Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 1.\ m=-1\\ 2.m=-3\\ 3.B( -1;4)\\ 4.m=\{-1;-7\} \ \\ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 1.\ d//d'\Rightarrow m-1=-2\ và\ m+3\neq 1\\ \Rightarrow m=-1\\ 2.A( 1;-4) \in \ ( d) \Rightarrow m-1+m+3=-4\\ \Rightarrow m=-3\\ 3.\ Ta\ có\ -( m-1) +m+3=4\\ \Rightarrow B( -1;4) \in ( d)\\ 4.Ox\cap ( d) =A\Rightarrow A\left(\frac{m+3}{1-m} ;0\right)\\ Oy\cap ( d) =B\Rightarrow B( 0;m+3)\\ Để\ S_{OAB} =1\Rightarrow |\frac{m+3}{1-m} ||m+3|=2\\ \Rightarrow ( m+3)^{2} =2|1-m|\\ \Rightarrow m^{2} +6m+9=2|1-m|\\ TH1:\ m\leqslant 1\\ m^{2} +6m+9=2-2m\\ \Rightarrow m=-1\ ( TM) ;\ m=-7\ ( TM)\\ TH\ 2:m\geqslant 1\\ m^{2} +6m+9=2m-2\\ \Rightarrow m=\diameter \end{array}$
