How do I find the slope of an angle bisector, given the equations of the two lines that form the angle? The equation for the first line is $y = \frac{1}{2}x - 2$, and the equation for the second line is $y = 2x + 1$. They intersect at $(-2, -3)$.
Someone told me I can just average the slopes of the two lines to find the slope of the bisector, but I'm not sure if it's right.

Take the unitary vectors parallel to the lines: their sum will be parallel to one of the bisecting line, their difference will be parallel to the other one.
To determine which is the one bisecting the acute /obtuse angle, just take the dot product of the two vectors:


*

*if that is positive, then their sum will be parallel to the acute
bisectrix and their difference to the obtuse one

*viceversa if the dot product is negative.


note: this method is valid also in 3D. 
In 2D you can apply the same method to the normal (instead than parallel) unitary vectors.
in your example (using parallel vectors)  

Rewrite the line equations in the proportional form
$$
\left\{ \begin{gathered}
  y = \frac{1}
{2}x - 2\quad  \Rightarrow \quad \frac{{x - 0}}
{2} = \frac{{y - \left( { - 2} \right)}}
{1} \hfill \\
  y = 2x + 1\quad  \Rightarrow \quad \frac{{x - 0}}
{{1/2}} = \frac{{y - 1}}
{1}\quad  \Rightarrow \quad \frac{{x - 0}}
{1} = \frac{{y - 1}}
{2} \hfill \\ 
\end{gathered}  \right.
$$
then the parallel unitary vectors are
$$
\mathbf{u} = \frac{1}
{{\sqrt 5 }}\left( {2,1} \right)\quad \mathbf{v} = \frac{1}
{{\sqrt 5 }}\left( {1,2} \right)\quad 0 < \mathbf{u} \cdot \mathbf{v} = \frac{4}
{5}
$$
and the angle between them is acute. Their sum and difference is
$$
\mathbf{u} + \mathbf{v} = \frac{1}
{{\sqrt 5 }}\left( {3,3} \right)\quad \mathbf{u} - \mathbf{v} = \frac{1}
{{\sqrt 5 }}\left( {1, - 1} \right)
$$
and the equations of the bisecting lines will be:
$$
\left\{ \begin{gathered}
  \frac{{x - x_{\,c} }}
{3} = \frac{{y - y_{\,c} }}
{3}\quad  \Rightarrow \quad x + 2 = y + 3\quad  \Rightarrow \quad y = x - 1\;\;acute \hfill \\
  \frac{{x - x_{\,c} }}
{1} = \frac{{y - y_{\,c} }}
{{ - 1}}\quad  \Rightarrow \quad x + 2 =  - y - 3\quad \quad  \Rightarrow \quad y =  - x - 5\;\;obtuse \hfill \\ 
\end{gathered}  \right.
$$
