a: Để (d)//y=3x+1 thì \(\left\{{}\begin{matrix}m-3=3\\m+2< >1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m=6\\m< >-1\end{matrix}\right.\)
=>m=6
b: (d): y=(m-3)x+m+2
=mx-3x+m+2
=m(x+1)-3x+2
Tọa độ điểm mà (d) luôn đi qua là:
\(\left\{{}\begin{matrix}x+1=0\\y=-3x+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\cdot\left(-1\right)+2=3+2=5\end{matrix}\right.\)
c: y=(m-3)x+m+2
=>(m-3)x-y+m+2=0
Khoảng cách từ O đến (d) là:
\(d\left(O;\left(d\right)\right)=\dfrac{\left|0\left(m-3\right)+0\cdot\left(-1\right)+m+2\right|}{\sqrt{\left(m-3\right)^2+\left(-1\right)^2}}=\dfrac{\left|m+2\right|}{\sqrt{\left(m-3\right)^2+1}}\)
Để d(O;(d))=1 thì \(\dfrac{\left|m+2\right|}{\sqrt{\left(m-3\right)^2+1}}=1\)
=>\(\sqrt{\left(m-3\right)^2+1}=\left|m+2\right|\)
=>\(\sqrt{\left(m-3\right)^2+1}=\sqrt{\left(m+2\right)^2}\)
=>\(\left(m-3\right)^2+1=\left(m+2\right)^2\)
=>\(m^2-6m+9+1=m^2+4m+4\)
=>-6m+10=4m+4
=>-10m=-6
=>\(m=\dfrac{3}{5}\left(nhận\right)\)
