I am given a triangle with vertices $A=(0,0)$, $B=(-7,0)$ and $C=(0,4)$. This is a right triangle. One might think from graphing that the angle bisector cuts from point B and point $(0,2)$ but this is not the case. I am having trouble coming up with the equation of the bisector given that the bisector goes through point B and has angle of $29.7$ degrees. How does one go about find the slope for this? I know once I find the slope I can plug it into the point slope formula to get my equation. I have searched this problem on here but all I found was something that said I had to get the inverse of $tan$ and subtract this from my $29.7$ degree that I previously found but when I do this, I get something that is not even close to the correct answer.
EDIT I found this but I am still getting the wrong answer.
This bisector is one of the two straight lines of points which are at the same distance from line $(BC)$ and line $(BA)$.
Now the distance from point $(x,y)$ to line $(BA)$ is just $|y|$. As to the line $(BC)$, its equation is $$-\frac x7+\frac y4=1\iff 4x-7y+28=0,$$ so the distance from point $(x,y)$ to the hypotenuse is $$\frac{|4x-7y+28|}{\sqrt{4^2+(-7)^2}}=\frac{|4x-7y+28|}{\sqrt{65}}.$$ So we have to solve $$\frac{|4x-7y+28|}{\sqrt{65}}=|y|\iff (4x-7y+28)^2=65y^2$$ and the equation of the pair of angle bisectors is $$\bigl(4x+(\sqrt{65}-7)y+28\bigr)\bigl(4x-(\sqrt{65}+7)y+28\bigr)=0$$ The internal angle bisector has a positive $y$-intercept, so its equation is $$4x-(\sqrt{65}+7)y+28=0.$$
Added: a short approach based on geometry:
Denote $I$ the intersection of the angle bisector through $B$ with side $AC$ and apply the angle bisector theorem: $$\frac{AI}{BA}=\frac{IC}{AC}=\frac{AI+IC}{BA+AC}=\frac 4{7+\sqrt{65}},\quad\text{so }\enspace AI=\frac{4BA}{7+\sqrt{65}}=\frac{28}{7+\sqrt{65}}.$$ Now, we have the $x$ and $y$-intercepts of the angle bisector, and there is a formula for the equation of a line given its $x$ and $y$-intercepts: $$\frac x{-7}+\frac{7+\sqrt{65}}{28}y=1\iff -4x+(7+\sqrt{65})y=-28\iff\dotsm$$