In-finding-the-equations-of-the-bisectors-of-the-angles-between-two-lines-a-1-x-b-1-y-c-1-0-and-a-2-x-b-2-y-c-2-0-why-we-observe-a-1-a-2-b-1-b-2-gt-0-or-lt-0-for-obtuse-and-acute-angle-bisectors

Question Number 25971 by Tinkutara last updated on 17/Dec/17

I n f i n d i n g t h e e q u a t i o n s o f t h e

b i s e c t o r s o f t h e a n g l e s b e t w e e n t w o

l i n e s a_{1} x + b_{1} y + c_{1} = 0 a n d a_{2} x + b_{2} y + c_{2} = 0,

w h y w e o b s e r v e a_{1} a_{2} + b_{1} b_{2} > 0 o r < 0

f o r o b t u s e a n d a c u t e a n g l e b i s e c t o r s ?

Commented by Tinkutara last updated on 17/Dec/17

Commented by Tinkutara last updated on 17/Dec/17

F o r e x a m p l e h e r e, w h y w e a r e

c a l c u l a t i n g a_{1} a_{2} + b_{1} b_{2} ? W h y n o t

a_{1} b_{1} + a_{2} b_{2} o r s o m e t h i n g e l s e ?

Answered by ajfour last updated on 17/Dec/17

t o c h e c k i f ∣ θ_{2} - θ_{1} ∣ i s o b t u s e o r

a c u t e w e l o o k f o r s i g n o f

\cos (θ_{2} - θ_{1}) .

∣ θ_{2} - θ_{1} ∣ i s a c u t e o n l y i f

\cos (θ_{2} - θ_{1}) > 0, a n d

\cos θ_{1} = \frac{- b_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}, \sin θ_{1} = \frac{a_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}

\cos θ_{2} = \frac{- b_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}, \sin θ_{2} = \frac{a_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}

s o \cos (θ_{2} - θ_{1}) > 0

\Rightarrow \cos θ_{2} \cos θ_{1} + \sin θ_{2} \sin θ_{1} > 0

o r b_{1} b_{2} + a_{1} a_{2} > 0 .

Commented by Tinkutara last updated on 17/Dec/17

W h a t a r e θ_{1} a n d θ_{2} ?

Commented by ajfour last updated on 17/Dec/17

A n g l e s o f l i n e s w i t h + v e x a x i s,

e q u a t i o n s o f w h o s e b i s e c t o r s w e

s e e k .

Commented by Tinkutara last updated on 17/Dec/17

Sir can you prove exactly Step 3?