Step 1

Let the given vertex be $A(1,1)$

Also given that the equation of one side $BC$ is $x+y+1=0$

$\Rightarrow\:$slope of $BC=m=-1$

Let the slope of $AB=m_1$ and slope of $AC=m_2$

step 2

Since the $\Delta\:ABC$ is equilateral triangle,

angle between the lines $AB$ and $BC$ is $60^{\circ}$ and

angle between $AC$ and $BC$ is also $60^{\circ}$.

We know that angle between any two lines $(\theta)$ with slopes $m_1$ and $m_2$

is given by $tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$

$\therefore\:tan60^{\circ}=\bigg|\large\frac{m-m_1}{1+m.m_1}\bigg|$

$\Rightarrow\:\sqrt 3=\bigg|\large\frac{-1-m_1}{1+(-1).m_1}\bigg|$

$\Rightarrow\:\sqrt 3(1-m_1)=-1-m_1$

$\Rightarrow\:m_1(1-\sqrt 3)=-1-\sqrt 3$

$\Rightarrow\:m_1=\large\frac{\sqrt 3+1}{\sqrt 3-1}$

Rationalising the denominator we get $m_1=2+\sqrt 3$

Step 3

Similarly angle between $AC$ and $BC$ is given by

$tan60^{\circ}=\bigg|\large\frac{m_2-m}{1+m.m_2}\bigg|$

$\Rightarrow\:\sqrt 3=\bigg|\large\frac{m_2-(-1)}{1+(-1).m_2}\bigg|$

$\Rightarrow\:\sqrt 3(1-m_2)=1+m_2$

$\Rightarrow\:m_2(1+\sqrt 3)=\sqrt 3-1$

$\Rightarrow\:m_2=\large\frac{\sqrt 3-1}{\sqrt 3+1}$

Rationalising the denominator, we get $m_2=2-\sqrt 3$

Step 4

Hence equation of AB is $y-1=(2+\sqrt 3)(x-1)$

and

Equation of $AC$ is $y-1=(2-\sqrt 3)(x-1)$

Hence (a) is the correct answer.