Answer:
Solution Given:
let ABC be an equilateral triangle with the vertex A(2,-1) and slope =-1.
and
∡ABC=∡BAC=∡ACB=60°
slope of BC[tex][ m_1]=-1[/tex]
we have
[tex]\theta[/tex]=60°
Slope of AB=[tex][ m_2]=a[/tex]
now
we have
angle between two lines is
[tex]Tan\theta =±\frac{m_1-m_2}{1+m_1m_2}[/tex]
now substituting value
tan 60°= ± [tex]\frac{-1-a}{1+-1*a}[/tex]
[tex]\sqrt{3}=±\frac{-1-a}{1-a}[/tex]
doing criss-cross multiplication;
[tex](1-a)\sqrt{3}=±-(1+a)[/tex]
[tex]\sqrt{3}-a\sqrt{3}[/tex]=±(1+a)
taking positive
[tex]-a-a\sqrt{3}=1-\sqrt{3}[/tex]
[tex]-a(1+\sqrt{3})=1-\sqrt{3}[/tex]
a=[tex]\frac{1-\sqrt{3}}{-(1+\sqrt{3})}=\frac{\sqrt{3}-1}{1+\sqrt{3}}[/tex]
taking negative
[tex]a-a\sqrt{3}=-1-\sqrt{3}[/tex]
[tex]a(1-\sqrt{3})=-1-\sqrt{3}[/tex]
a=[tex]\frac{-(1+\sqrt{3})}{(1-\sqrt{3})}=\frac{\sqrt{3}+1}{-1+\sqrt{3}}[/tex]
Equation of a line when a =[tex]\frac{\sqrt{3}-1}{1+\sqrt{3}}[/tex]
and passing through (2,-1),we have
[tex] y-y_1=m(x-x_1)[/tex]
y+1=[tex]\frac{\sqrt{3}-1}{1+\sqrt{3}}[/tex](x-2)
y=[tex]\frac{\sqrt{3}-1}{1+\sqrt{3}}x-2\frac{\sqrt{3}-1}{1+\sqrt{3}}[/tex]-1
[tex]\boxed{\bold{\green{y=\frac{\sqrt{3}-1}{1+\sqrt{3}}x-5+2\sqrt{3}}}}[/tex] is a first side equation of line.
again
Equation of a line when a =[tex]\frac{\sqrt{3}+1}{-1+\sqrt{3}}[/tex]
and passing through (2,-1),we have
[tex] y-y_1=m(x-x_1)[/tex]
y+1=[tex]\frac{\sqrt{3}+1}{-1+\sqrt{3}}[/tex](x-2)
y+1=[tex]\frac{\sqrt{3}+1}{-1+\sqrt{3}}x -2\frac{\sqrt{3}+1}{-1+\sqrt{3}}[/tex]
y=[tex]\frac{\sqrt{3}+1}{-1+\sqrt{3}}x -2\frac{\sqrt{3}+1}{-1+\sqrt{3}}-1[/tex]
[tex]\boxed{\bold{\blue{y=\frac{\sqrt{3}+1}{-1+\sqrt{3}}x -5+2\sqrt{3}}}}[/tex] is another equation of line.
