geometry
Let $L_1$ be a line, $a_1X + b_1Y+ c_1 =0$,
$L_2$ be a line, $a_2X + b_2Y+ c_2= 0$.
Then prove that the equation of line(s) passing through the intersection of these two line is of type $\mathbf{L_1 + KL_2 = 0}$.
We can prove it by multiply two equations with different constants and then adding them…, this proof didn't worked for me,therefore i need a best proof, prove it directly or give me some awesome hints.
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.
$\displaystyle\frac{|a_1x + b_1y + c_1|}{\sqrt{a_1^2+b_1^2}}=\frac{|a_2x + b_2y + c_2|}{\sqrt{a_2^2+b_2^2}}$
$\displaystyle\Rightarrow\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2+b_1^2}}=\pm\ \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2+b_2^2}}$
Put $(0,0)$ into both sides of your equation:
$\displaystyle\frac{c_1}{\sqrt{a_1^2+b_1^2}}=\pm \ \frac{c_2}{\sqrt{a_2^2+b_2^2}}$
You can see that both sides should yield nonnegative values, so we only choose $+$ in $\pm$.
Best Answer
Let $C(x_0,y_0)$ be the common point $L_1 \cap L_2$. Thus
$$\begin{cases}a_1x_0 + b_1y_0+ c_1 =0\\a_2x_0 + b_2y_0+ c_2 =0\end{cases}$$
By difference with the initial equations, we obtain the new equivalent equations:
$$\tag{1}\begin{cases}a_1(x-x_0) + b_1(y-y_0)=0 \ \ (L_1)\\a_2(x-x_0) + b_2 (y-y_0)=0 \ \ (L_2)\end{cases} \ \ \implies \ \underbrace{(a_1+Ka_2)(x-x_0) + (b_1+Kb_2)(y-y_0)=0}_{\text{line} (L_1+KL_2)}$$
As any line passing through C has (for fixed coefficients $u$ and $v$) the following equation:
$$\tag{2}u(x-x_0) + v(y-y_0)=0,$$ it suffices to check that there can exist a value of $K$ such that
$$\tag{3}\frac{a_1+Ka_2}{u}= \frac{b_1+Kb_2}{v}$$
(equations (1) and (2) should not necessarily be the same, but they must have proportionnal coefficients).
(3) is a first degree equation that has a single solution, except for a special case where $a_2v-b_2u=0$ (i.e., line $(L_2)$ : see my comment just after the question.)
Remarks:
a) In (3) we have assumed $u\neq0$ and $v\neq0$. This little technical difficulty is removed if one uses, instead of fractions, a determinant equal to zero.
b) What we have done is equivalent to an origin shift.