$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{a.\ \Delta \ đi\ qua\ A( 2;5) \ nhận\ \overrightarrow{n_{AB}}( 2;-3) \ là\ vtpt}\\ \mathrm{\Rightarrow \Delta \ :\ 2( x-2) -3( y-5) =0}\\ \mathrm{hay\ \Delta \ :2x-3y+11=0}\\ \mathrm{b.\ \Delta \ đi\ qua\ M( 3;-4) \ nhận\ \overrightarrow{n_{d}}( 3;1) \ là\ vtpt}\\ \mathrm{\Rightarrow \Delta \ :3( x-3) +( y+4) =0\ }\\ \mathrm{hay\ \Delta \ :3x+y-5=0}\\ \mathrm{c.\ \Delta \ đi\ qua\ M( 3;-4) \ nhận\ \overrightarrow{u_{d}}( 1;-3) \ là\ vtpt}\\ \mathrm{\Rightarrow \Delta \ :1( x-3) -3( y+4) =0\ }\\ \mathrm{hay\ \Delta \ :x-3y-15=0}\\ \mathrm{Gọi\ \Delta \cap Ox=A;\ \Delta \cap Oy=B}\\ \mathrm{d.\ \Delta \ \ đi\ qua\ M\ ( 3,-4) \ chắn\ 2\ trục\ tọa\ độ\ các\ đoạn\ bằng\ nhau}\\ \mathrm{\Rightarrow \Delta OAB\ vuông\ cân\ tại\ O.}\\ \mathrm{\Rightarrow Hệ\ số\ góc\ k=\ tan45^{o} =1}\\ \mathrm{\Delta \ \ đi\ qua\ M\ ( 3,-4) \ có\ hệ\ số\ góc\ là\ 1\ có\ dạng:}\\ \mathrm{y+4=x-3\ hay\ x-y-7=0}\\ \mathrm{e.\ Giả\ sử:\ A( a;0) \ B( 0;b)}\\ \mathrm{\Rightarrow \Delta \ :\ ax+by=1}\\ \mathrm{Mà\ M( 3;-4) \in \Delta \ \ \Rightarrow 3a-4b=1\ ( 1)}\\ \mathrm{S_{OAB} =4\ \Rightarrow \frac{1}{2} .|a||b|=4}\\ \mathrm{\Leftrightarrow ab=8\ hoặc\ ab=-8}\\ \mathrm{TH1:\ ab=8\Rightarrow a=\frac{8}{b} \ thay\ vào\ ( 1) :}\\ \mathrm{\frac{24}{b} -4b=1\Rightarrow b=\frac{-1\pm \sqrt{385}}{8} \Rightarrow a=\frac{1\pm \sqrt{385}}{6}}\\ \mathrm{\Rightarrow \Delta :\frac{1+\sqrt{385}}{6} x+\frac{-1+\sqrt{385}}{8} y=1}\\ \mathrm{hay\ \Delta :\frac{1-\sqrt{385}}{6} x+\frac{-1-\sqrt{385}}{8} y=1}\\ \mathrm{TH2:\ ab=-8\Rightarrow a=-\frac{8}{b} \ thay\ vào\ ( 1) :}\\ \mathrm{-\frac{24}{b} -4b=1( \ vô\ nghiệm)}\\ \mathrm{f.\Delta :\ 3x+y+c=0}\\ \mathrm{Do\ d( M,\Delta ) =\frac{|3.2+5+c|}{\sqrt{3^{2} +1^{2}}} =\sqrt{10}}\\ \mathrm{\Leftrightarrow |c+11|=10}\\ \mathrm{\Leftrightarrow c=-1\ hoặc\ x=-21}\\ \mathrm{Vậy\ \Delta :\ 3x+y-1=0}\\ \mathrm{hoặc\ \Delta :\ 3x+y-11=0}\\ \mathrm{g.\ Lấy\ D( 3;0) \in d}\\ \mathrm{\Delta :\ 3x+y+c=0}\\ \mathrm{Do\ \Delta //d\ nên\ khoảng\ cách\ từ\ d\ đến\ \Delta \ bằng\ khoảng\ cách\ từ\ D\ đến\ \Delta }\\ \mathrm{d( D,\Delta ) =\frac{|3.3+1.0+c|}{\sqrt{3^{2} +1^{2}}} =\sqrt{10}}\\ \mathrm{\Leftrightarrow |c+9|=10}\\ \mathrm{\Leftrightarrow c=1\ hoặc\ x=-19}\\ \mathrm{Vậy\ \Delta :\ 3x+y+1=0}\\ \mathrm{hoặc\ \Delta :\ 3x+y-19=0}\\ \mathrm{h.\ \Delta \ đi\ qua\ M\ ( 1,2) \ và\ cách\ đều\ 2\ điểm\ A\ ( 2,3) \ và\ B( 4,-5)}\\ \mathrm{Gỉa\ sử\Delta :\ ax+by-1=0}\\ \mathrm{M\in \Delta \Rightarrow a+2b=1( 1)}\\ \mathrm{Ta\ có\ d( A,\Delta ) =d( B,\Delta )}\\ \mathrm{\Leftrightarrow \frac{|2a+3b-1|}{\sqrt{a^{2} +b^{2}}} =\frac{|4a-5b-1|}{\sqrt{a^{2} +b^{2}}}}\\ \mathrm{\Leftrightarrow |2a+3b-1|=|4a-5b-1|}\\ \mathrm{\Leftrightarrow 2a+3b-1=4a-5b-1\ \ hoặc\ 2a+3b-1=-4a+5b+1\ }\\ \mathrm{TH1:2a+3b-1=4a-5b-1\ }\\ \mathrm{\Leftrightarrow 2a-5b=0\ ( 2)}\\ \mathrm{Từ\ ( 1) \ ( 2) \Rightarrow a=\frac{5}{9} \ \ \ b=\frac{2}{9}}\\ \mathrm{Vậy\ \ \Delta :\ 5x+2y-9=0}\\ \mathrm{TH2:2a+3b-1=-4a+5b+1}\\ \mathrm{\Leftrightarrow 6a-2b=2\ ( 2')}\\ \mathrm{Từ\ ( 1) \ ( 2') \Rightarrow a=\frac{3}{7} \ \ \ b=\frac{2}{7}}\\ \mathrm{Vậy\ \ \Delta :\ 3x+2y-7=0}\\ \end{array}$