JEE Mains Maths · Straight Lines

Angle Bisectors and Pairs of Lines

The bisectors of the angles between two lines, the angle bisector theorem in a triangle, and a pair of lines through one point written as a single second-degree equation.

All Straight Lines notes JEE Mains Maths strategy

Why this matters

Ten PYQs, eight of them multiple choice, and three from 2026. Six find an angle bisector, between two lines or at a vertex of a triangle, some with the angle bisector theorem. Four write a pair of lines through one point as a single equation and read off its angle or its bisectors. Two ideas cover the page.

Concept 1 of 2: Angle bisectors

A point on a bisector is equally far from the two lines, so the bisectors come from setting the two distances equal, with a ± sign. With both constants positive, the + sign gives the bisector of the angle that contains the origin. In a triangle, the bisector of angle

B

cuts

AC

D

with

AD:DC=BA:BC

. The image of

A

in that bisector lies on line

BC

, which often gives the side

BC

directly.

Definition

Bisectors: $\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$ .
With $c_1,c_2>0$ , the + sign gives the bisector of the angle containing the origin.
Internal bisector of $B$ meets $AC$ at $D$ with $AD:DC=BA:BC$ .
The image of $A$ in the bisector of angle $B$ lies on $BC$ .

Angle bisectors

\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}}

Worked example

Find the bisectors of the angles between

3x-4y+1=0

and

4x+3y+2=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q63Moderate

Example 1 · Straight Lines · Angle Bisectors and Pairs of Lines

Let

A(1,0),B(2, - 1)

and

C\left( \frac{7}{3},\frac{4}{3} \right)

be three points. If the equation of the bisector of the angle ABC is

\alpha x + \beta y = 5

, then the value of

\alpha^{2}+\beta^{2}

Origin angle is not always the acute one

The + sign picks the angle containing the origin, which may be acute or obtuse. For the acute bisector, check the angle it makes with one of the lines: it must be less than

45^\circ

Drill 5 more on angle bisectors

Concept 2 of 2: A pair of lines in one equation

Two lines

L_1=0

and

L_2=0

together are

L_1L_2=0

, one second-degree equation. Through the origin,

ax^2+2hxy+by^2=0

is two lines whose slopes are the roots of

bm^2+2hm+a=0

. The points equidistant from two lines satisfy one such equation too: squaring the bisector condition gives both bisectors at once. To get the lines joining the origin to the points where a line meets a curve, make the curve's equation homogeneous using the line.

Definition

$ax^2+2hxy+by^2=0$ : $m_1+m_2=-\frac{2h}b$ , $m_1m_2=\frac ab$ .
Angle: $\tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}$ ; perpendicular when $a+b=0$ .
Its bisectors: $\frac{x^2-y^2}{a-b}=\frac{xy}h$ .
Equidistant from two lines: $\frac{(a_1x+b_1y+c_1)^2}{a_1^2+b_1^2}=\frac{(a_2x+b_2y+c_2)^2}{a_2^2+b_2^2}$ .
Homogenise with $lx+my=n$ : replace each 1 in the curve by $\frac{lx+my}n$ .

Pair of lines through the origin

ax^2+2hxy+by^2=0:\quad \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}

Worked example

Find the angle between the lines

x^2-4xy+y^2=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q157Moderate

Example 2 · Straight Lines · Angle Bisectors and Pairs of Lines

x^{2}-y^{2}+ 2hxy + 2gx + 2fy + c = 0

is the locus of a point, which moves such that it is always equidistant from the lines

x + 2y + 7 = 0

and

2x - y + 8 = 0

, then the value of

g + c + h - f

equals

h is half the xy coefficient

ax^2+2hxy+by^2

, the coefficient of

xy

2h

. For

x^2-4xy+y^2

h=-2

, not

-4

Drill 3 more on a pair of lines in one equation

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Angle bisectors
Angle bisectors
$\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$
A pair of lines in one equation
Pair of lines through the origin
$ax^2+2hxy+by^2=0:\quad \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}$

Watch out for (2)

Origin angle is not always the acute one
→ Angle bisectors
h is half the xy coefficient
→ A pair of lines in one equation

Test yourself on Straight Lines

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Straight Lines chapter test Drill all 10 questions