Question

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 $3x+y=5$.
The radius of circle circumscribing triangle $ABC$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. 2
• D. 1

Answer 1

Answer: (B)

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, $△AMB\cong △AM{B}^{\prime }$, so $B^{\prime }$ is image of $B$ in line $AD$.
As slope of $AD$ is -3 , so slope of line $B{B}^{\prime }=\frac{1}{3}=\tan \theta$ (say)

$\therefore$ Equation of $B{B}^{\prime }$ 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 $y=2x$ ....(1)
Also, equation of $AD$ is $3x+y=5$ (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 $△ABC$ is right angled at $A$. So, by using sine law, we get
$\frac{a}{\sin A}=2R\Rightarrow \frac{5}{\sin 90^{\circ }}=2R\Rightarrow R=\frac{5}{2}$