Thay n=2 vào (d), ta được:
y=(k-1)x+2
Tọa độ C là:
\(\left\{{}\begin{matrix}y=0\\\left(k-1\right)x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{2}{k-1}\end{matrix}\right.\)
Vậy: \(C\left(-\dfrac{2}{k-1};0\right)\)
\(S_{OAC}=2\cdot S_{OAB}\)
=>AC=2AB
=>\(AC^2=4AB^2\)
=>\(\left(-\dfrac{2}{k-1}-0\right)^2+\left(0-2\right)^2=4\left[\left(-1-0\right)^2+\left(0-2\right)^2\right]\)
=>\(\dfrac{4}{\left(k-1\right)^2}+4=4\left(1+4\right)\)
=>\(\dfrac{4}{\left(k-1\right)^2}=4\cdot5-4=20-4=16\)
=>\(\left(k-1\right)^2=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}k-1=\dfrac{1}{2}\\k-1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k=\dfrac{3}{2}\\k=\dfrac{1}{2}\end{matrix}\right.\)
