Find bisector equation using a point the bisector goes through and an angle 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.
$BC=\sqrt{AB^2+AC^2}$

$BC=\sqrt{(-7)^2+4^2}=\sqrt{65}$

$BCC'$ is an isosceles triangle (by construction) $\implies BC=BC'$.

$C'=(\sqrt{65}-7,0)$

Let $M$ be the midpoint of $CC'$.

$M=\left(\dfrac{\sqrt{65}-7}{2},2\right)$



Bisector $\overleftrightarrow{BM}$

$\left| \begin{array}{} x &amp; y &amp; 1 \\ -7 &amp; 0 &amp; 1 \\ \dfrac{\sqrt{65}-7}{2} &amp; 2 &amp; 1 \\ \end{array} \right| =0$

$4x-(\sqrt{65}+7)y+28=0$
$$ a=\sqrt{49+16}= \sqrt{65},\, b=4,\, c=7\,$$

Using a known formula bisector length =  Product/AM

$$d= (2 \sqrt{65}\, 7)/ (7+\sqrt{65})\approx 7.49367$$

$$ B/2=\cos^{-1} \frac{7}{d}\approx 20.9133 ^{\circ}$$

EDIT1:

Equation of bisector

$$ \frac{x}{-7}+\frac{y}{d}=1$$