What is can understand from this image is that the equation of the bisector of L1 and L2 is derived by equating their perpendicular distance. The table gives the sign (+/-) for the bisector at different condition(which is also given in the image). But I am not able to derive this. Please help me! Edit: I am talking about the conditions. My books says-
Shortcut Method for Identifying Acute and Obtuse Angle Bisectors
The equations of the bisectors of the lines $L_{1}: a_{1} x+b_{1} y+c_{1}$ $=0$ and $L_{2}: a_{2} x+b_{2} y+c_{2}=0,\left(a_{1} b_{2} \neq a_{2} b_{1}\right)$ where $c_{1}>0$ and $c_{2}>0$, are $$ \frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}= \pm\frac{a_{2} x+b_{2} y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}} $$ $$ \begin{array}{|c|c|c|} \hline \text { Conditions } & \text { Acute angle bisector } & \text { Obtuse angle bisector } \\ \hline a_{1} a_{2}+b_{1} b_{2}>0 & - & + \\ \hline a_{1} a_{2}+b_{1} b_{2}<0 & + & -\\ \hline \end{array} $$
Given a line with equation $ax+by+c=0$, then $(a,b)$ is a vector perpendicular to the line and $$ \vec{n}=\left({a\over\sqrt{a^2+b^2}},{b\over\sqrt{a^2+b^2}}\right) $$ is a unit vector perpendicular to the line. If you have now two lines $L_1$ and $L_2$ and their associated perpendicular unit vectors $\vec n_1$ and $\vec n_2$, then it is easy to see that vectors $\vec n_1+\vec n_2$ and $\vec n_1-\vec n_2$ are perpendicular to the bisectors of $L_1$ and $L_2$, as it is confirmed by the cartesian equations given in the question ($\vec n_1+\vec n_2$ is associated to the equation with $-$, while $\vec n_1-\vec n_2$ is associated to the equation with $+$).
To assign the right bisector to each angle, observe that $\vec n_1+\vec n_2$ is associated with the line bisecting the angle which is the same as the angle formed by $\vec n_1$ and $\vec n_2$ (pink angle in figure below), whereas $\vec n_1-\vec n_2$ is associated with the line bisecting the other angle.
Finally, we can check if the the angle $\theta$ formed by $\vec n_1$ and $\vec n_2$ is acute or obtuse by computing the dot product $$ \vec n_1\cdot\vec n_2={a_1a_2+b_1b_2\over \sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}. $$ If $\vec n_1\cdot\vec n_2>0$ (i.e. $a_1a_2+b_1b_2>0$) then $\theta$ is acute, if $\vec n_1\cdot\vec n_2<0$ (i.e. $a_1a_2+b_1b_2<0$) then $\theta$ is obtuse. This follows from the property:
$$\vec n_1\cdot\vec n_2=|\vec n_1|\ |\vec n_2|\ \cos\theta.$$
Putting all things together, you end up with the table given in the question.