Đáp án:
$\begin{array}{l}
\left( d \right)y = a.x + b\\
a)\left( d \right)//y = - \dfrac{3}{5}x + 7\\
\Rightarrow \left\{ \begin{array}{l}
a = - \dfrac{3}{5}\\
b \ne 7
\end{array} \right.\\
\Rightarrow \left( d \right):y = - \dfrac{3}{5}x + b\\
Do:P\left( { - 3;5} \right) \in \left( d \right)\\
\Rightarrow 5 = - \dfrac{3}{5}.\left( { - 3} \right) + b\\
\Rightarrow b = \dfrac{{16}}{5}\left( {tm} \right)\\
\text{Vậy}\,\left( d \right):y = - \dfrac{3}{5}x + \dfrac{{16}}{5}\\
b)\left( d \right) \bot y = - 6x - 8\\
\Rightarrow a.\left( { - 6} \right) = - 1\\
\Rightarrow a = \dfrac{1}{6}\\
\Rightarrow \left( d \right):y = \dfrac{1}{6}x + b\\
E\left( {4; - 3} \right) \in \left( d \right)\\
\Rightarrow - 3 = \dfrac{1}{6}.4 + b\\
\Rightarrow b = - 3 - \dfrac{2}{3} = \dfrac{{ - 11}}{3}\\
\text{Vậy}\,y = \dfrac{1}{6}x - \dfrac{{11}}{3}\\
c)\left( {0;4} \right);\left( {5;0} \right) \in \left( d \right)\\
\Rightarrow \left\{ \begin{array}{l}
4 = a.0 + b\\
0 = a.5 + b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
b = 4\\
5a = - b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
b = 4\\
a = - \dfrac{4}{5}
\end{array} \right.\\
\Rightarrow \left( d \right):y = - \dfrac{4}{5}x + 4\\
d)\left( d \right)//y = - 4x + 11\\
\Rightarrow \left\{ \begin{array}{l}
a = - 4\\
b \ne 11
\end{array} \right.\\
\Rightarrow \left( d \right):y = - 4x + b\\
\left( {2;0} \right) \in \left( d \right)\\
\Rightarrow 0 = - 4.2 + b\\
\Rightarrow b = 8\left( {tm} \right)\\
\Rightarrow \left( d \right):y = - 4x + 8\\
e)\left( d \right)//y = 0,35.x - 12\\
\Rightarrow \left\{ \begin{array}{l}
a = 0,35\\
b \ne - 12
\end{array} \right.\\
\Rightarrow \left( d \right):y = 0,35.x + b\\
Khi:y = 5 \Rightarrow 3x - 1 = 5 \Rightarrow x = 2\\
\Rightarrow \left( {2;5} \right) \in \left( d \right)\\
\Rightarrow 5 = 0,35.2 + b\\
\Rightarrow b = 4,3\left( {tm} \right)\\
\Rightarrow \left( d \right):y = 0,35x + 4,3\\
g)A\left( { - 3;4} \right);B\left( {2; - 5} \right) \in \left( d \right)\\
\Rightarrow \left\{ \begin{array}{l}
4 = - 3a + b\\
- 5 = 2a + b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \dfrac{{ - 9}}{5}\\
b = - \dfrac{7}{5}
\end{array} \right.\\
\text{Vậy}\,y = - \dfrac{9}{5}x - \dfrac{7}{5}
\end{array}$
