Question

14. Determinar o ponto médio do segmento AB̅̅̅̅ , quando:
a) A(1, 7) e B(11, 3)

Answer 1

✅ Após resolver os cálculos, concluímos que o ponto médio do referido segmento retilíneo é:

           $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf M_{\overline{AB}} = (6,\,5)\:\:\:}}\end{gathered} }$    

 

Sejam os Pontos:

                       $\Large\begin{cases} A(1, 7)\\B(11, 3)\end{cases}$

Se estamos calcular o ponto médio "M" do segmente AB, então, temos:

     $\Large\displaystyle\text{ \begin{gathered} M_{\overline{AB}} = \bigg(\frac{X_{A} + X_{B}}{2},\,\frac{Y_{A} + Y_{B}}{2}\bigg)\end{gathered} }$

                 $\Large\displaystyle\begin{gathered} = \bigg(\frac{1 + 11}{2},\,\frac{7 + 3}{2}\bigg)\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \bigg(\frac{12}{2},\,\frac{10}{2}\bigg)\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = (6,\,5)\end{gathered}$

✅ Portanto, o ponto médio procurado é:

      $\Large\displaystyle\text{ \begin{gathered} M_{\overline{AB}} = (6, \,5)\end{gathered} }$

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Veja a solução gráfica representada na figura:

Answer 2

$\sf M=\left(\sf\dfrac{x_A+x_B}{2},\dfrac{y_A+y_B}{2}\right)$

$\sf M=\left(\sf\dfrac{1+11}{2},\dfrac{7+3}{2}\right)$

$\sf M=\left(\sf\dfrac{12}{2},\dfrac{10}{2}\right)$

$\red{\sf M= \big( 6,5\big)}$