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

Answer

c
$(\sqrt{50})^{2}=(\sqrt{45})^{2}+(\sqrt{5})^{2}$

$\angle \mathrm{B}=90^{\circ}$

$\text { Circum-center }=\left(\frac{1}{2}, \frac{11}{2}\right)$

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 through $\left(0, \frac{\alpha}{2}\right)$

$\frac{\alpha}{2}=\frac{9}{2} \Rightarrow \alpha=9$

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