Calculation error the length of an angle bisector How to find the length of an angle bisector ($BK$) in a triangle $A(1;4),~B(7;8),~C(9;2)$.

I use this formula:
$\frac{A_{1} \cdot x+B_{1} \cdot y+C_{1}}{\sqrt{A_{1}^{2}+ B_{1}^{2}}}=\frac{A_{2} \cdot x+B_{2} \cdot y+C_{2}}{\sqrt{A_{2}^{2}+ B_{2}^{2}}}$.
And, my result: $\frac{2x−3y+10}{\sqrt{4+9}}=\frac{3x+y−29}{\sqrt{9+1}}$, but it is not equation of angle bisector.
What's my mistake?
Update: $\frac{x-7}{\frac{1+\frac{\sqrt{13}}{\sqrt{10}} \cdot 9}{1+\frac{\sqrt{13}}{\sqrt{10}}} - 7} - \frac{y-8}{\frac{4+\frac{\sqrt{13}}{\sqrt{10}} \cdot 2}{1+\frac{\sqrt{13}}{\sqrt{10}}} - 8}=0$ - is it bad?

Maybe you might find interesting this approach:
Once you have all tree points you have all sides ($AB,AC,BC$) and then you can use the angle bisector theorem: 
$$\frac{AB}{AK}=\frac{BC}{KC}$$
and using that $AK+KC=AC$ you can find $AK$ and $KC$. After that you can use Stewart's theorem in order to find $BK$:
$$\frac{AB^2}{AK\cdot AC}-\frac{BK^2}{AK\cdot KC}+\frac{BC^2}{KC\cdot AC}=1$$
