geometry
The angle bisector of
$L_1: a_1 x + b_1 y + c_1=0$
and
$ L_2 : a_2 x + b_2 y + c_2=0$
$(a_i,b_i,c_i) \in \Bbb R$
can be found be solving the equation
$\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2+b_1^2}}=\pm \ \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2+b_2^2}}$
But our teacher told us that the equation of the angle bisector pasing through the region containing the origin can be obtained by solving only the positive case of the equation given that $(c_1,c_2)\gt 0$.How can we prove this?
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.
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.
Best Answer
$\displaystyle\frac{|a_1x + b_1y + c_1|}{\sqrt{a_1^2+b_1^2}}=\frac{|a_2x + b_2y + c_2|}{\sqrt{a_2^2+b_2^2}}$
$\displaystyle\Rightarrow\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2+b_1^2}}=\pm\ \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2+b_2^2}}$
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 $\pm$.