SASTRA University B. Tech Application Form 2025 – Apply Online @sastra.edu

Equation of the Bisectors - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Equation of the Bisectors is considered one of the most asked concept.
28 Questions around this concept.

Solve by difficulty

The equation of the bisector of angle between the lines $\mathrm{x+y=1}$ and $\mathrm{7 x-y=3}$ that contains the point $\mathrm{(2,3)}$ is

$\mathrm{3 x+y-1=0}$

$\mathrm{x-3 y+1=0}$

$\mathrm{2 x+y-1=0}$

$\mathrm{x-2 y+1=0}$

View Solution

Concepts Covered - 0

Equation of the Bisectors

Equation of the Bisectors

The equation of the angle bisectors between the two lines

$\mathrm{L}_{1}=\mathrm{a}_{1} \mathrm{x}+\mathrm{b}_{1} \mathrm{y}+\mathrm{c}_{1}=0$ and $\mathrm{L}_{2}=\mathrm{a}_{2} \mathrm{x}+\mathrm{b}_{2} \mathrm{y}+\mathrm{c}_{2}=0$ is

$\mathbf{\frac{\left(a_{1} x+b_{1} y+c_{1}\right)}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{\left(a_{2} x+b_{2} y+c_{2}\right)}{\sqrt{a_{2}^{2}+b_{2}^{2}}}}$

$\\\mathrm{Given\;equations\;of\;lines}\\\mathrm{L_1:AB:a_1x+b_1y+c_1=0}\\\mathrm{L_2:CD:a_2x+b_2y+c_2=0}\\\text{RR' and SS' are two bisectors of the angle between the line}\\\text{AB and CD. And, P (x,y) be any point on the line RR', then}\\\text{length of perepndicular from P on AB }\\\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= length of perepndicular from P on CD }\\\\\mathrm{\therefore\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{|a_1x+b_1y+c_1|}{\sqrt{a_1^2+b_1^2}}=\frac{|a_2x+b_2y+c_2|}{\sqrt{a_2^2+b_2^2}}}\\\\\mathrm{or\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\left (a_1x+b_1y+c_1 \right )}{\sqrt{a_1^2+b_1^2}}=\pm\frac{\left (a_2x+b_2y+c_2 \right )}{\sqrt{a_2^2+b_2^2}}}$

Bisector of the Angle Containing the Origin

Rewrite the equation of the line a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 such that the constant term c₁ and c₂ are positive.

Then, the equation

$\frac{\left (a_1x+b_1y+c_1 \right )}{\sqrt{a_1^2+b_1^2}}=\frac{\left (a_2x+b_2y+c_2 \right )}{\sqrt{a_2^2+b_2^2}}$

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

$\frac{\left (a_1x+b_1y+c_1 \right )}{\sqrt{a_1^2+b_1^2}}=-\frac{\left (a_2x+b_2y+c_2 \right )}{\sqrt{a_2^2+b_2^2}}$

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

Distinguish between obtuse and acute angle bisector

Distinguish between obtuse and acute angle bisector

$\\\mathrm{Let,\;\;\;L_1:a_1x+b_1y+c_1=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;L_2:a_2x+b_2y+c_2=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)}\\\mathrm{where,\;\;c_1>0,\;c_2>0}\\\\\mathrm{Equation\;of\;bisectors\;are}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\left (a_1x+b_1y+c_1 \right )}{\sqrt{a_1^2+b_1^2}}=\frac{\left (a_2x+b_2y+c_2 \right )}{\sqrt{a_2^2+b_2^2}}\;\;\;\;\;\;\;\;\;\;\;\ldots(iii)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\left (a_1x+b_1y+c_1 \right )}{\sqrt{a_1^2+b_1^2}}=-\frac{\left (a_2x+b_2y+c_2 \right )}{\sqrt{a_2^2+b_2^2}}\;\;\;\;\;\;\;\;\;\ldots(iv)}$

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,

Ө < π/2

⇒ Ө/2 < π/4

⇒ |tan (Ө/2)| < 1

⇒ tan (∠ROB) < 1

Similarly, RO is the bisector of an obtuse angle if, | tan (Ө/2) | > 1

Shortcut Method for Identifying Acute Obtuse Angle Bisectors

$\\\text{Equation of two non-parallel lines are}\\\mathrm{L_1:AB:a_1x+b_1y+c_1=0}\\\mathrm{L_2:CD:a_2x+b_2y+c_2=0}\\\text {Then equation of bisectors are }\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\left (a_1x+b_1y+c_1 \right )}{\sqrt{a_1^2+b_1^2}}=\pm\frac{\left (a_2x+b_2y+c_2 \right )}{\sqrt{a_2^2+b_2^2}}}\\\\\\\begin{array}{c| c ccc}\hline \text{Conditions.} & &\text{Acute angle bisector} &\text { Obtuse angle bisector } \\ \hline\hline\\ a_1a_2+b_1b_2>0 && -&+\\ \\ \hline\\a_1a_2+b_1b_2<0 && +&-\\ \\\hline \end{array}$