# Angle Bisector

Angle bisector is locus of point $P(h,k)$ which move such that perpendicular distance of point from lines $L_1=0$ and $L_2=0$ are equal. If the lines $L_1\,:\,a_1x+b_1y+c_1=0$ and $L_2\,:\,a_2x+b_2y+c_2=0$ intersect at point $Q$

i.e $\vert \cfrac{a_1h+b_1k+c_1}{\sqrt{a_1^2+b_1^2}} \vert = \vert \cfrac{a_2h+b_2k+c_2}{\sqrt{a_2^2+b_2^2}} \vert$

Now Replace $(h,k)\rightarrow (x,y)$ we get bisectors $B_1=0\;,\;B_2=0$.

$$\Rightarrow \Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)=\pm \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$$

Let Bisector $B_1 :: \Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)= \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$

Bisector $B_2 :: \Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)=- \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$

## Bisector containing origin.

Let the equation of two lines are $L_1\,:\,a_1x+b_1y+c_1=0$ and $L_2\;:\;a_2x+b_2y+c_2=0$ where $c_1c_2 \gt 0$

Let angle bisector $B_2$ is in the same region as that of origin with respect to line $L_1$ and $L_2$ .

Since $c_1c_2\gt 0$ , hence every point $P(x,y)$ on angle bisector $B_2$ satisfy the same condition i.e $(a_1x+b_1y+c_1)(a_2x+b_2y+c_2 )\gt 0$ for every $(x,y)$ on $B_2$.

Therefore equation of Bisector $B_2$ is

$$\Bigl(\cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)= \Bigl(\cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$$

is the bisector containing origin.

## Equation of bisector containing the given point $P(\alpha,\beta)$

Let $L_1\,:\,a_1x+b_1y+c_1=0$ and $L_2\,:\,a_2x+b_2y+c_2=0$ are two intersecting lines and $B_1=0$ and $B_2=0$ are angle bisectors of line $L_1=0$ and $L_2=0$.

(a) If $(a_1\alpha +b_1\beta +c_1)(a_2\alpha + b_2\beta +c_2)\gt 0$ then for every point $(x,y)$ on bisector $(a_1x +b_1y +c_1)(a_2x + b_2y +c_2)\gt 0$

Hence angle bisector will be $\Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)= \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$

(b) If $(a_1\alpha +b_1\beta +c_1)(a_2\alpha + b_2\beta +c_2)\lt 0$ then for every point $(x,y)$ on bisector $(a_1x +b_1y +c_1)(a_2x + b_2y +c_2)\lt 0$

Hence angle bisector will be $\Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)=- \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$

## Acute angle or Obtuse angle bisector.

Let $L_1\,:\,a_1x+b_1y+c_1=0$ and $L_2\,:\,a_2x+b_2y+c_2=0$ are two intersecting lines and $B_1=0$ and $B_2=0$ are angle bisectors of line $L_1=0$ and $L_2=0$.

Let slope of line $L_1=0$ be $m_1$ and that of angle bisector $B_1=0$ be $m_2$ , then angle between lines $L_1=0$ and $B_1=0$ is given by $$tan(\frac{\theta}{2})=\vert \cfrac{m_1-m_2}{1+m_1m_2} \vert$$

then (i) If $tan(\frac{\theta}{2})\lt 1 \;\Rightarrow \theta \in (0,\cfrac{\pi}{2})$

Then Bisector $B_1=0$ is acute angle bisector.

(ii) If $tan(\frac{\theta}{2}=1 \;\Rightarrow \theta = \cfrac{\pi}{2}$ i.e lines $L_1=0$ and $L_2=0$ are perpendicular to each other.

(iii) If $tan(\frac{\theta}{2})\gt 1\;\Rightarrow \theta \in (\cfrac{\pi}{2} , \pi)$

Hence bisector $B_1=0$ is obtuse angle bisector between line $L_1=0$ and $L_2=0$.

Note :- Let equation of line $L-1 \,:\,a_1x+b_1y+c_1=0$ and $L_2\,:\,a_2x+b_2y+c_2=0$ where $c_1.c_2 \gt 0$

Equation of angle bisectors are $B_1 :: \Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)= \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$

Bisector $B_2 :: \Bigl( \cfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}\Bigr)=- \Bigl( \cfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}\Bigr)$

\begin{matrix} Condition&Acute Angle Bisector &Obtuse Angle Bisector \\if a_1a_2+b_1b_2\gt 0&B_2=0&B_1=0\\ if a_1a_2+b_1b_2\lt 0&B_1=0&B_2=0\\ \end{matrix}

## Properties of angle bisector.

(1) Lines parallel to angle bisectors are equally inclined to both lines.

Lines $B_1=0’$ and $B_2=0’$ are lines parallel to the two bisectors $B_1=0$ and $B_2=0$ are equally inclined to both intersecting lines $L_1=0$ and $L_2=0$ .

(2) Two bisectors are always perpendicular to each other.

$$\angle AQB=\theta \;\;,\;\;\angle AQC=\pi -\theta$$

$$\therefore \angle AQP =\frac{A}{2} \;\;.\;\;\angle AQR =\cfrac{\pi}{2}-\cfrac{\theta}{2}$$

$$\therefore \angle PQR=(\cfrac{\pi}{2}-\cfrac{\theta}{2})+\cfrac{\theta}{2}=\cfrac{\pi}{2}$$

(3) Image of any point on one of the line with respect to bisectors lie on other line.

In Figure $\Delta AQP$ and $\Delta BQO$.

$$\angle QAP = \angle QBP = 90^0$$

$$OP=QP\;\;and\;\;PA=PB\;\;(\,By\, definition\, of\, angle\, bisector\,)$$

$$\therefore \Delta AQP \cong \Delta BQP$$

$$\Rightarrow QA=QB\,—–(1)$$

In $\Delta QBM\;and\;\Delta QAM$

$$QA=QB\;\;(from (1))$$

$$\angle AQM=\angle BQM$$

$$QM=QM \;\; (\,Common \,side )$$

$$\therefore \Delta QBM \cong \Delta QAM$$

$$\Rightarrow AM=MB$$

Hence M is mid – point of AB and $\angle AMQ=\angle BMQ$

and $\angle AMQ + \angle BMQ =180^0$

$$\Rightarrow \angle AMQ=\angle BMQ = 90^0$$

Hence A is image of B with respect to bisector and vice – versa B is image of A with respect to Bisector $B_1$.

NOTE :- The above property is also called as reflection property of bisector.