ĐKXĐ: m<>1, m<>0
a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)
=>\(-2m=\sqrt{3}m-\sqrt{3}\)
\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)
hay \(m=-3+2\sqrt{3}\)
tana=căn 3
nên a=60 độ
b:
\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)
=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)
\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)
\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)
\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)
Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất
\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)
\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)
\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)
=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)
Dấu = xảy ra khi m=1/5