Matemática, perguntado por jtcgrzyfd7, 4 meses atrás

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

Soluções para a tarefa

Respondido por solkarped
2

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

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

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

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

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

  • Recuperar o coeficiente angular da reta "s":

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

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

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

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

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

  • Comparar os coeficientes angulares das retas:

        Verificando se as retas são paralelas:

        Dizemos que:

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

         Então:

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

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

         Verificando se as retas são perpendiculares:

          Dizemos que:

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

          Então:

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

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

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

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

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

✅ Portanto, as retas são:

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

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

   

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

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