline $\overline{AB} =\sqrt{(4-1)^2+(3\sqrt{2}-\sqrt{2})^2}=\sqrt{(3)^2+(2\sqrt{2})^2}=\sqrt{9+(2)^2)(\sqrt{2})^2}=\sqrt{9+(4)(2)}=\sqrt{9+8}=\sqrt{17}$
\newline $\overline{BC} =\sqrt{(10-4)^2+(6\sqrt{2}-3\sqrt{2})^2}=\sqrt{(6)^2+(3\sqrt{2})^2}=\sqrt{36+(3)^2(\sqrt{2})^2}=\sqrt{36+(9)(2)}=\sqrt{36+18}=\sqrt{54}$
\newline $\overline{AC} =\sqrt{(10-1)^2+(6\sqrt{2}-3\sqrt{2})^2}=\sqrt{(9)^2+(3\sqrt{2})^2}=\sqrt{81+(3)^2(\sqrt{2})^2}=\sqrt{81+(9)(2)}=\sqrt{81+18}=\sqrt{99}=\sqrt{9}\sqrt{11}=3\sqrt{11}$
\\
\newline \textit{In Problems 15-18, determine whether the three given points are the vertices of a right triangle}
\begin{quote}
[Given 3 points A, B, C, these points create a triangle if $\overline{AC} < \overline{AB}+\overline{BC}$]
\end{quote}
29. Show that a triangle with vertices $(x1, y1)$, $(x2,y2)$, $(x3, y3)$ has area
\begin{align}
\frac{1}{2}
\begin{vmatrix}
x_1y_2 +x_2y_3 +x_3y_1 - x_1y_3 -x_2y_1 -x_3y_2
\end{vmatrix}
=
\begin{vmatrix}
\frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1\\
x_2 & y_2 & 1\\
x_3 & y_3 & 1
\end{vmatrix}
\end{vmatrix}
\end{align}
[Hint: Consider the rectangle with sides parallel to the coordinate axes and containing the vertices of the triangle.]
30. Prove analytically that if the diagonals of a parallelogram are equal, then the parallelogram is a rectangle.\\
[Hint: Place the axes as shown in fig 1.10 and show that $\overflow{AC} = \overflow{BD}$ implies that $A$ is the origin]
31. Prove analytically that the sum of the lengths of two sides of a triangle is greater than the length of the third side.
\end{document}
The log output
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Underfull \hbox (badness 10000) in paragraph at lines 108--167
[]
Underfull \hbox (badness 10000) in paragraph at lines 108--167
[]
Underfull \hbox (badness 10000) in paragraph at lines 108--167
[]
[3]
! Missing number, treated as zero.
<to be read again>
H
l.185 ...flow{BD}$ implies that $A$ is the origin]
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)
! Illegal unit of measure (pt inserted).
<to be read again>
H
l.185 ...flow{BD}$ implies that $A$ is the origin]
Dimensions can be in units of em, ex, in, pt, pc,
cm, mm, dd, cc, nd, nc, bp, or sp; but yours is a new one!
I'll assume that you meant to say pt, for printer's points.
To recover gracefully from this error, it's best to
delete the erroneous units; e.g., type `2' to delete
two letters. (See Chapter 27 of The TeXbook.)
! Undefined control sequence.
<recently read> \overflow
l.185 ...flow{BD}$ implies that $A$ is the origin]
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
! Undefined control sequence.
<argument> ...ow that $\overflow {AC} = \overflow
{BD}$ implies that $A$ is ...
l.185 ...flow{BD}$ implies that $A$ is the origin]
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
Underfull \hbox (badness 10000) in paragraph at lines 184--186
[]
[4] (Section1.2_Problems.aux) )
Here is how much of TeX's memory you used:
3376 strings out of 478927
45318 string characters out of 2859123
389184 words of memory out of 3000000
21221 multiletter control sequences out of 15000+600000
405783 words of font info for 37 fonts, out of 8000000 for 9000
1141 hyphenation exceptions out of 8191
57i,20n,66p,283b,230s stack positions out of 5000i,500n,10000p,200000b,80000s
<C:/Applications/Editors/MiKTeX/fonts/type1/publ
ic/amsfonts/cm/cmbx10.pfb><C:/Applications/Editors/MiKTeX/fonts/type1/public/am
sfonts/cm/cmex10.pfb><C:/Applications/Editors/MiKTeX/fonts/type1/public/amsfont
s/cm/cmmi10.pfb><C:/Applications/Editors/MiKTeX/fonts/type1/public/amsfonts/cm/
cmr10.pfb><C:/Applications/Editors/MiKTeX/fonts/type1/public/amsfonts/cm/cmr7.p
fb><C:/Applications/Editors/MiKTeX/fonts/type1/public/amsfonts/cm/cmsy10.pfb><C
:/Applications/Editors/MiKTeX/fonts/type1/public/amsfonts/cm/cmsy7.pfb><C:/Appl
ications/Editors/MiKTeX/fonts/type1/public/amsfonts/cm/cmti10.pfb><C:/Applicati
ons/Editors/MiKTeX/fonts/type1/public/amsfonts/symbols/msam10.pfb>
Output written on W:\PROJECTS\math\analyticgeometry\build\Section1.2_Problems.p
df (4 pages, 127594 bytes).
PDF statistics:
60 PDF objects out of 1000 (max. 8388607)
0 named destinations out of 1000 (max. 500000)
1 words of extra memory for PDF output out of 10000 (max. 10000000)
Best Answer
Your first error is due to this in your code:
LaTeX assumes the the bracket
[starts the optional argument of\\and so looks for a number. Add for example an empty brace group to stop this\\{}.And don't use so many
\\, you get tons of underfull box messages in your code (and your output looks rather ugly):Use empty lines to separate paragraphs, or use a list.