Para hallar la pendiente recordemos el siguiente teorema:
[tex]\underset{\boxed{\begin{array}{c}\begin{array}{l}\rm{Si\ P_1(x_1,y_1)\ y \ P_2(x_,y_2)\ son\ dos\ puntos\ diferentes\ cualesquiera\ de\ una\ recta,}\\\vphantom{\bigg|}\rm{la\ pendiente\ de\ la\ recta\ es:}\end{array}\\\begin{array}{c}\sf{m=\dfrac{y_1 - y_2}{x_1 - x_2},\quad x_1\not= x_2}\end{array}\end{array}}}{\vphantom{\big|}\underline{\boldsymbol{\sf{Teorema}}}}[/tex]
Nuestros puntos son:
[tex]\begin{array}{cccccccccc}\boldsymbol{\bigcirc \kern-8pt \triangleright} \quad\quad\sf{(\underbrace{1}_{\boldsymbol{\sf{x_1}}},\overbrace{2}^{\boldsymbol{\sf{y_1}}})}&&&&&&&&&\boldsymbol{\bigcirc \kern-8pt \triangleright} \quad\quad\sf{(\underbrace{-2}_{\boldsymbol{\sf{x_2}}},\overbrace{3}^{\boldsymbol{\sf{y_2}}})}\end{array}[/tex]
Reemplazamos
[tex]\begin{array}{c}\sf{m=\dfrac{y_2 -y_1}{x_2 - x_1}}\\\\\sf{m=\dfrac{\left(3\right)-\left(2\right)}{\left(-2\right)-\left(1\right)}}\\\\\sf{m=\dfrac{3\ -2}{-2\ -1}}\\\\\sf{m=\dfrac{1\ }{-3\ }}\\\\\boxed{\boxed{\boldsymbol{\sf{m=-\dfrac{1}{3}}}}}\end{array}[/tex]
Rpta. La pendiente es -1/3. Alternativa a).
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[tex]\boxed{\sf{{R}}\quad\raisebox{10pt}{$\sf{\red{O}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{O}}$}\quad\raisebox{15pt}{$\sf{{G}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{G}}$}\quad\raisebox{15pt}{$\sf{\red{H}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{H}}$}\quad\raisebox{10pt}{$\sf{{E}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{E}}$}\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]
