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.
Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align}
x \cos\theta + y \sin\theta - p &= 0 \\
x \cos\phi + y \sin\phi - q &= 0 \\
x \cos\psi + y \sin\psi - r &= 0
\end{align}$$
with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:
Then
$$C_1 = \left|\begin{array}{cc}
\cos\phi & \sin\phi \\
\cos\psi & \sin\psi
\end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$
Likewise,
$$C_2 = \sin B \qquad C_3 = \sin C$$
Moreover,
$$D := \left|\begin{array}{ccc}
\cos\theta & \sin\theta & - p \\
\cos\phi & \sin\phi & - q \\
\cos\psi & \sin\psi & - r
\end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$
Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
Therefore,
$$\begin{align}
D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt]
&= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt]
&= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt]
&= -\frac{2\;|\triangle ABC|}{d}
\end{align}$$
Also,
$$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$
Finally:
$$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$
Best Answer
If two lines do not intersect--i.e., they have no points in common--then the system of equations $$\begin{align*} a_1 x + b_1 y + c_1 &= 0 \\ a_2 x + b_2 y + c_2 &= 0 \end{align*}$$ will have no solution for $(x,y)$. Thus, if we solve the system, we find $$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_2 c_1 - a_1 c_2}{a_1 b_2 - a_2 b_1}.$$ This solution does not exist or is indeterminate if $a_1 b_2 - a_2 b_1 = 0$.
However, some care is required: two lines coincide if $$(a_1, b_1, c_1) = k(a_2, b_2, c_2)$$ for some nonzero scalar constant $k$, and in this case, both the numerators and denominator of the aforementioned solution are zero, meaning that there are infinitely many points that the two equations share in common. So while it is a sufficient condition for $a_1 b_2 - a_2 b_1 \ne 0$ to imply that the lines intersect, it is not a strictly necessary condition, and for that reason, the question should have been better phrased by saying "two lines...intersect if..." rather than "only if"; alternatively, it might be phrased "two distinct lines...intersect only if...."