Equation of the Bisectors - Practice Questions & MCQ

Equation of the Bisectors

The equation of the angle bisectors between the two lines

$
\begin{aligned}
& \mathrm{L}_1=\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0 \text { and } \mathrm{L}_2=\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0 \text { is }^{{ }^{-}} \\
& \frac{\left(\mathbf{a}_1 \mathbf{x}+\mathbf{b}_1 \mathbf{y}+\mathbf{c}_1\right)}{\sqrt{\mathbf{a}_1^2+\mathbf{b}_1^2}}= \pm \frac{\left(\mathbf{a}_2 \mathbf{x}+\mathbf{b}_2 \mathbf{y}+\mathbf{c}_2\right)}{\sqrt{\mathbf{a}_2^2+\mathbf{b}_2^2}}
\end{aligned}
$

Given equations of lines

$
\begin{aligned}
& \mathrm{L}_1: \mathrm{AB}: \mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0 \\
& \mathrm{~L}_2: \mathrm{CD}: \mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0
\end{aligned}
$
RR' and SS' are two bisectors of the angle between the line $A B$ and $C D$. And, $P(x, y)$ be any point on the line $R R$ ', then length of perepndicular from P on AB

$
\begin{array}{ll} 
& \quad \text { length of perepndicular from } \mathrm{P} \text { on } \mathrm{CD} \\
\therefore & \frac{\left|\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1\right|}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2}}=\frac{\left|\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2\right|}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2}} \\
\text { or } & \frac{\left(\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1\right)}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2}}= \pm \frac{\left(\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2\right)}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2}}
\end{array}
$
Bisector of the Angle Containing the Origin

Rewrite the equation of the line $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$ such that the constant term $\mathrm{c}_1$ and $\mathrm{c}_2$ are positive.
Then, the equation

$
\frac{\left(a_1 x+b_1 y+c_1\right)}{\sqrt{a_1^2+b_1^2}}=\frac{\left(a_2 x+b_2 y+c_2\right)}{\sqrt{a_2^2+b_2^2}}
$

gives the equation of the bisector of the angle containing the origin and

$
\frac{\left(a_1 x+b_1 y+c_1\right)}{\sqrt{a_1^2+b_1^2}}=-\frac{\left(a_2 x+b_2 y+c_2\right)}{\sqrt{a_2^2+b_2^2}}
$

gives the equation of the bisector of the angle not containing the origin.

Distinguish between obtuse and acute angle bisector

Let, $\quad \mathrm{L}_1: \mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$

$
\mathrm{L}_2: \mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0
$

where, $c_1>0, c_2>0$
Equation of bisectors are

$
\begin{aligned}
& \frac{\left(a_1 x+b_1 y+c_1\right)}{\sqrt{a_1^2+b_1^2}}=\frac{\left(a_2 x+b_2 y+c_2\right)}{\sqrt{a_2^2+b_2^2}} \\
& \frac{\left(a_1 x+b_1 y+c_1\right)}{\sqrt{a_1^2+b_1^2}}=-\frac{\left(a_2 x+b_2 y+c_2\right)}{\sqrt{a_2^2+b_2^2}}
\end{aligned}
$
To distinguish between acute angles and obtuse angle bisectors, choose one of the equations of bisector, say eq (iii). Let the angle between this bisector and one of the given line be Ө/2, where Ө is an angle between lines containing these bisectors.   

RO is the bisector of an acute angle if,

$
\begin{aligned}
& \ominus<\pi / 2 \\
& \Rightarrow \Theta / 2<\pi / 4 \\
& \Rightarrow|\tan (\Theta / 2)|<1 \\
& \Rightarrow \tan (\angle \mathrm{ROB})<1
\end{aligned}
$
Similarly, RO is the bisector of an obtuse angle if, $|\tan (\Theta / 2)|>1$