Đáp án:
$I(1;1); J(2;1)$
Giải thích các bước giải:
$I(x_I;y_I); J(x_J;y_J)\\ y_A=y_B=1\\ =>AB \in y=1\\ \left\{\begin{array}{l} I,J \in AB\\ AB \in y=1 \end{array} \right.\\ =>I,J \in y=1\\ =>y_I=y_J=1\\ AB=\sqrt{(3-0)^2+(1-1)^2}=3\\ AI=IJ=JB=\dfrac{AB}{3}=1\\ AI=\sqrt{(x_I-0)^2+(1-1)^2}=1=>\left[\begin{array}{l} x_I=-1\\ x_I=1 \end{array} \right.(1)\\ JB=\sqrt{(x_J-3)^2+(1-1)^2}=1=>\left[\begin{array}{l} x_J=2\\ x_J=4 \end{array} \right.(2)\\ IJ=\sqrt{(x_I-x_J)^2+(1-1)^2}=1=>x_I-x_J=1(3)\\ (1)(2)(3)=>\left\{\begin{array}{l} I(1;1)\\ J(2;1) \end{array} \right.$
