Phương trình \(\left(d\right)\) đi qua \(M\left(1;2\right)\) có dạng :
\(y-2=k\left(x-1\right)\) hay \(y=kx+2-k\)
\(A\in\left(Ox\right)\Rightarrow A\left(a;0\right);B\in\left(Oy\right)\Rightarrow B\left(0;b\right)\)
\(A\in\left(d\right)\Rightarrow0=ka+2-k\Rightarrow a=\dfrac{k-2}{k}\left(k\ne0\right)\Rightarrow A\left(\dfrac{k-2}{k};0\right)\)
\(B\in\left(d\right)\Rightarrow b=k.0+2-k\Rightarrow b=2-k\Rightarrow B\left(0;2-k\right)\)
Tam giác \(OAB\) cân tại \(O\) khi và chỉ khi
\(\Leftrightarrow OA=OB\)
\(\Leftrightarrow\left|\dfrac{k-2}{k}\right|=\left|2-k\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{k-2}{k}=2-k\\\dfrac{k-2}{k}=-2+k\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}k^2-k-2=0\\k^2-3k+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}k=-1\\k=1\\k=2\end{matrix}\right.\)
\(TH_1:k=-1\Rightarrow y=-x+3\Rightarrow x+y-3=0\)
\(TH_2:k=1\Rightarrow y=x+1\Rightarrow x-y+1=0\)
\(TH_3:k=2\Rightarrow y=2x\Rightarrow2x-y=0\)
Vậy \(\left(d\right):\left[{}\begin{matrix}x+y-3=0\\x-y+1=0\\2x-y=0\end{matrix}\right.\)
