Do d vuông góc d' nên pt d có dạng: \(x+2y-c=0\)
\(\Rightarrow\) d cắt Ox và Oy lần lượt tại \(A\left(c;0\right)\) ; \(B\left(0;\frac{c}{2}\right)\)
a/ \(\overrightarrow{AB}=\left(-c;\frac{c}{2}\right)\Rightarrow c^2+\left(\frac{c}{2}\right)^2=1\)
\(\Leftrightarrow\frac{5c^2}{4}=1\Rightarrow c=\pm\frac{2\sqrt{5}}{5}\)
\(\Rightarrow d:\left[{}\begin{matrix}x+2y+\frac{2\sqrt{5}}{5}=0\\x+2y-\frac{2\sqrt{5}}{5}=0\end{matrix}\right.\)
b/ \(OA=\left|c\right|;OB=\left|\frac{c}{2}\right|\)
\(S_{OAB}=\frac{1}{2}OA.OB=\frac{1}{4}\left|c\right|^2=\frac{1}{4}c^2\)
\(\Leftrightarrow\frac{1}{4}c^2=1\Rightarrow c=\pm2\)
\(\Rightarrow d:\left[{}\begin{matrix}x+2y+2=0\\x+2y-2=0\end{matrix}\right.\)
c/ \(\frac{2}{c^2}+\frac{1}{\left(\frac{c}{2}\right)^2}=1\)
\(\Leftrightarrow\frac{2}{c^2}+\frac{4}{c^2}=1\Leftrightarrow\frac{6}{c^2}=1\Rightarrow c=\pm\sqrt{6}\)
\(\Rightarrow d:\left[{}\begin{matrix}x+2y+\sqrt{6}=0\\x+2y-\sqrt{6}=0\end{matrix}\right.\)
