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Q. Consider a triangle $ABC$ whose vertices are $B (-1,3)$ and $C (3,6)$. Also the equation of internal angle bisector of $\angle BAC$ is $3 x + y =5$.
The radius of circle circumscribing triangle $ABC$ is equal to

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Solution:

Let $AD$ be internal angle bisector of $\angle BAC$. Drop perpendicular from $B$ on $AD$ and 10439-40-41/st. line produce it to meet AC at B'.
Clearly, $\triangle AMB \cong \triangle AMB '$, so $B ^{\prime}$ is image of $B$ in line $AD$.
As slope of $AD$ is -3 , so slope of line $BB ^{\prime}=\frac{1}{3}=\tan \theta$ (say)
image
$\therefore$ Equation of $BB ^{\prime}$ in parametric form is $\frac{ x +1}{\frac{3}{\sqrt{10}}}=\frac{ y -3}{\frac{1}{\sqrt{10}}}= r$
$\therefore$ Co-ordinates of ' ' $^{\text {are }}\left(-1+\frac{3}{\sqrt{10}} \times 2 \times \frac{5}{\sqrt{10}}, 3+\frac{1}{\sqrt{10}} \times 2 \times \frac{5}{\sqrt{10}}\right)=(2,4)$
(As distance of line $AD$ from $B$ is $\frac{5}{\sqrt{10}}$ )
Clearly, equation of $AC$ is $y =2 x$ ....(1)
Also, equation of $AD$ is $3 x + y =5$ (given) .....(2)
$\therefore$ On solving (1) and (2), we get $A (1,2)$
As $AB =\sqrt{5}, AC =2 \sqrt{5}$ and $BC =5$
So, clearly $\triangle ABC$ is right angled at $A$. So, by using sine law, we get
$\frac{a}{\sin A }=2 R \Rightarrow \frac{5}{\sin 90^{\circ}}=2 R \Rightarrow R =\frac{5}{2}$