\[\begin{array}{l} A\left( { - 2;\,\,1} \right)\\ 5)\,\,\,Goi\,\,\,\Delta :\,\,\,y = ax + b\,\,\,\,\left( {a \ne 0} \right)\\ A \in \Delta \Rightarrow - 2a + b = 1 \Leftrightarrow b = 1 + 2a.\\ \Rightarrow \Delta :\,\,\,y = ax + 2a + 1\\ Goi\,\,\,Ox \cap \Delta = \left\{ B \right\} \Rightarrow B\left( { - \frac{{2a + 1}}{a};\,\,\,0} \right)\\ Oy \cap \Delta = \left\{ C \right\} \Rightarrow C\left( {0;\,\,2a + 1} \right).\\ \Delta OBC\,\,\,vuong\,\,\,can\,\,\,tai\,\,\,O \Rightarrow OB = OC\\ \Leftrightarrow \left| {{x_B}} \right| = \left| {{y_C}} \right| \Leftrightarrow \left| {\frac{{2a + 1}}{a}} \right| = \left| {2a + 1} \right|\\ \Leftrightarrow \left| {2a + 1} \right|\left( {\frac{1}{{\left| a \right|}} - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} \left| {2a + 1} \right| = 0\\ \frac{1}{{\left| a \right|}} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2a + 1 = 0\\ \left| a \right| = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} a = - \frac{1}{2} \Rightarrow b = 0\\ a = 1 \Rightarrow b = 3\\ a = - 1 \Rightarrow b = - 1 \end{array} \right.\\ 6)\,\,\,Goi\,\,\,\Delta :\,\,\,y = ax + b\,\,\,\,\left( {a \ne 0} \right)\\ A \in \Delta \Rightarrow - 2a + b = 1 \Leftrightarrow b = 1 + 2a.\\ \Rightarrow \Delta :\,\,\,y = ax + 2a + 1\\ Goi\,\,\,Ox \cap \Delta = \left\{ B \right\} \Rightarrow B\left( { - \frac{{2a + 1}}{a};\,\,\,0} \right)\\ Oy \cap \Delta = \left\{ C \right\} \Rightarrow C\left( {0;\,\,2a + 1} \right).\\ \left\{ \begin{array}{l} {x_B} > 0\\ {y_C} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - \frac{{2a + 1}}{a} > 0\\ 2a + 1 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2a + 1 > 0\\ a < 0 \end{array} \right. \Leftrightarrow - \frac{1}{2} < a < 0.\\ {S_{BOC}} = 16 \Leftrightarrow \frac{1}{2}OB.OC = 16\\ \Leftrightarrow {x_B}.{y_C} = 32\\ \Leftrightarrow - \frac{{2a + 1}}{a}.\left( {2a + 1} \right) = 32\\ \Leftrightarrow {\left( {2a + 1} \right)^2} + 32a = 0\\ \Leftrightarrow 4{a^2} + 4a + 1 + 32a = 0\\ \Leftrightarrow 4{a^2} + 36a + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} a = \frac{{ - 9 + 4\sqrt 5 }}{2}\,\,\left( {tm} \right)\\ a = \frac{{ - 9 - 4\sqrt 5 }}{2}\,\,\,\left( {ktm} \right) \end{array} \right. \Rightarrow b = 2a + 1 = - 8 + 4\sqrt 5 . \end{array}\] Em kết luận phương trình đường thẳng delta nhé em.