Warning: this is an htmlized version! The original is here, and the conversion rules are here.
% (find-angg "LATEX/2017-2-GA-P1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2017-2-GA-P1.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2017-2-GA-P1.pdf"))
% (defun b () (interactive) (find-zsh "bibtex 2017-2-GA-P1; makeindex 2017-2-GA-P1"))
% (defun e () (interactive) (find-LATEX "2017-2-GA-P1.tex"))
% (find-xpdfpage "~/LATEX/2017-2-GA-P1.pdf")
% (find-sh0 "cp -v  ~/LATEX/2017-2-GA-P1.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2017-2-GA-P1.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2017-2-GA-P1.pdf
%               file:///tmp/2017-2-GA-P1.pdf
%           file:///tmp/pen/2017-2-GA-P1.pdf
% http://angg.twu.net/LATEX/2017-2-GA-P1.pdf

\documentclass[oneside]{book}
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{pict2e}
\usepackage{color}                % (find-LATEX "edrx15.sty" "colors")
\usepackage{colorweb}             % (find-es "tex" "colorweb")
%\usepackage{tikz}
%
% (find-dn6 "preamble6.lua" "preamble0")
%\usepackage{proof}   % For derivation trees ("%:" lines)
%\input diagxy        % For 2D diagrams ("%D" lines)
%\xyoption{curve}     % For the ".curve=" feature in 2D diagrams
%
\usepackage{edrx15}               % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxchars.tex              % (find-LATEX "edrxchars.tex")
\input edrxgac2.tex               % (find-LATEX "edrxgac2.tex")
%
% (find-angg ".emacs.papers" "latexgeom")
% (find-latexgeomtext "total={6.5in,8.75in},")
\usepackage[%total={6.5in,4in},
%textwidth=4in,  paperwidth=4.5in,
%textheight=5in, paperheight=4.5in,
a4paper,
top=1.5in, left=1.5in%, includefoot
]{geometry}
%
\begin{document}

\catcode\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
%\directlua{output(preamble1)}
\def\expr#1{\directlua{output(tostring(#1))}}
\def\eval#1{\directlua{#1}}
\def\pu{\directlua{pu()}}

\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
\directlua{dofile "edrxpict.lua"} % (find-LATEX "edrxpict.lua")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end

\def\ang{\operatorname{ang}}

\def\V(#1){\VEC{#1}}

%   ____      _                    _ _
%  / ___|__ _| |__   ___  ___ __ _| | |__   ___
% | |   / _ | '_ \ / _ \/ __/ _ | | '_ \ / _ \
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
%  \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/
%

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2017.2
\par P1 - 22/nov/2017 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.
% \par Versão: 14/mar/2016
% \par \url{http://angg.twu.net/2015.2-C2.html} (página do curso)
% \par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
% \par {\tt eduardoochs@gmail.com} (meu e-mail)

}

\bsk
\bsk

\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B       (#1 pts){{\bf(#1 pts)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar

1) \T(Total: 1.0 pts) Prove que

a) \B(0.2 pts) se $\vv⊥\ww$ então $||\vv+\ww||^2 = ||\vv||^2 + ||\ww||^2$,

b) \B(0.3 pts) $||\vv+\ww||^2 = ||\vv||^2 + ||\ww||^2$ é falso em geral.

c) \B(0.5 pts) se $k<0$ então $\cos(\ang(k\vv,\ww)) = -\cos(\ang(\vv,\ww))$.

Dica: $\uu·\vv = \cos(\ang(\uu,\vv))\,||\uu||\,||\vv||$.

\bsk
\bsk

2) \T(Total: 2.0 pts) Sejam
%
$$\begin{array}{rcl} r_0 &=& \setofxyst{x=2}, \\ r &=& \setofxyst{\frac{x}{4} + \frac{y}{3}=1}, \\ S_0 &=& \setofst{P∈\R^2}{d(P,r_0)=1}, \\ S &=& \setofst{P∈\R^2}{d(P,r)=1}. \\ \end{array}$$

a) \B(0.1 pts) Represente graficamente $r_0$ e $S_0$.

b) \B(0.4 pts) Dê as coordenadas de um ponto de $S$.

c) \B(0.5 pts) O conjunto $S$ é a união de duas retas, $S=r_1∪r_2$.

Represente graficamente $r$, $r_1$, $r_2$.

d) \B(1.0 pts) Dê equações --- ou parametrizações --- para $r_1$ e
$r_2$.

\bsk
\bsk

3) \T(Total: 1.5 pts) Sejam $A=(0,1)$, $B=(3,1)$ e $C=(5,3)$.

Qual é o ângulo mais agudo do triângulo $ΔABC$?

\bsk
\bsk

4) \T(Total: 1.5 pts) Vamos definir as coordenadas $ab$'' da
seguinte forma: $(a,b)_Σ = (0,2) + a\V(2,-1) + b\V(0,1)$.

a) \B(0.3 pts) Represente graficamente $(0,0)_Σ$, $(1,0)_Σ$,
$(2,0)_Σ$, $(0,1)_Σ$, $(1,1)_Σ$, $(2,1)_Σ$, $(0,2)_Σ$, $(1,2)_Σ$ e
$(2,2)_Σ$.

b) \B(0.6 pts) Dê a equação da reta que contém $(0,0)_Σ$, $(1,0)_Σ$ e
$(2,0)_Σ$.

c) \B(0.6 pts) Dê a equação da reta que contém $(0,1)_Σ$, $(1,1)_Σ$ e
$(2,1)_Σ$.

\bsk
\bsk

5) \T(Total: 2.0 pts) Sejam $A=(0,1)$, $B=(0,4)$, $C=(2,0)$.

a) \B(0.2 pts) Calcule $\area(\Vec{AB}, \Vec{AC})$ (a área de um paralelogramo).

b) \B(0.3 pts) Calcule $\area(ΔABC)$ (a área de um triângulo).

c) \B(1.5 pts) Encontre uma fórmula para $\area(Δ(x,y)BC)$ e teste a
sua fórmula em pelo menos três pontos para os quais essa área seja
fácil de calcular no olhômetro. Dica: teste-a usando pontos que estão
tanto abaixo quanto acima da reta que contém $B$ e $C$.

\bsk
\bsk

6) \T(Total: 2.0 pts) Sejam $C$ o círculo de centro $(0,0)$ e raio 2 e
$C'$ o círculo de centro $(2,2)$ e raio 2. Encontre os dois pontos
$I_1,I_2∈C∩C'$ usando o método de subtrair as equações dos dois
círculos. Obs: com os círculos nestas posições as contas são bem
simples --- o importante é você deixar elas claras e nomear os objetos
que você construir.

\newpage

%   ____       _                _ _
%  / ___| __ _| |__   __ _ _ __(_) |_ ___
% | |  _ / _ | '_ \ / _ | '__| | __/ _ \
% | |_| | (_| | |_) | (_| | |  | | || (_) |
%  \____|\__,_|_.__/ \__,_|_|  |_|\__\___/
%

\msk

\unitlength=5pt
\def\closeddot{\circle*{0.6}}
\def\pictpoint#1{\put(#1){\closeddot}}
\def\pictline#1{{\linethickness{1.0pt}\expr{Line.new(#1):pict()}}}
\def\pictlinethin#1{{\linethickness{0.2pt}\expr{Line.new(#1):pict()}}}

%  _
% / |
% | |
% | |
% |_|
%
1a) Se $\vv⊥\ww$ então $\vv·\ww=0$, $||\vv+\ww||^2 = (\vv+\ww)·(\vv+\ww)$

$= \vv·\vv + 2\vv·\ww + \ww·\ww = \vv·\vv + \ww·\ww = ||\vv||^2 + ||\ww||^2$.

1b) $||\VEC{2,0}+\VEC{3,0}||^2 = 25$, $||\VEC{2,0}||^2 + ||\VEC{3,0}||^2 = 4+9 = 13$.

1c) $k<0$ então
$\cos(\ang(k\vv,\ww)) = \frac {k\vv·\ww} {||k\vv||\,||\ww||} = \frac {k\vv·\ww} {-k||\vv||\,||\ww||} = - \frac {\vv·\ww} {||\vv||\,||\ww||} = -\cos(\ang(\vv,\ww))$.

\bsk

%  ____
% |___ \
%   __) |
%  / __/
% |_____|
%

\unitlength=10pt

2a)
%
$\vcenter{\hbox{% \beginpicture(-1,-1)(4,3)% \pictaxes% \Line(1,-1)(1,3) \Line(2,-1)(2,3) \Line(3,-1)(3,3) \end{picture}% }}$

2b) $r$ passa por $(0,3)$ e tem $m=-\frac{3}{4}$; $\sqrt{1+m^2} = \frac{5}{4}$; ponto: $(0,3+\frac45)=(0,3.8)$.

2c)
%
$\vcenter{\hbox{% \beginpicture(-1,-1)(4,4)% \pictaxes% \pictline{v(0, 3), v(1, -3/4), 0, 4} % u = -1 \pictline{v(0, 3.8), v(1, -3/4), 0, 4} % u = -1 \pictline{v(0, 2.2), v(1, -3/4), 0, 4} % u = -1 \end{picture}% }}$

2d) $y=3.8-\frac34 x$, $y=2.2-\frac34 x$.

\bsk

%  _____
% |___ /
%   |_ \
%  ___) |
% |____/
%

\unitlength=10pt

3)
%
$\vcenter{\hbox{% \beginpicture(0,0)(5,3)% \pictaxes% \Line(0,1)(3,1)(5,3)(0,1) % \mygrid % \pictline{v(4, 0), v(0,1), -0.5, 4} % u = 0 % \pictline{v(6, 0), v(0,1), 0.5, 5} % u = 1 \end{picture}% }}$
%
$\begin{array}{cc} \cosang(B\hat A C) = \frac {\VEC{3,0}·\VEC{5,2}} {|| \VEC{3,0} || \, || \VEC{5,2} ||} = \frac {15} {3\sqrt{29}} \\ % \cosang(A\hat C B) = \frac {\VEC{-5,-2}·\VEC{-2,-2}} {|| \VEC{-5,-2} || \, || \VEC{-2,-2} ||} = \frac {14} {\sqrt{29}\sqrt{8}} \\ \end{array}$

$\cosang(B\hat A C) > \cosang(A\hat C B)$ portanto $B\hat A C$ é mais
agudo que $A\hat C B$.

% (/ 15 (sqrt (* 29 9)))
% (/ 14 (sqrt (* 29 8)))

\bsk

%  _  _
% | || |
% | || |_
% |__   _|
%    |_|
%
4a)
%
\def\closeddot{\circle*{0.4}}
\def\pictpoint(#1){\put(#1){\closeddot}}
%
$\vcenter{\hbox{% \beginpicture(-2,-1)(6,5)% \pictaxes% \Line(-2,3)(6,-1) \Line(-2,4)(6,0) \Line(-2,5)(6,1) \pictpoint(0,4) \pictpoint(2,3) \pictpoint(4,2) \pictpoint(0,3) \pictpoint(2,2) \pictpoint(4,1) \pictpoint(0,2) \pictpoint(2,1) \pictpoint(4,0) \end{picture}% }}$

4b) $y=2-\frac x 2$

4c) $y=3-\frac x 2$

\bsk

%  ____
% | ___|
% |___ \
%  ___) |
% |____/
%
5a) $\area(\Vec{AB}, \Vec{AC}) = \area(\VEC{0,3}, \VEC{2,-1}) = |(\vsm{0 & 3 \\ 2 & -1})| = |-6| = 6$

5b) $\area(ΔABC) = \frac12 \area(\Vec{AB}, \Vec{AC}) = 3$

5c) $\area(Δ(x,y)BC) = \frac12 \area(\Vec{(x,y)B}, \Vec{(x,y)C}) = \frac12 \area(\VEC{-x,4-y}, \VEC{2-x,-y})$

$= \frac12 |( \vsm{-x & 4-y \\ 2-x & -y } )| = \frac12 |xy - (4-y)(2-x)| = \frac12 |xy - 8 + 4x + 2y -xy| = \frac12 |4x + 2y - 8|$

$= |2x + y - 4|$

$\area(Δ(0,1)BC)=3$

$\area(Δ(2,0)BC)=0$

$\area(Δ(0,4)BC)=0$

\bsk

%   __
%  / /_
% | '_ \
% | (_) |
%  \___/
%
6) $C = \setofxyst{x^2 + y^2 - 4 = 0}$,

$C' = \setofxyst{(x-2)^2 + (y-2)^2 - 4 = 0} = \setofxyst{x^2-4x+4 + y^2-4y + 4 - 4 = 0} = \setofxyst{x^2-4x+ y^2-4y + 4 = 0}$

$r = \setofxyst{4x + 4y = 8} = \setofxyst{y = 2-x}$

As soluções de $x^2 + (2-x)^2 - 4 = 0$ são $x_1=0$ e $x_2=2$, daí
$(x_1,y_1)=(0,2)$ e $(x_2,y_2)=(2,0)$.

\end{document}

% Local Variables:
% coding: utf-8-unix
% End:
`