Question

Calcule k de modo que a reta que passa por A(1,1) e B(k + 1, 2k) tenha inclinação α = 60º, relativamente ao eixo x.

Answer 1

Equação de uma reta :

$\text y-\text y_\text o =\text{tg}(\alpha)( \text x-\text x_o) \\\\ \underline{\text{onde}}: \\ \alpha = \text{inclina{\c c}{\~a}o da reta} \\\ \text x_\text o,\text y_\text o=\text{um ponto qualquer por onde a reta passa}$

Temos :

$\displaystyle \alpha = 60^\circ \\\\ \text A(1,1) \ ; \ \text B(\text k+1,\ 2\text k) \\\\ \underline{\text{Equa{\c c}{\~a}o da reta}}: \\\\ \text y-1=\text{tg}(60^\circ)(\text x-1) \\\\ \text y-1=\sqrt{3}.\text x-\sqrt{3} \\\\ \underline{\text{Substituindo o ponto B}} : \\\\ 2\text k-1=\sqrt{3}(\text k+1)-\sqrt{3} \\\\ 2\text k -1 = \sqrt{3}.\text k+\sqrt{3}-\sqrt{3} \\\\ \text k(2-\sqrt{3})=1$

$\displaystyle \text k = \frac{1}{2-\sqrt{3}} \to \text k =\frac{1}{(2-\sqrt{3})}\frac{(2+\sqrt{3})}{(2+\sqrt3)} \\\\\\ \text k = \frac{2+\sqrt{3}}{4-3} \to \text k = \frac{2+\sqrt{3}}{1}\\\\\\ \huge\boxed{\text k = 2+\sqrt{3}\ } \checkmark$

Answer 2

✅ Após resolver todos os cálculos, concluímos que o valor numérico do parâmetro "k" que verifica a inclinação da reta dada é:

            $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf k = 2 + \sqrt{3}\:\:\:}}\end{gathered} }$

Sejam os dados:

         $\Large\begin{cases} A = (1, 1)\\B = (k + 1,\,2k)\\\alpha = 60^{\circ}\\k = \:?\end{cases}$

Sabemos que a inclinação - declividade -  de uma reta é o ângulo que esta forma com o eixo das abscissas em seu sentido positivo. Para calcular a medida deste ângulo, devemos obter a medida do arco cuja tangente vale a medida do coeficiente angular, ou seja:

$\Large\displaystyle\begin{gathered} \bf I\end{gathered}$                $\Large\displaystyle\begin{gathered} \alpha = \arctan(m_{r})\end{gathered}$

Desenvolvendo a equação "I", temos:

                  $\Large\displaystyle\begin{gathered} \alpha = \arctan(\tan \alpha)\end{gathered}$

                  $\Large\displaystyle\begin{gathered} \alpha = \arctan\bigg(\frac{\sin\alpha}{\cos\alpha}\bigg)\end{gathered}$

$\Large\displaystyle\begin{gathered} \bf II\end{gathered}$               $\Large\displaystyle\text{ \begin{gathered} \alpha = \arctan\bigg(\frac{y_{B} - y_{A}}{x_{B} - x_{A}}\bigg)\end{gathered} }$

Substituindo os dados na equação "II", temos:

                 $\Large\displaystyle\begin{gathered} 60^{\circ} = \arctan\bigg(\frac{2k - 1}{(k + 1) - 1}\bigg)\end{gathered}$

                 $\Large\displaystyle\begin{gathered} 60^{\circ} = \arctan\bigg(\frac{2k - 1}{k + 1 - 1}\bigg)\end{gathered}$

$\Large\displaystyle\begin{gathered} \bf III\end{gathered}$           $\Large\displaystyle\begin{gathered} 60^{\circ} = \arctan\bigg(\frac{2k - 1}{k}\bigg)\end{gathered}$

Sabendo que:

    $\Large\displaystyle\begin{gathered} \tan 60^{\circ} = \sqrt{3} \:\:\:\Longrightarrow \:\:\:60^{\circ} = \arctan(\sqrt{3})\end{gathered}$

Então, temos:

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

                   $\Large\displaystyle\begin{gathered} \sqrt{3} = \frac{2k - 1}{k}\end{gathered}$

Para facilitar a visualização dos cálculos, podemos inverter os membros sem perda alguma de generalidade. Então temos:

           $\Large\displaystyle\begin{gathered} \frac{2k - 1}{k} = \sqrt{3}\end{gathered}$

            $\Large\displaystyle\begin{gathered} 2k - 1 = \sqrt{3}k\end{gathered}$

     $\Large\displaystyle\begin{gathered} 2k - \sqrt{3}k = 1\end{gathered}$

    $\Large\displaystyle\begin{gathered} (2 - \sqrt{3})k = 1\end{gathered}$

                        $\Large\displaystyle\text{ \begin{gathered} k = \frac{1}{2 - \sqrt{3}}\end{gathered} }$

                        $\Large\displaystyle\text{ \begin{gathered} k = \frac{1}{2 - \sqrt{3}}\cdot\frac{2 + \sqrt{3}}{2 + \sqrt{3}}\end{gathered} }$

                        $\Large\displaystyle\text{ \begin{gathered} k = \frac{2 + \sqrt{3}}{2^{2} - (\sqrt{3})^{2}}\end{gathered} }$

                         $\Large\displaystyle\text{ \begin{gathered} k = \frac{2 + \sqrt{3}}{4 - 3}\end{gathered} }$

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

                         $\Large\displaystyle\begin{gathered} k = 2 + \sqrt{3}\end{gathered}$

✅ Portanto, "k" é:

                         $\Large\displaystyle\begin{gathered} k = 2 + \sqrt{3}\end{gathered}$

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

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