a) vì \(\left(d\right)\backslash\backslash\left(d'\right)\) \(\Rightarrow\left(d\right)\) có dạng \(\left(d\right):y=-2x+b\)
ta có : \(A\in\left(d\right)\Rightarrow2=-2\left(-3\right)+b\Rightarrow b=-4\)
vậy \(\left(d\right):-2x-4\)
b) gọi \(\left(d_1\right):y=ax+b\)
ta có : \(\left(d_1\right)\perp\left(d_2\right)\Rightarrow a=-1\) \(\Rightarrow\left(d_1\right)y=-x+b\)
ta có : \(A\in\left(d_1\right)\) \(\Rightarrow\) \(2=-\left(-3\right)+b\Leftrightarrow b=-1\)
vậy \(\left(d_1\right):-x-1\)