Trong mặt phẳng Oxy, cho hai đường thẳng $d_{1}: 3x+y+5=0 ,  d_{2}: 3x+y+1=0  và  điểm  I(1;2). $ Viết phương trình đường thẳng $\Delta  qua  I  và  cắt  d_{1},d_{2}$ lần lượt tại A và B sao cho AB=$\frac{2}{\sqrt{2}}$.
pt dt $\triangle co dang :y=ax+b        hay     ax-y+b=0$
Ta co  :$d_1//d_2    ,M(0;-5) \in d_1\Rightarrow d(d_1;d_2)=d(M;d_2)=\frac{4}{\sqrt{10}}$
$sin (d_1;\triangle )=\frac{d(d_1;d_2)}{AB}=\frac{2}{\sqrt{5}}\Rightarrow cos(d_1;\triangle )=\sqrt{1-(\frac{2}{\sqrt5})^2}=\frac{1}{\sqrt5}$
Mat khac :$cos(d_1;\triangle )=\frac{|3a-1|}{\sqrt{a^2+1}\sqrt{3^2+1}}=\frac{1}{\sqrt5}\Leftrightarrow a=1$hoac $a=-1/7$
$\Rightarrow \triangle_1 :x-y+b_1=0;\triangle _2:(-1/7)x-y+b_2=0$ 
+)$I(1:2)\in \triangle_1 \Rightarrow 1-2+b_1=0\Leftrightarrow b=1$
Vay ptdt $\triangle_1: x-y+1=0 $
+)$I(1;2)\in \triangle _2\Rightarrow (-1/7)-2+b_2=0\Leftrightarrow b_2=15/7$
Vay ptdt  $\triangle _2: (-1/7)x-y+15/7=0 $ hay $x+7y-15=0$

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