Find a specific vector equation of a line that divides a angle in half.

Question

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

Answer 1

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

$$ \beta = \frac{\alpha}{2} $$

Answer 2

I would say that

• normalized vectors $\vec{AB},\vec{AC}$
• their vector sum be the direction vector of the line search

Answer 3

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

1. Intersect lines $AB$ and $AC$ with a circle of radius $1$ around $A$. Call the resulting points of intersection $B'$ and $C'$.
2. Construct the midpoint between $B'$ and $C'$.
3. Connect that mitpoint with $A$.