Two lines - a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are given(adsbygoogle = window.adsbygoogle || []).push({});

I know that the equation of its bisectors is

a1x + b1y + c1 / √a1^2 + b1^2 = +- a2x + b2y + c2 / √a2^2 + b2^2

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 c1 , c2 both are of same sign

My text says if a1a2 + b1b2 > 0 and if we take the positve sign we get the obtuse angle bisector - and after many examples i think its true

But i want to prove it using general equation of line

I tried to find the angle between bisector adn original line

i.e. tan θ = m1 - m2 / 1+ m1m2 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

Any simple proof of the following statement:

Assuming c1 , c2 both are of same sign

IF a1a2 + b1b2 > 0 then if we take positve sing we get the obtuse angle bisector

Thank you!

**Physics Forums - The Fusion of Science and Community**

# Coordinate geometry - bisector of two lines

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Coordinate geometry - bisector of two lines

Loading...

**Physics Forums - The Fusion of Science and Community**