Question

6 Sendo G o baricentro do triângulo de vértices A(23, 22), B(4, 9) e C(8, 21), determine o ponto médio do segmento AG.

Answer 1

coordenadas:

X = 52/3

Y = 59/3

Answer 2

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

                 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf M_{\overline{AG}}\bigg(\frac{52}{3},\,\frac{59}{3}\bigg)\:\:\:}}\end{gathered} }$

Sejam os dados:

                              $\Large\begin{cases} A(23, 22)\\B(4, 9)\\C(8, 21)\\G = \:?\\M_{\overline{AG}} = \:?\end{cases}$

Para resolver esta questão devemos:

• Calcular o baricentro do triângulo. Para isso, devemos calcular a média aritmética das abscissas dos vértices do triângulo, bem como a média aritmética das ordenadas dos vértices. Então, temos:

             $\Large\displaystyle\begin{gathered} G = (X_{G},\,Y_{G})\end{gathered}$

                  $\Large\displaystyle\text{ \begin{gathered} = \bigg(\frac{X_{A} + X_{B} + X_{C}}{3},\,\frac{Y_{A} + Y_{B} + Y_{C}}{3}\bigg)\end{gathered} }$

                  $\Large\displaystyle\begin{gathered} = \bigg(\frac{23 + 4 + 8}{3},\,\frac{22 + 9 + 21}{3}\bigg)\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \bigg(\frac{35}{3},\,\frac{52}{3}\bigg)\end{gathered}$

       Portanto, o baricentro "G" é:

                   $\Large\displaystyle\begin{gathered} G \bigg(\frac{35}{3},\,\frac{52}{3}\bigg)\end{gathered}$

• Calcular o ponto médio do segmento AG.

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

                       $\Large\displaystyle\text{ \begin{gathered} = \left(\frac{23 + \dfrac{35}{3}}{2},\,\frac{22 + \dfrac{52}{3}}{2}\right)\end{gathered} }$

                        $\Large\displaystyle\text{ \begin{gathered} = \left(\frac{\dfrac{69 + 35}{3}}{2},\,\frac{\dfrac{66 + 52}{3}}{2}\right)\end{gathered} }$

                        $\Large\displaystyle\text{ \begin{gathered} = \left(\frac{\dfrac{104}{3}}{2},\,\frac{\dfrac{118}{3}}{2}\right)\end{gathered} }$

                        $\Large\displaystyle\begin{gathered} = \bigg(\frac{104}{3}\cdot\frac{1}{2},\,\frac{118}{3}\cdot\frac{1}{2}\bigg)\end{gathered}$

                         $\Large\displaystyle\begin{gathered} = \bigg(\frac{104}{6},\,\frac{118}{6}\bigg)\end{gathered}$

                         $\Large\displaystyle\begin{gathered} = \bigg(\frac{52}{3},\,\frac{59}{3}\bigg)\end{gathered}$

✅ Portanto, o ponto médio do segmento "AG" é:

                     $\Large\displaystyle\text{ \begin{gathered} M_{\overline{AG}}\bigg(\frac{52}{3},\,\frac{59}{3}\bigg)\end{gathered} }$

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

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