Equation of the bisector of the angle between two lines containing the given point 
The general form of the equation of the angle bisector of two lines :
$$ \begin{align}L_1 &=a_1x+b_1y+ c_1=0
\\L_2 &= a_2x+b_2y+ c_2=0 \end{align}$$
Given as:
$$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}= \pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$

Now, we have a point $(\alpha, \beta)$ lying in one of the angles between these two.
I have been instructed to do as follows:

*

*Find the sign of the expression of $L_1$ and $L_2$.

*If they are opposite then, choose the negative sign from the general
form otherwise choose the positive

but I want to know this method's proof for better understanding. Or, if there's an alternate/ better method to deal with it, please let me know (with proof).

There is a much more elegant and geometric view point for deriving this equation. Let the lines be $L_1 = a_1 x + b_1 y + c_1=0$ and $L_2 = a_2 x + b_2 y + c_2=0$, the unit normal for these lines are given as :
$$ \begin{align} n_1 &= \frac{\nabla L_1}{|\nabla L_1|} \\ n_2 &= \frac{|\nabla L_2|}{|\nabla L_2|} \end{align}$$
Now, the geometric interpretation of subtract two unit vectors is finding the vector which is equi-angle from the two unit vectors similar proof i.e: midpoint of the arc of the two vectors.
$$ \chi= n_1 -n_2$$
The above is the vector in direction to the midpoint of the arc cut by the two unit vectors on the unit circle. It is also the normal of one of the bisectors. How do we in the other normal? We rotate it by ninty degrees.
If we denote $\phi(v)$ to be the map which associates a vector with components $(v_x,v_y)$ in the cartesian plane to a complex number with components $z=x+iy$, then we can rotate the previously obtained vector by ninty degrees to find the normal of the other bisector (Relevant theorem):
$$ \tilde{\chi} = \phi^{-1} \left[ i\phi(\chi) \right]$$
Hence, we can write the equation of two bisector as:
$$ \begin{align} \chi \cdot (x-\alpha,y -\beta) &=0  \\ \tilde{\chi} \cdot (x-\alpha,y- \beta) &=0\end{align}$$
Now, check the angle between $n_1$ and $n_2$, if it is greater than ninty degrees then it means $\chi$ is the obtuse bisector and if not, it $\chi$ is the acute bisector. To understand why, check the linked in equi-angle form.
