Angle Bisectors

The concept of angle bisector is an important head under straight lines. Questions are often framed on this topic in various competitions like the IIT JEE. Hence, it is crucial to understand the acute angle bisector and the obtuse angle bisector in order to be successful in such exams. In this section, we shall discuss the angle bisector definition, followed by the concept of bisector of an angle.
 

What is an Angle Bisector?

Angle bisector of two lines i.e. the line which bisects the angle between the two lines is the locus of a point which is equidistant from the two lines. In other words, an angle bisector has equal perpendicular distance from the two lines. 

Suppose we have two lines

AP is the bisector of an acute angle if,

tan (∠PAN) = tan (θ/2) is such that |tan θ/2| < 1.

AP is an obtuse angle bisector if,

tan (∠PAN) = tan (θ/2) is such that |tan θ/2| > 1.

Note:

1. When both c₁ and c₂ are of the same sign, evaluate a₁a₂ + b₁b₂. If negative, then acute angle bisector is (a₁x + b₁y + c₁)/√(a₁² + b₁²) = + (a₂x + b₂y + c₂)/√(a₂² + b₂²).

2. When both c₁ and c₂ are of the same sign, the equation of the bisector of the angle which contains the origin is (a₁x + b₁y + c₁)/√(a₁² + b₁²) = + (a₂x + b₂y + c₂)/√(a₂² + b₂²).

3. Bisectors of the angle containing the point (α, ß) is (a₁x + b₁y + c₁)/√(a₁² + b₁²) = + (a₂x + b₂y + c₂)/√(a₂² + b₂²) if a₁α + b₁ß + c₁ and a₂α + b₂ß + c₂ have the same sign.

4. Bisectors of the angle containing the point (α, ß) is (a₁x + b₁y + c₁)/√(a₁² + b₁²) = + (a₂x + b₂y + c₂)/√(a₂² + b₂²) if a₁α + b₁ß + c₁ and a₂α + b₂ß + c₂ have the opposite sign.

Another way of identifying the Acute Angle Bisectors and Obtuse Angle Bisectors

Suppose we have two lines L₁ = 0 and L₂ = 0 and e₁ = 0 and e₂ = 0 are the bisectors between these two lines. We take a point R on one of these lines and draw perpendiculars on e₁ and e₂. 

If |p| < |q|, then e₁ is the acute angle bisector.

If |p| > |q|, then e₁ is the obtuse angle bisector.

If |p| = |q|, then the lines L₁ and L₂ are perpendicular.

Note: 

Equation of straight lines passing through P(x₁, y₁) and which are equally inclined with the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are the ones which are aparllel to teh bisectors between these two lines and they pass through the point P.

Illustration:

For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0 , find the equation of the

• bisector of the obtuse angle between them

• bisector of the acute angle between them,

• bisector of the angle which contains (1, 2)

Solution:

Equations of bisectors of the angles between the given lines are

(4x + 3y – 6)/√(42 + 32) = + (5x + 12y + 9)/√(52 +122)

⇒9x – 7y – 41 = 0 and 7x + 9y – 3 = 0.

If θ is the angle between the line 4x + 3y – 6 = 0 and the bisector 9x – 7y – 41 = 0, then tan θ ≥ 1.

Hence, the bisector of the obtuse angle is 9x – 7y – 41 = 0 and the bisector of the acute angle is 7x + 9y – 3 = 0.

For the point (1, 2), we have

4x + 3y – 6 = 4 × 1 + 3 × 2 – 6 > 0\

5x + 12y + 9 = 12 × 12 + 9 > 0.

Hence equation of the bisector of the angle containing the point (1, 2) is 4x + 3y – 6/5 = 5x + 12y + 9/13 ⇒ 9x – 7y – 41 = 0.