Réponses

2014-03-02T22:48:36+01:00
Bonsoir,

1) a) M(0 ; -3)  ;  N(2 ; 3)  ;  P(-9 ; 0) ; Q(-1 ; -1)

\vec{NA}=\dfrac{1}{2}\vec{MN}\\\\(x_A-x_N;y_A-y_N)=\dfrac{1}{2}(x_N-x_M;y_N-y_M)}\\\\(x_A-2;y_A-3)=\dfrac{1}{2}(2-0;3+3)\\\\(x_A-2;y_A-3)=\dfrac{1}{2}(2;6)\\\\(x_A-2;y_A-3)=(1;3)\\\\\left\{\begin{matrix}x_A-2=1 \\y_A-3=3\end{matrix}\right.\ \ \ \ \ \left\{\begin{matrix}x_A=1+2 \\y_A=3+3\end{matrix}\right.\ \ \ \ \ \left\{\begin{matrix}x_A=3 \\y_A=6\end{matrix}\right.

D'où A (3 ; 6)

b) 
\vec{MB}=3\vec{MQ}\\\\(x_B-x_M;y_B-y_M)=3(x_Q-x_M;y_Q-y_M)}\\\\(x_B-0;y_B+3)=3(-1-0;-1+3)\\\\(x_B;y_B+3)=3(-1;2)\\\\(x_B;y_B+3)=(-3;6)\\\\\left\{\begin{matrix}x_B=-3 \\y_B+3=6\end{matrix}\right.\ \ \ \ \ \left\{\begin{matrix}x_B=-3 \\y_B=6-3\end{matrix}\right.\ \ \ \ \ \left\{\begin{matrix}x_B=-3 \\y_B=3\end{matrix}\right.

D'où B(-3 ; 3)

2) \vec{PA}:(x_A-x_P;y_A-y_P)\\\\\vec{PA} : (3+9;6-0)\\\\\vec{PA} : (12;6)\\\\\\\vec{PB}:(x_B-x_P;y_B-y_P)\\\\\vec{PB} : (-3+9;3-0)\\\\\vec{PB} : (6;3)


3) Les points P, A et B sont colinéaires car :

12\times3 - 6\times6 = 0