(m-1)x+2m+1=y
=>(m-1)x-y+2m+1=0
\(d\left(O;d1\right)=\dfrac{\left|0\cdot\left(m-1\right)+0\cdot\left(-1\right)+2m+1\right|}{\sqrt{\left(m-1\right)^2+1}}=\dfrac{\left|2m+1\right|}{\sqrt{\left(m-1\right)^2+1}}\)
Để (d) lớn nhất thì \(\sqrt{\left(m-1\right)^2+1}_{Min}\)
=>m=1