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

We have $\mathrm{A}(-2,3),\mathrm{B}(1,9)\text{ and }\mathrm{C}(3,8)\cdot \mathrm{AB}=\sqrt{45},\mathrm{BC}=\sqrt{5}\text{ and }\mathrm{AC}=\sqrt{50}$
Clearly, ${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2\text{ So, }\mathrm{\angle }\mathrm{B}={90}^{\circ }$  So, 
circumcenter is mid-point of $\mathrm{AC},\mathrm{P}\left(\frac{1}{2},\frac{11}{2}\right)$
Now, midpoint of BC is $\mathrm{Q}\left(2,\frac{17}{2}\right)$ So, 
equation of line passing through points P and Q is $\mathrm{y}=2\mathrm{x}+\frac{9}{2}$
Line passes though $\left(0,\frac{\mathrm{\alpha }}{2}\right)\text{ if }\mathrm{\alpha }=9$