$a)(d')=T_{\vec{v}}(d)⇒d'//d$
$⇒d':2x-y+c=0$
$A(-2;1)∈d$
$⇒A'=T_{\vec{v}}A$
$⇒$$\begin{cases}x_{A'}=x_A+2\\y_{A'}=y_A-3\end{cases}$
$⇔$$\begin{cases}x_{A'}=-2+2=0\\y_{A'}=1-3=-2\end{cases}$
$⇒A'(0;-2)$
$A'∈d'⇒0+2+c=0$
$⇔c=-2$
$⇒d':2x-y-2=0$
$b)(d')=T_{\vec{v}}(d)⇒d'//d$
$⇒d':2x-y+c=0$
$A(-2;1)∈d$
$⇒A'=T_{\vec{v}}A$
$⇒$$\begin{cases}x_{A'}=x_A+2\\y_{A'}=y_A+1\end{cases}$
$⇔$$\begin{cases}x_{A'}=-2+2=0\\y_{A'}=1+1=2\end{cases}$
$⇒A'(0;2)$
$A'∈d'⇒0-2+c=0$
$⇔c=2$
$⇒d':2x-y+2=0$
$c)(d')=T_{\vec{v}}(d)⇒d'//d$
$⇒d':2x-y+c=0$
$A(-2;1)∈d$
$⇒A'=T_{\vec{v}}A$
$⇒$$\begin{cases}x_{A'}=x_A-2\\y_{A'}=y_A+1\end{cases}$
$⇔$$\begin{cases}x_{A'}=-2-2=-4\\y_{A'}=1+1=2\end{cases}$
$⇒A'(-4;2)$
$A'∈d'⇒2.(-4)-2+c=0$
$⇔c=10$
$⇒d':2x-y+10=0$
