Réponse :
Bonjour,
[tex]A(7;-1), \ B(-4; 10), \ C(-4;5), \ M(x ; y)[/tex]
1)
[tex]\overrightarrow{MA} \left(\begin{array}{c}x_A-x_M\\y_A-y_M\end{array}\right) \ ; \ \overrightarrow{MA} \left(\begin{array}{c}7-x_M\\-1-y_M\end{array}\right)\\\\\\\overrightarrow{MB} \left(\begin{array}{c}x_B-x_M\\y_B-y_M\end{array}\right)\ ; \ \overrightarrow{MB} \left(\begin{array}{c}-4-x_M\\10-y_M\end{array}\right)\\\\\\\overrightarrow{MC} \left(\begin{array}{c}x_C-x_M\\y_C-y_M\end{array}\right)\ ; \ \overrightarrow{MC} \left(\begin{array}{c}-4-x_M\\5-y_M\end{array}\right)[/tex]
2)
[tex]\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} = \vec0\\\\\left(\begin{array}{cc}7-x_M - 4 - x_M - 4 - x_M\\-1-y_m+10-y_M+5-y_M\end{array}\right) = \vec0\\\\\\\left(\begin{array}{cc}-3x_M - 1\\-3y_M+14\end{array}\right) = \vec0[/tex]
[tex]\begin{cases} -3x_M -1 = 0 \\ -3y_M + 14 = 0 \end{cases}\\\\\\\begin{cases} -3x_M = 1 \\ -3y_M = -14 \end{cases}\\\\\\\begin{cases} x_M = -\dfrac{1}{3} \\ y_M = \dfrac{14}{3} \end{cases}[/tex]
[tex]D'o\`u \ M \left(-\dfrac{1}{3};\dfrac{14}{3} \right)[/tex]
