Do C thuộc d nên tọa độ C có dạng \(C\left(c;2c+3\right)\)
\(\left\{{}\begin{matrix}\overrightarrow{AC}=\left(c+1;2c+1\right)\\\overrightarrow{BC}=\left(c+3;2c+1\right)\end{matrix}\right.\)
\(AC=BC\Leftrightarrow\left(c+1\right)^2+\left(2c+1\right)^2=\left(c+3\right)^2+\left(2c+1\right)^2\)
\(\Leftrightarrow2c+1=6c+9\Rightarrow c=-2\)
\(\Rightarrow C\left(-2;-1\right)\)