Cho tam giác ABC có \(C\left( { - 1; - 1} \right);\,AB = \sqrt 5 .\) Phương trình đường thẳng AB : \(x + 2y - 3 = 0\). Trọng tâm \(G \in \left( \Delta \right):\,\,x + y - 2 = 0.\) Tìm A, B.
A.\(\left[ \begin{array}{l}A\left( {4; \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {4; \frac{1}{2}} \right)\end{array} \right.\)
B.\(\left[ \begin{array}{l}A\left( {4; - \frac{1}{2}} \right);\,\,B\left( {6; \frac{3}{2}} \right)\\A\left( {6; \frac{3}{2}} \right);\,\,B\left( {4; - \frac{1}{2}} \right)\end{array} \right.\)
C.\(\left[ \begin{array}{l}A\left( {4; - \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {4; - \frac{1}{2}} \right)\end{array} \right.\)
D.\(\left[ \begin{array}{l}A\left( {-4; - \frac{1}{2}} \right);\,\,B\left( {6; - \frac{3}{2}} \right)\\A\left( {6; - \frac{3}{2}} \right);\,\,B\left( {-4; - \frac{1}{2}} \right)\end{array} \right.\)