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

Triangle ABC

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?

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    That is the equation for the perpendicular distance from a point to a line. The angle bisector of $\angle ABC$ is not perpendicular to the side AC so that formula will not give you the correct distance. – 2017-01-24
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    Too bad! Thank you for answer. – 2017-01-24
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    @JohnWaylandBales thank you for graph! – 2017-01-24
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    Since you can use the distance formula to find the length of the three sides you can use Law of Sines and Law of Cosines but that gets tedious. Perhaps an approach using vectors would be easier? – 2017-01-24
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    Try using the angle bisector theorem: http://www.varsitytutors.com/hotmath/hotmath_help/topics/triangle-angle-bisector-theorem – 2017-01-24
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    I add new calc, but is it bad too? – 2017-01-24

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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 $BK$:

$$\frac{AB^2}{AK\cdot AC}-\frac{BK^2}{AK\cdot KC}+\frac{BC^2}{KC\cdot AC}=1$$

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    you are very welcome. – 2017-01-24