How do you find the equation for the angle bisecting line given three coordinates that make up an angle?

Start with a translation -- i.e., shift things to a slightly modified coordinate system so that $B$ is the origin $(0, 0)$:

$$ A \mapsto \left[A_x-B_x, A_y-B_y\right] = A'\\ B \mapsto \left[B_x-B_x, B_y-B_y\right] = B'\\ C \mapsto \left[C_x-B_x, C_y-B_y\right] = C'$$

Compute the angles formed by $A'$ and $C'$ with respect to the positive $x$ axis:

$$\theta_A = \tan^{-1} \frac{A_y-B_y}{A_x-B_x}$$ $$\theta_C = \tan^{-1} \frac{C_y-B_y}{C_x-B_x}$$

Compute the angle between $A'$ and $C'$:

$$\theta = \cos^{-1} \left( \frac{A'\cdot C'}{\|A'\|\|C'\|} \right)$$

Divide by two, and add this angle to the smaller of $\theta_A$ and $\theta_C$:

$$\phi = \min \left\{ \theta_A, \theta_C\right\}+\frac{\theta}{2}$$

Find a unit vector that has this angle in the shifted coordinate system, i.e.

$$\begin{align*} \tan \phi &= \frac{D_y}{D_x} \\ D_x^2+D_y^2 &= 1 \end{align*}$$

Alternatively, ignore the fact that it must be a unit vector, and just set $D_x = 1$ and compute $D_y$--it will still be on the same line, no matter what.

Finally, shift things back to original coordinates:

$$D = \left[ D_x + B_x, D_y + B_y \right].$$

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Example:

$$A = [3,3] \\ B = [2,2] \\ C = [1,3]$$

Then,

$$A' = [3-2,3-2] = [1,1] \\ C' = [1-2,3-2] = [-1,1],$$ $$ \theta = \cos^{-1} \frac{ 1\cdot (-1)+1\cdot 1}{\sqrt{1^2+1^2}\sqrt{(-1)^2+1^2}} = \cos^{-1} 0 = \pi/2\ \;\;\; (\text{90 degrees}),$$ $$ \theta_A = \tan^{-1} 1 = \pi/4\ \;\;\; (\text{45 degrees})$$ $$ \theta_C = \tan^{-1} -1 = 3\pi/4\ \;\;\; (\text{135 degrees})$$

The smaller of $\theta_A$, $\theta_C$ is $\theta_A$, so finally,

$\phi = \frac{\pi}{4}+\frac{1}{2}\theta = \frac{\pi}{4}+\frac{\pi}{4} = \frac{\pi}{2}$

as expected.

If $\phi = \pi/2$, then this is a vector pointing along the y-axis, so let $D' = (0,1)$, which leads to $D = D'+B = (0+2,1+2) = (2,3).$

Plot $A$, $B$, $C$ and D$ and you will see that this is the desired result.