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 $ABC$, bisects line $BC$, and intersects $y$-axis at point $\left(0,\frac{\alpha }{2}\right)$, then the value of real number $\alpha$ is ________.

Answer 1

Answer: 9

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