Two lines: a 1 x + b 1 y + c 1 = 0 and...

babyagelesszj

babyagelesszj

Answered

2022-07-07

Two lines: a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are given. I know that the equation of its bisectors is a 1 x + b 1 y + c 1 ( a 1 2 + b 1 2 ) = ± a 2 x + b 2 y + c 2 ( a 2 2 + b 2 2 ) But I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming c 1 , c 2 both are of same sign, I know if a 1 a 2 + b 1 b 2 > 0 and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e. t a n θ = m 1 m 2 1 + m 1 m 2 and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: "Assuming c 1 , c 2 both are of same sign IF a 1 a 2 + b 1 b 2 > 0 then if we take positive sign we get the obtuse angle bisector".

Answer & Explanation

Dobermann82

Dobermann82

Expert

2022-07-08Added 15 answers

We have two lines :
L 1 : a 1 x + b 1 y + c 1 = 0 , L 2 : a 2 x + b 2 y + c 2 = 0
and the angle bisectors :
L ± : a 1 x + b 1 y + c 1 a 1 2 + b 1 2 = ± a 2 x + b 2 y + c 2 a 2 2 + b 2 2
If we let θ be the (smaller) angle between L + and L 1 , then we have
cos θ = | a 1 ( a 1 a 1 2 + b 1 2 a 2 a 2 2 + b 2 2 ) + b 1 ( b 1 a 1 2 + b 1 2 b 2 a 2 2 + b 2 2 ) | a 1 2 + b 1 2 ( a 1 a 1 2 + b 1 2 a 2 a 2 2 + b 2 2 ) 2 + ( b 1 a 1 2 + b 1 2 b 2 a 2 2 + b 2 2 ) 2
= | a 1 2 + b 1 2 a 1 a 2 + b 1 b 2 a 2 2 + b 2 2 | a 1 2 + b 1 2 2 2 a 1 a 2 + b 1 b 2 ( a 1 2 + b 1 2 ) ( a 2 2 + b 2 2 ) × 2 1 a 1 2 + b 1 2 2 1 a 1 2 + b 1 2 = 1 a 1 a 2 + b 1 b 2 ( a 1 2 + b 1 2 ) ( a 2 2 + b 2 2 ) 2
Hence, we can see that
a 1 a 2 + b 1 b 2 > 0 cos θ < 1 / 2 θ > 45 L +  is the obtuse angle bisector
as desired.

(Note that " c 1 , c 2 both are of same sign" is irrelevant.)

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