<s>[INST] 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.
 [/INST] Line 1 is $y = \frac 12 x - 2$.  Line 2 is $y = 2x +1$.  They intersect at $(-2-3)$.
If on line 1 you go over on $x$ $2$ units you will go up on $y$ by $1$ unit.  This will put you at $(0, -2)$.  The distance traveled is $\sqrt {1^2 + 2^2} = \sqrt{5}$.
If on line 2 you go over on $x$ $1$ units you will go up on $y$ by $2$ unit.  This will put you at $(-1, -1)$.  The distance traveled  is $\sqrt {2^2 + 1^2} = \sqrt{5}$.
The angle bisector will go through the midpoint of $(0,-2)$ and $(-1,-1)$$*$.  So the angle bisector will go through $(-\frac 12, -1\frac 12)$.  So the angle bisector goes through the point $(-2,-3)$ and $(-\frac 12, -1\frac 12)$ so the slope is $\frac{-1\frac 12 - (-3)}{-1\frac 12 -(-2)} = \frac {1\frac 12}{1\frac 12} = 1$. 
$*$ because... $A = (-2,-3); B= (0,-2); C=(-1,-1);$ and $AB$ = $AC= \sqrt{5}$ so $\triangle BAC$ is isoceles, and the angle bisector of $\angle BAC$ passes through the midpoint of $BC$.
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You friend is not quite right.  You can average the angles but slopes are not angles and there is not a linear conversion between them.  (There is a trigonometric conversion between them.  But not a linear conversion.)
Bear with me.
Suppose the two lines intersect at $(u,v)$ and line $1$ has slope $m$ and line $2$  has slope $n$. 
Move along the line $1$ from $(u,v)$ a distance of $1$ unit.  You will have move $\delta $ in the $x$ direction and $m*\delta $ in the $y$ direction so your total distance is $\sqrt{\delta^2 + m^2\delta^2} = 1$.
So $\delta\sqrt{1 + m^2} = 1$ so $\delta = \frac 1{\sqrt{1 + m^2}}$.
So the point on line $1$ that is one unit away form $(u, v)$ is the point $(x_1, y_1) = (u + \frac 1{\sqrt{1 + m^2}}, v + \frac m{\sqrt{1 + m^2}})$.
Likewise the point on line $2$ that is one unit away from $(u,v)$ will be the point $(x_2, y_2) = (u + \frac 1{\sqrt{1 + n^2}}, v + \frac n{\sqrt{1 + n^2}})$
The angle bisector will contain the midpoint of $(x_1, x_2)$ and $(x_2, y_2)$.
The midpoint is $(x_m, y_m) = (u + \frac{[\frac 1{\sqrt{1 + m^2}}]+[\frac 1{\sqrt{1 + n^2}}]}2, w + \frac{[\frac m{\sqrt{1 + m^2}}]+[\frac n{\sqrt{1 + n^2}}]}2)$.
So.... the slope of the angle bisector will be:
$\frac {y_m - v}{x_m - u}= \frac{\frac{[\frac m{\sqrt{1 + m^2}}]+[\frac n{\sqrt{1 + n^2}}]}2}{\frac{[\frac 1{\sqrt{1 + m^2}}]+[\frac 1{\sqrt{1 + n^2}}]}2}=$
$\frac{[\frac m{\sqrt{1 + m^2}}]+[\frac n{\sqrt{1 + n^2}}]}{[\frac 1{\sqrt{1 + m^2}}]+[\frac 1{\sqrt{1 + n^2}}]}=\frac{m\sqrt{1 + n^2}+n\sqrt{1 + m^2}}{\sqrt{1 + m^2}+\sqrt{1 + n^2}}$
$[\frac{m\sqrt{1 + n^2}+n\sqrt{1 + m^2}}{\sqrt{1 + m^2}+\sqrt{1 + n^2}}=\frac{\frac 12\sqrt{1 + 2^2}+2\sqrt{1 + \frac 12^2}}{\sqrt{1 + \frac 12^2}+\sqrt{1 + 2^2}}=\frac{\sqrt{5}/2+ \sqrt{5}}{\sqrt{5}+ \sqrt{5}/2}=1]$
Which... I must confess is a formula I never learned and would never expect anyone to memorize.  I'd expect if one needs to find the slope of an angle bisector one would calculate it for the specific lines.
You might be able to simplify that equation further. 
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That equation is sort of an "average"; just not a standard arithmetic  average.
You can calculate the average of the angles of the lines.  But slopes are not angles and do not have a linear conversion.
If you know trigonemetry:
Slope = $\frac{rise}{run} = \frac{\sin \theta}{\cos \theta} = \tan \theta$ so the angle of a line is $\theta = \arctan m$.
So the angle of the angle bisector is $\psi = \frac {\theta + \omega}2 = \frac {\arctan(m) + \arctan(n)}2$
And the slope of the angle bisector is $k = \tan(\psi) = \tan(\frac {\arctan(m) + \arctan(n)}2)=\frac{m\sqrt{1 + n^2}+n\sqrt{1 + m^2}}{\sqrt{1 + m^2}+\sqrt{1 + n^2}}$
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