Expanding my comment above.
For the second part of your question, which is the easier one. Two straight
lines $$a_{1}x+b_{1}y=c_{1}\qquad (1)\qquad\text{ and }a_{2}x+b_{2}y=c_{2}\qquad(2)$$ are parallel if
and only if $a_{1}b_{2}-a_{2}b_{1}=0$, because only then their slope $%
m=-a_{1}/b_{1}=-a_{2}/b_{2}$ is the same (in other words the system of
linear equations (1) and (2) has no solutions, its determinant vanishes).
Let $b_{1}b_{2}\neq 0$. From $(1)$ and $(2)$ we get, respectively, $y=-\frac{
a_{1}}{b_{1}}x+\frac{c_{1}}{b_{1}}$ and $y=-\frac{a_{2}}{b_{2}}x+\frac{c_{2}
}{b_{2}}$. The first line crosses the $y$-axe at $(c_{1}/b_{1},0)$, while the
second, at $(c_{2}/b_{2},0)$. Since the straight line parallel to these two
and equidistant to them crosses the $y$-axe at $\left( \left(
c_{1}/b_{1}+c_{2}/b_{2}\right) /2,0\right) $, and has the same slope $m$,
its equation is $$y=-\frac{a_{1}}{b_{1}}x+\frac{1}{2}\left( \frac{c_{1}}{b_{1}}+\frac{c_{2}}{b_{2}}\right) ,\qquad (3)$$ which is equivalent to $$a_{1}x+b_{1}y-\frac{\ c_{1}b_{2}+c_{2}b_{1}}{2b_{2}}=0 .\qquad (4)$$
Without loss of generality assume that $b_{1}=0$ and $a_{1}\neq 0$. Then $(1)$
becomes $x=c_{1}/a_{1}$ and $(2)$ should be of the form $x=c_{2}/a_{2}$, if
both lines are parallel. The line equidistant to both is given by the
equation $x=\left( c_{1}/a_{1}+c_{2}/a_{2}\right) /2$.
If your equations are $y=c_{1}/b_{1}$ and $y=c_{2}/b_{2}$, the line
equidistant to them is given by $y=\left( c_{1}/b_{1}+c_{2}/b_{2}\right) /2$.
Added. As for the main question I got a different solution, namely, the lines whose equations are
$$\left( a_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-a_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}%
\right) x+\left( b_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-b_{2}\sqrt{%
a_{1}^{2}+b_{1}^{2}}\right) y$$
$$=c_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-c_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}\qquad
\left( 5\right) $$
and
$$\left( a_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}+a_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}%
\right) x+\left( b_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}+b_{2}\sqrt{%
a_{1}^{2}+b_{1}^{2}}\right) y$$
$$=c_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}+c_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}.\qquad
\left( 6\right) $$
The distance $d$ from a point $M(x_{M},y_{M})$ to a straight line $r$ whose
equation is $Ax+By+C=0$ can be derived algebraically as follows:
i) Find the equation of the straight line $s$ passing through $M$ and being
orthogonal to $r$. Call $N$ the intersecting point of $r$ and $s$;
ii) Find the co-ordinates of $N(x_{N},y_{N})$;
iii) Find the distance from $M$ to $N$. This distance is $d$;
after which we get the formula
$$d=\frac{\left\vert Ax_{M}+By_{M}+C\right\vert }{\sqrt{A^{2}+B^{2}}}.\qquad
(\ast )$$
The distances from $M$ to lines $(1)$ and $(2)$ are thus given by
$$d_{i}=\frac{\left\vert a_{i}x_{M}+b_{i}y_{M}-c_{i}\right\vert }{\sqrt{
a_{i}^{2}+b_{i}^{2}}}.\qquad i=1,2$$
The points $P(x,y)$ that are equidistant to lines (1) and (2) define two
lines which are the solutions of $d_{1}=d_{2}$:
$$\frac{\left\vert a_{1}x+b_{1}y-c_{1}\right\vert }{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{\left\vert a_{2}x+b_{2}y-c_{2}\right\vert }{\sqrt{a_{2}^{2}+b_{2}^{2}}}.
\qquad (\ast \ast )$$
Therefore, RHS and LHS should have the same or opposite sign:
$$\frac{a_{1}x+b_{1}y-c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{a_{2}x+b_{2}y-c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}.\qquad (\ast \ast \ast )$$
Equations $(5)$ and $(6)$ for the two angle bisectors follow.
Example: For $a_{1}=b_{1}=b_{2}=c_{1}=1,a_{2}=c_{2}=2$, we have $x+y=1$ and $2x+y=2$. The equidistant lines are
$$\left( \sqrt{5}-2\sqrt{2}\right) x+\left( \sqrt{5}-\sqrt{2}\right) y=\sqrt{5%
}-2\sqrt{2}$$
and
$$\left( \sqrt{5}+2\sqrt{2}\right) x+\left( \sqrt{5}+\sqrt{2}\right) y=\sqrt{5}+2\sqrt{2}.$$
Graph of $x+y=1$, $2x+y=2$ and angle bisectors.
Best Answer
The slope of the bisector, can be calculated using the slopes of two given lines. The slope is a tangent so we can calculate the slope in question using something like this: $$\text{slope of a bisector}=\tan(\frac{\tan^{−1}slope_1+\tan^{−1}slope_2}{2})$$ Keep in mind that there are two possible bisectors which are perpendicular to each other so the other slope will be negative reciprocal of this slope. You can also see that if lines are parallel, they will have the same slope to find the equation of the bisecting line you can just take arithmetic mean of two given lines. Finally, to find point of intersection if lines are not parallel, just equate the expressions of the given lines and solve the resulting equation for $x$.