Bonjour,
Soit [tex]D(x_D,y_D)[/tex]
a) ABCD est un parallélogramme donc [tex]\vec{AB}=\vec{DC}[/tex] :
[tex](x_B-x_A,y_B-y_A)=(x_C-x_D,y_C-y_D)\\\\\Longleftrightarrow \left \{ {{x_B-x_A=x_C-x_D} \atop {y_B-y_A=y_C-y_D}} \right. \\\\
\Longleftrightarrow \left \{ {{-2-0=3-x_D} \atop {1-4=-1-y_D}} \right. \\\\
\Longleftrightarrow \left \{ {{x_D=5} \atop {y_D=2}} \right.\\\\
\Longleftrightarrow \boxed{D(5,2)} [/tex]
b)
[tex]I( \frac{x_C+x_D}{2}, \frac{y_C+y_D}{2})\\\\
\Longleftrightarrow I( \frac{3+5}{2}, \frac{-1+2}{2})\\\\
\Longleftrightarrow \boxed{I\left(4, \frac{1}{2}\right)} [/tex]
c)
[tex]\vec{BJ}}=2\vec{BI}\\\\
\Longleftrightarrow \left \{ {{x_J-x_B=2(x_I-x_B)} \atop {y_J-y_B=2(y_I-y_B)}} \right. \\\\
\Longleftrightarrow \left \{ {{x_J-(-2)=2(4-(-2))} \atop {y_J-1=2(0,5-1)}} \right. \\\\
\Longleftrightarrow \left \{ {{x_J=10} \atop {y_J=0}} \right. \\\\
\Longleftrightarrow J(10,0)[/tex]
d)
[tex]\vec{AJ}=k\vec{AD}\\\\
\Longleftrightarrow \left \{ {{x_J-x_A=k(x_D-x_A)} \atop {y_J-y_A=k(y_D-y_A)}} \right. \\\\
\Longleftrightarrow \left \{ {{10-0=k(5-0)} \atop {0-4=k(2-4)}} \right. \\\\
\Longleftrightarrow \boxed{k= 2}[/tex]
