`A(1;2);B(-1;2)`
`(d): x-2y+1=0<=>x=2y-1`
Vì `C\in (d)=>C(2a-1;a)`
`\vec{CA}=(1-2a+1;2-a)=(2-2a;2-a)`
`=>CA^2=(2-2a)^2+(2-a)^2`
`\vec{CB}=(-1-2a+1;2-a)=(-2a;2-a)`
`=>CB^2=(-2a)^2+(2-a)^2`
Để $CA=CB$
`=>CA^2=CB^2`
`<=>(2-2a)^2+(2-a)^2=(-2a)^2+(2-a)^2`
`<=>(2-2a)^2=(-2a)^2`
$⇔\left[\begin{array}{l}2-2a=-2a\\2-2a=2a\end{array}\right.$
$⇔\left[\begin{array}{l}2=0(vô \ lý)\\a=\dfrac{1}{2}\end{array}\right.$
`a=1/ 2=>2a-1=2. 1/ 2 -1=0`
`=>C(0; 1/ 2)`
Vậy `C(0; 1/ 2)`
