MHB Calculating the Length of an Angle Bisector: Is My Solution Correct?

[divisor]

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

I calc $$BK$$:
[math]\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[/math]

Is it right?

 

[Euge]

Welcome, divisor! (Wave)

Is the equation you wrote supposed to be the equation of line containing $BK$? I haven't checked it for correctness, but you usually find lengths in the coordinate plane using the distance formula. Since $BK$ is an angle bisector of $\triangle ABC$, the angle bisector theorem gives $\frac{AK}{KC} = \frac{AB}{AC}$. Find $AB$ and $AC$ using the distance formula. Next, use the equation in Stewart's theorem to solve for $BK$.

 

[divisor]

Euge, thank you. I want to check my solution by Stewart's theorem.