Question

Consider a triangle having vertices $A(-2,3), B(1,9)$ and $C(3,8)$. If a line $L$ passing through the circum-centre of triangle $A B C$, bisects line $B C$, and intersects $y$-axis at point $\left(0, \frac{\alpha}{2}\right)$, then the value of real number $\alpha$ is ________.

Answer 1

$(\sqrt{50})^{2}=(\sqrt{45})^{2}+(\sqrt{5})^{2}$
$\angle B=90^{\circ}$
Circum-center $=\left(\frac{1}{2}, \frac{11}{2}\right) $
Mid point of $B C=\left(2, \frac{17}{2}\right)$
Line :$\left(y-\frac{11}{2}\right)=2\left(x-\frac{1}{2}\right) $
$\Rightarrow y=2 x+\frac{9}{2}$
Passing though $\left(0, \frac{\alpha}{2}\right)$
$\frac{\alpha}{2}=\frac{9}{2} \Rightarrow \alpha=9$