`@`Cho \(x=0\Rightarrow y=m\)
`@`Cho \(y=0\Rightarrow x=-m\)
\(A\left(0;m\right)\in\left(d\right)\); \(B\left(-m;0\right)\in\left(d\right)\)
Vậy \(\left(d\right)\) cắt \(Oy\) tại \(A\); \(\left(d\right)\) cắt \(Ox\) tại \(B\)
\(\Rightarrow OA=\left|m\right|\); \(O\left|-m\right|\)
Ta có:
\(S_{AOB}=12\)
\(\Leftrightarrow12=\dfrac{1}{2}.OA.OB\)
\(\Leftrightarrow12=\dfrac{1}{2}.\left|m\right|.\left|-m\right|\)
\(\Leftrightarrow24=m^2\)
\(\Leftrightarrow m=\pm2\sqrt{6}\)
Vậy \(m=\left\{\pm2\sqrt{6}\right\}\) thì \(S_{AOB}=12\)