Đáp án:
$\begin{array}{l}
d:y = a.x + b\\
a){d_1}:x + 2y = 1\\
\Leftrightarrow 2y = - x + 1\\
\Leftrightarrow y = - \dfrac{1}{2}x + \dfrac{1}{2}\\
d//{d_1}\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - \dfrac{1}{2}\\
b \ne \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow d:y = \dfrac{{ - 1}}{2}x + b\left( {b \ne \dfrac{1}{2}} \right)\\
M\left( {1; - 2} \right) \in d\\
\Leftrightarrow - 2 = - \dfrac{1}{2}.1 + b\\
\Leftrightarrow b = - 2 + \dfrac{1}{2} = - \dfrac{3}{2}\left( {tmdk} \right)\\
Vậy\,d:y = - \dfrac{1}{2}x - \dfrac{3}{2}\\
b){d_3}:y = 3 - x = - x + 3\\
d \bot {d_3}\\
\Leftrightarrow a.\left( { - 1} \right) = - 1\\
\Leftrightarrow a = 1\\
\Leftrightarrow d:y = x + b\\
{d_2}:x - y + 1 = 0\\
Khi:y = 2 \Leftrightarrow x = 1\\
\Leftrightarrow d \cap {d_2} = \left( {1;2} \right)\\
\Leftrightarrow 2 = 1 + b\\
\Leftrightarrow b = 1\\
Vậy\,d:y = x + 1\\
c)O \in d\\
\Leftrightarrow b = 0\\
\Leftrightarrow d:y = a.x\\
{d_4};{d_5}\\
Xet:4x - 3 = - x + 3\\
\Leftrightarrow 5x = 6\\
\Leftrightarrow x = \dfrac{6}{5}\\
\Leftrightarrow y = - x + 3 = - \dfrac{6}{5} + 3 = \dfrac{9}{5}\\
\Leftrightarrow {d_4} \cap {d_5} = \left( {\dfrac{6}{5};\dfrac{9}{5}} \right)\\
\Leftrightarrow \left( {\dfrac{6}{5};\dfrac{9}{5}} \right) \in d\\
\Leftrightarrow \dfrac{9}{5} = a.\dfrac{6}{5}\\
\Leftrightarrow a = \dfrac{3}{2}\\
Vậy\,d:y = \dfrac{3}{2}x\\
d) + khi:y = 0 \Leftrightarrow x = 5\\
+ M\left( {2;3} \right) \in d\\
\Leftrightarrow \left\{ \begin{array}{l}
0 = a.5 + b\\
3 = a.2 + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5a + b = 0\\
2a + b = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3a = - 3\\
b = - 5a
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - 1\\
b = 5
\end{array} \right.\\
Vậy\,d:y = - x + 5
\end{array}$
