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BISECTORS OF ANGLES BETWEEN STRAIGHT LINES:
THEOREM:
To find the equations of the bisectors of the angles between the straight lines
a1x + b1x + c1 = 0
(1)
a2x + b2x + c2 = 0
(2)
Solution:

Let the two straight lines be AL1 and AL2 and the bisectors of the angles between
them be AM1 and AM2.
Let P be a point on either of these bisectors and PN1 and PN2 be two perpendiculars
to the given lines.
The triangles PAN1 and PAN2 are equal in all respects, so that perpendiculars
PN1 and PN2 are equal in magnitude.
Let the equations to the straight lines be written so that c1 and c2 are both
negative and to the quantities
let the positive sign be prefixed.
let the positive sign be prefixed.
If P lie on AM1, i.e. on the bisector of the angle between the two straight lines in which the origin
lies, the point P and the origin lie on the same side of each of the two lines. Hence the two quantities
(1) have the same sign as c1 and c2 respectively.
In this case since c1 and c2 have the same sign, the quantities (1) have the same sign and hence
But this is the condition that the point (h, k) may lie on the straight line
Hence this is the equation of AM1.
If P lies on the other bisector AM2 then the two quantities (1) will have opposite signs so that the equation AM2 becomes
Hence the equation of the bisectors (whether both the constant terms be positive or negative) will be
The upper sign gives the bisector of the angle in which the origin lies.
Example:
Find the equations to the bisectors of the angles between the straight lines
3x - 4y + 7 = 0 and 12x - 5y - 8 = 0.
Solution:
If the constant terms are both positive then the given equations will be written as
3x - 4y + 7 = 0 and -12x + 5y + 8 = 0
The equation to the bisector of the angle in which the origin lies is
Or 13(3x - 4y + 7) = 5(-12x + 5y + 8)
Or 99x - 77y + 51 = 0
The equation to the other bisector is
Or 13(3x - 4y + 7) = -5(-12x + 5y + 8)
Or 13(3x - 4y + 7) + 5(-12x + 5y + 8) = 0
Or 21x + 27y - 131 = 0. (Answer)
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