Trong mặt phẳng $Oxy$, cho $A\left( { - 2;0} \right),\;B\left( {5; - 4} \right),\;C\left( { - 5;1} \right)$. Tọa độ điểm $D$ để tứ giác $BCAD$ là hình bình hành là:

\(\left( {i;j} \right)\)

\(\left( {\overrightarrow i ;\overrightarrow j } \right)\)

\(\left( {\overrightarrow m ;\overrightarrow n } \right)\)

\(\left( {m;n} \right)\)

$\left\{ \begin{array}{l}{u_1} = {u_2}\\{v_1} = {v_2}\end{array} \right.$.

$\left\{ \begin{array}{l}{u_1} =  - {v_1}\\{u_2} =  - {v_2}\end{array} \right.$

$\left\{ \begin{array}{l}{u_1} = {v_1}\\{u_2} = {v_2}\end{array} \right.$.

$\left\{ \begin{array}{l}{u_1} = {v_2}\\{u_2} = {v_1}\end{array} \right.$.

\(\overrightarrow u  =  - \overrightarrow i \)

\(\overrightarrow u  =  - \overrightarrow j \)

\(\overrightarrow u  = \overrightarrow i \)

\(\overrightarrow u  = \overrightarrow j \)

$\overrightarrow {AB}  = \left( {{y_A} - {x_A};{y_B} - {x_B}} \right)$.

$\overrightarrow {AB}  = \left( {{x_A} + {x_B};{y_A} + {y_B}} \right)$

$\overrightarrow {AB}  = \left( {{x_A} - {x_B};{y_A} - {y_B}} \right)$.

$\overrightarrow {AB}  = \left( {{x_B} - {x_A};{y_B} - {y_A}} \right)$.

\(\overrightarrow {OM}  = \left( { - 3; - 1} \right)\)

\(\overrightarrow {OM}  = \left( {3;1} \right)\)         

\(\overrightarrow {MO}  = \left( { - 3;1} \right)\)

\(\overrightarrow {OM}  = \left( { - 3;1} \right)\)

$\left( {2;4} \right)$.

$\left( {5;6} \right)$.

$\left( {15;10} \right)$.

$\left( {50;6} \right)$.

\(\overrightarrow {OM}  =  - 2\overrightarrow i  + 4\overrightarrow j \)

\(\overrightarrow {OM}  = 2\overrightarrow i  - 4\overrightarrow j \)

\(\overrightarrow {OM}  = 2\overrightarrow j  - 4\overrightarrow i \)

\(\overrightarrow {OM}  =  - 2\overrightarrow i  - 4\overrightarrow j \)