Angular bisector is the locus of a point which moves in such a way so that its distance from two intersecting lines remains same.

The equations of the two bisectors of the angles between the lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are

$ \displaystyle \frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}$

If the two given lines are not perpendicular i.e. a_{1} a_{2} + b_{1} b_{2} ≠ 0, then one of these equations is the equation of the bisector of the acute angle and the other that of the obtuse angle.

Note:

∎ Whether both lines are perpendicular or not but the angular bisectors of these lines will always be mutually perpendicular.

__The bisectors of the acute and the obtuse angles:__

Take one of the lines and let its slope be m_{1} and take one of the bisectors and let its slope be m_{2}. If θ be the acute angle between them, then find .

$ \displaystyle tan\theta = |\frac{m_1 – m_2}{1 + m_1 m_2} | $

If tanθ > 1 then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle.

If 0 < tanθ < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angle.

If two lines are a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x+ b_{2}y + c_{2} = 0, then

$ \displaystyle \frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = \frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}$

will represent the equation of the bisector of the acute or obtuse angle between the lines according as c_{1}c_{2}(a_{1}a_{2} + b_{1}b_{2}) is negative or positive.

__The equation of the bisector of the angle containing the origin :__

Write the equations of the two lines so that the constants c_{1} and c_{2} become positive. Then the equation

$ \displaystyle \frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = \frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}$

is the equation of the bisector containing the origin.

Remarks:

∎ If a_{1}a_{2} + b_{1}b_{2} < 0, then the origin will lie in the acute angle and if a_{1}a_{2} + b_{1}b_{2} > 0, then origin will lie in the obtuse angle.

∎ The remark (i) is helpful in finding the equation of bisector of the obtuse angle or acute angle directly.

__The equation of the bisector of the angle which contains a given point__

The equation of the bisector of the angle between the two lines containing the point (α,β) is

$ \displaystyle \frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = \frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}$

or , $ \displaystyle \frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = – \frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}$

according as a_{1} α + b_{1} β +c_{1} and a_{2} α + b_{2} β + c_{2} are of the same signs or of opposite signs.

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

(i) bisector of the obtuse angle between them.

(ii) bisector of the acute angle between them.

(iii) bisector of the angle which contains (1, 2).

Solution: Equations of bisectors of the angles between the given lines are

$ \displaystyle \frac{4 x + 3 y -6}{\sqrt{4^2 + 3^2}} = \pm \frac{5 x + 12 y + 9}{\sqrt{5^2 + 12^2}}$

=> 9x − 7y − 41 = 0 and 7x + 9y − 3 = 0.

If θ is the acute angle between the line 4x + 3y − 6 = 0 and the bisector 9x − 7y − 41 = 0, then

$ \displaystyle tan\theta = |\frac{(-4/3)-9/7}{1+(-4/3)(9/7)} | $

= 11/3 > 1

Hence

(i) The bisector of the obtuse angle is 9x − 7y − 41 = 0

(ii) The bisector of the acute angle is 7x + 9y − 3 = 0

(iii) The bisector of the angle containing the origin

$ \displaystyle \frac{-4 x -3 y + 6}{\sqrt{(-4)^2 + (-3)^2}} = \pm \frac{5 x + 12 y + 9}{\sqrt{5^2 + 12^2}}$

=> 7x + 9y − 3 = 0

(i) For the point (1 , 2) , 4x + 3y − 6 = 4 × 1 + 3×2 − 6 > 0

5x + 12y + 9 = 5 × 1 + 12 × 2 + 9 > 0

Hence equation of the bisector of the angle containing the point (1 , 2) is

$ \displaystyle \frac{4 x +3 y – 6}{5} = \pm \frac{5 x + 12 y + 9}{13}$

=> 9x − 7y − 41 = 0.

Alternative:

Making C_{1} and C_{2} positive in the given equations, we get

−4x − 3y + 6 = 0 and 5x + 12y + 9 = 0

Since a_{1}a_{2} + b_{1}b_{2} = −20 −36 = −56 < 0, so the origin will lie in the acute angle. Hence bisector of the acute angle is given by

$ \displaystyle \frac{-4 x -3 y + 6}{\sqrt{(-4)^2 + (-3)^2}} = \pm \frac{5 x + 12 y + 9}{\sqrt{5^2 + 12^2}}$

i.e. 7x + 9y − 3 = 0

Similarly bisector of obtuse angle is 9x − 7y − 41 = 0.