a, Thay \(x=-1;y=1\Leftrightarrow1=-3+4\left(đúng\right)\)
Thay \(x=-1;y=1\Leftrightarrow-1-2\cdot1=0\left(vô.lí\right)\)
Vậy \(A\in d_1;A\notin d_2\)
b, \(\left(d_2\right)x-2y=0\Leftrightarrow\left(d_2\right):y=\dfrac{1}{2}x\)
PTHDGD \(\left(d_1\right);\left(d_2\right)\) là \(3x+4=\dfrac{1}{2}x\Leftrightarrow\dfrac{5}{2}x=-4\Leftrightarrow x=-\dfrac{8}{5}\Leftrightarrow y=-\dfrac{4}{5}\)
\(\Leftrightarrow B\left(-\dfrac{8}{5};-\dfrac{4}{5}\right)\) là giao của 2 đths
c, \(\left(d_3\right)\) đồng quy với \(\left(d_1\right);\left(d_2\right)\Leftrightarrow B\in\left(d_3\right)\)
\(\Leftrightarrow-\dfrac{8}{5}\left(m-1\right)-\dfrac{4}{5}\left(m-2\right)+m+1=0\\ \Leftrightarrow-\dfrac{8}{5}m+\dfrac{8}{5}-\dfrac{4}{5}m+\dfrac{8}{5}+m+1=0\\ \Leftrightarrow-\dfrac{7}{5}m+\dfrac{21}{5}=0\Leftrightarrow-\dfrac{7}{5}m=-\dfrac{21}{5}\Leftrightarrow m=3\)
