Gradient of angle bisector

By Richard Mycroft (P2053) on Sunday, May 6, 2001 - 12:33 pm :

If you have a line OA with gradient x, and a line OB with gradient y, what is the gradient of the line that bisects angle AOB?

I have been thinking about this for a while and haven't come up with an answer, so thanks for any help you can give me.

By Kerwin Hui (Kwkh2) on Sunday, May 6, 2001 - 03:24 pm :

I'm afraid there is no nice way to do this question. There are essentially 2 ways of doing this. One way is by trig, and the other by vector.

If we let $θ_{1}$ and $θ_{2}$ be the angle line OA and OB make with the positive $x$ -axis, and $y = \frac{1}{2} (θ_{1} + θ_{2})$ . Then

$\tan θ_{1} = x$ , $\tan θ_{2} = y$ .

Now $\tan 2 y = x + y / (1 - x y)$

and now solve for $\tan y$ by the double angle formula. Note that the two solutions correspond to which of the two angles you bisect.

Alternatively, you can use the unit vectors along OA and OB added together will give you the angle bisector (by symmetry) and end up with the same answer.

Kerwin