Question

Estude a posíção relativa dos pares de retas. 3x-2y+1=0 e 4x+ 6y-1= 0
Alguém?

Answer 1

✅ Após realizar os cálculos e as devidas análises, concluímos que as referidas retas são:

                $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Perpendiculares\:\:\:}}\end{gathered} }$

Sejam as equações das retas:

                    $\Large\begin{cases} r: 3x - 2y + 1 = 0\\s: 4x + 6y - 1 = 0\end{cases}$

Para analisarmos a posição relativa dessas retas devemos comparar  seus coeficientes angulares. Para isso, fazemos:

• Recuperar o coeficiente angular da reta "r":

            $\Large\displaystyle\begin{gathered} 3x - 2y + 1 = 0\end{gathered}$

                             $\Large\displaystyle\begin{gathered} -2y = -3x - 1\end{gathered}$

                                 $\Large\displaystyle\begin{gathered} 2y = 3x + 1\end{gathered}$

                                   $\Large\displaystyle\begin{gathered} y = \frac{3}{2}x + \frac{1}{2}\end{gathered}$

                                 $\Large\displaystyle\begin{gathered} \therefore\:\:\:m_{r} = \frac{3}{2}\end{gathered}$

• Recuperar o coeficiente angular da reta "s":

             $\Large\displaystyle\begin{gathered} 4x + 6y - 1 = 0\end{gathered}$

                                  $\Large\displaystyle\begin{gathered} 6y = -4x + 1\end{gathered}$

                                     $\Large\displaystyle\begin{gathered} y = -\frac{4}{6}x + \frac{1}{6}\end{gathered}$

                                     $\Large\displaystyle\begin{gathered} y = -\frac{2}{3}x + \frac{1}{6}\end{gathered}$

                                    $\Large\displaystyle\begin{gathered} \therefore\:\:\:m_{s} = -\frac{2}{3}\end{gathered}$

• Comparar os coeficientes angulares das retas:

        Verificando se as retas são paralelas:

        Dizemos que:

                           $\Large\displaystyle\begin{gathered} r\parallel s \Longleftrightarrow m_{r} = m_{s}\end{gathered}$

         Então:

           $\Large\displaystyle\begin{gathered} \textrm{Se}\:\:m_{r} = \frac{3}{2}\:\:e\:\:m_{s} = -\frac{2}{3}\Longleftrightarrow m_{r} \neq m_{s}\end{gathered}$

                                       $\Large\displaystyle\begin{gathered} \therefore\:\:\:r\nparallel s\end{gathered}$

         Verificando se as retas são perpendiculares:

          Dizemos que:

                          $\Large\displaystyle\begin{gathered} r \perp s \Longleftrightarrow m_{r}\cdot m_{s} = -1\end{gathered}$

          Então:

                                      $\Large\displaystyle\begin{gathered} m_{r}\cdot m_{s} = -1\end{gathered}$

                               $\Large\displaystyle\begin{gathered} \frac{3}{2}\cdot\bigg(-\frac{2}{3}\bigg) = -1\end{gathered}$

                                                $\Large\displaystyle\begin{gathered} -\frac{6}{6} = -1\end{gathered}$

                                                 $\Large\displaystyle\begin{gathered} -1 = -1\end{gathered}$

                                                $\Large\displaystyle\begin{gathered} \therefore\:\:\:r \perp s\end{gathered}$

✅ Portanto, as retas são:

                                        $\Large\displaystyle\begin{gathered} \textrm{Perpendiculares}\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

   

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$