Answer:
[tex]\dfrac{72}{19}[/tex]
Step-by-step explanation:
Consider triangle ABC. Segment AX is angle A bisector. Its length can be calculated using formula
[tex]AX^2=\dfrac{AB\cdot AC}{(AB+AC)^2}\cdot ((AB+AC)^2-BC^2)[/tex]
Hence,
[tex]AX^2=\dfrac{9\cdot 10}{(9+10)^2}\cdot ((9+10)^2-17^2)=\dfrac{90}{361}\cdot (361-289)=\dfrac{90}{361}\cdot 72=\dfrac{6480}{361}[/tex]
By the angle bisector theorem,
[tex]\dfrac{AB}{AC}=\dfrac{BX}{XC}[/tex]
So,
[tex]\dfrac{9}{10}=\dfrac{BX}{17-BX}\Rightarrow 153-9BX=10BX\\ \\19BX=153\\ \\BX=\dfrac{153}{19}[/tex]
and
[tex]XC=17-\dfrac{153}{19}=\dfrac{170}{19}[/tex]
By the Pythagorean theorem for the right triangles AXY and CXY:
[tex]AX^2=AY^2+XY^2\\ \\XC^2=XY^2+CY^2[/tex]
Thus,
[tex]\dfrac{6480}{361}=XY^2+AY^2\\ \\\left(\dfrac{170}{19}\right)^2=XY^2+(10-AY)^2[/tex]
Subtract from the second equation the first one:
[tex]\dfrac{28900}{361}-\dfrac{6480}{361}=(10-AY)^2-AY^2\\ \\\dfrac{22420}{361}=100-20AY+AY^2-AY^2\\ \\\dfrac{1180}{19}=100-20AY\\ \\20AY=100-\dfrac{1180}{19}=\dfrac{1900-1180}{19}=\dfrac{720}{19}\\ \\AY=\dfrac{36}{19}[/tex]
Hence,
[tex]XY^2=\dfrac{6480}{361}-\left(\dfrac{36}{19}\right)^2=\dfrac{6480-1296}{361}=\dfrac{5184}{361}\\ \\XY=\dfrac{72}{19}[/tex]
