Find a specific vector equation of a line that divides a angle in half. I've been studying a little geometry on my own, and I just recently stumbled on this problem, that I'm unable to answer:
Given the points A=(2,-1), B=(5,4) and C=(-7,8), find a vector equation of a line that divides the angle BAC in half(Same angle for each size).
Thank you
-Dom
 A: I would say that


*

*normalized vectors $\vec{AB},\vec{AC}$

*their vector sum be the direction vector of the line search

A: Here is how you could construct this, in a way that can be turned into a computation as well.


*

*Intersect lines $AB$ and $AC$ with a circle of radius $1$ around $A$. Call the resulting points of intersection $B'$ and $C'$.

*Construct the midpoint between $B'$ and $C'$.

*Connect that mitpoint with $A$.

A: Using user georg's approach:
$$
\begin{align}
u &= \frac{B - A}{|B-A|} = (0.51450, 0.85749) \\
v &= \frac{C - A}{|C-A|} = (-0.70711, 0.70711) \\
z &= \frac{u+v}{|u+v|} = (-0.12218, 0.99251)
\end{align}
$$
where $z$ is the unit vector on the desired line
$$
g(t) = t z \quad (t \in \mathbb{R})
$$
Checking the angles gives:
$$
\begin{align}
u \cdot v &= \cos \alpha \iff \alpha = 75.964^\circ \\
u \cdot z &= \cos \beta \iff \beta = 37.982^\circ
\end{align}
$$
The interesting bit is to prove, that $\beta = \alpha /2$.
In the parallelogram of $u+v$ (draw in the other diagonal $v-u$ as well) one can read from the triangle with the sides $u$, $v-u$, $v$
$$
180^\circ = \alpha + 2 \gamma 
$$
and from the triangle with the sides $(u+v)/2$, $(v-u)/2$ and $v$
$$
90^\circ = \beta + \gamma
$$
which gives
$$
\beta = \frac{\alpha}{2}
$$
