Question

à equação da reta perpendicular à reta y = -2x + 5 e que contém o ponto médio entre (1, -9) e (7, -3) é:

a) y = x/2 + 2
b) y = -2x + 2
c) y = -x/3 + 5
d) y = -2x - 16
e) y = x/2 -8

Answer 1

✅ Após resolver todos os cálculos, concluímos que a equação reduzida da reta "ρ", perpendicular à reta "r" que contém o ponto médio do segmento retilíneo "AB" é:

                      $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \rho: y = \frac{x}{2} - 8\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

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

Sejam os dados:

                         $\Large\begin{cases} r: y = -2x + 5\\A(1, -9)\\B(7, -3)\\\end{cases}$

Para resolver esta questão devemos:

• Recuperar o coeficiente angular da reta "r".

        se a reta "r" é:

                      $\Large\displaystyle\begin{gathered} r: y = -2x + 5\end{gathered}$

       Então o coeficiente angular da reta "r" é:

                               $\Large\displaystyle\begin{gathered} m_{r} = -2\end{gathered}$    

• Determinar o coeficiente angular da reta "ρ" perpendicular a "r".

         $\Large\displaystyle\text{ \begin{gathered} \textrm{Se}\:\rho\perp r\Longrightarrow m_{\rho}\cdot m_{r} = -1 \longrightarrow m_{\rho} = -\frac{1}{m_{r}}\end{gathered} }$

       Então:

                          $\Large\displaystyle\begin{gathered} m_{\rho} = \frac{-1}{-2} = \frac{1}{2}\end{gathered}$

       Portanto, o coeficiente angular da reta "ρ" é:

                                  $\Large\displaystyle\begin{gathered} m_{\rho} = \frac{1}{2}\end{gathered}$

• Obter o ponto médio "M" do segmento retilíneo "AB".

                    $\Large\displaystyle\begin{gathered} M = (x_{M},\,y_{M})\end{gathered}$

                           $\Large\displaystyle\text{ \begin{gathered} = \bigg(\frac{x_{A} + x_{B}}{2},\,\frac{y_{A} + y_{B}}{2}\bigg)\end{gathered} }$

                           $\Large\displaystyle\begin{gathered} = \bigg(\frac{1 + 7}{2},\,\frac{-9 + (-3)}{2}\bigg)\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = \bigg(\frac{8}{2},\,-\frac{12}{2}\bigg)\end{gathered}$

                            $\Large\displaystyle\begin{gathered} = (4, -6)\end{gathered}$

       Portanto, o ponto médio do segmento é:

                            $\Large\displaystyle\begin{gathered} M(4, -6)\end{gathered}$

• Montar a equação da reta "ρ". Para isso, devemos utilizar a fórmula do ponto/declividade, ou seja:

                $\Large\displaystyle\begin{gathered} y - y_{M} = m_{\rho}\cdot(x - x_{M})\end{gathered}$

            $\Large\displaystyle\begin{gathered} y - (-6) = \frac{1}{2}\cdot(x - 4)\end{gathered}$  

                     $\Large\displaystyle\begin{gathered} y + 6 = \frac{x}{2} - \frac{4}{2}\end{gathered}$

                              $\Large\displaystyle\begin{gathered} y = \frac{x}{2} - \frac{4}{2} - 6\end{gathered}$

                              $\Large\displaystyle\begin{gathered} y = \frac{x - 4 - 12}{2}\end{gathered}$

                              $\Large\displaystyle\begin{gathered} y = \frac{x - 16}{2}\end{gathered}$

                              $\Large\displaystyle\begin{gathered} y = \frac{x}{2} - \frac{16}{2}\end{gathered}$

                              $\Large\displaystyle\begin{gathered} y = \frac{x}{2} - 8\end{gathered}$

✅ Portanto, a equação procurada da reta é:

                           $\Large\displaystyle\begin{gathered} \rho: y = \frac{x}{2} - 8\end{gathered}$

                         

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

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