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

$\begin{array}{l}\mathrm{AB}=\sqrt{(1+2{)}^2+(9-3{)}^2}=\sqrt{45}\\ \mathrm{BC}=\sqrt{(3-1{)}^2+(8-9{)}^2}=\sqrt{5}\\ \mathrm{AC}=\sqrt{(3+2{)}^2+(8-3{)}^2}=\sqrt{50}\\ \therefore (\sqrt{50}{)}^2=(\sqrt{45}{)}^2+(\sqrt{5}{)}^2\\ (\mathrm{AC}{)}^2=(\mathrm{AB}{)}^2+(\mathrm{BC}{)}^2\\ \therefore \mathrm{\angle }\mathrm{B}={90}^{\circ }\end{array}$

$\Rightarrow \mathrm{ABC}$ is right angled triangle.

Circumcentre = Mid-point of hypotenuse = Mid-point of AC

$=\left(\frac{1}{2},\frac{11}{2}\right)$

Mid-point of line $\mathrm{BC}=\left(2,\frac{17}{2}\right)$

Line passing through circumcentre and bisect line BC will be

$\begin{array}{l}\mathrm{y}-\frac{11}{2}=\frac{\frac{17}{2}-\frac{11}{2}}{2-\frac{1}{2}}\left(\mathrm{x}-\frac{1}{2}\right)\\ \Rightarrow \mathrm{y}-\frac{11}{2}=\frac{3\times 2}{3}\left(\mathrm{x}-\frac{1}{2}\right)\Rightarrow \mathrm{y}-\frac{11}{2}=2\left(\mathrm{x}-\frac{1}{2}\right)\end{array}$

It passes through $\left(0,\frac{\mathrm{\alpha }}{2}\right)$.

$\begin{array}{ll}\therefore & \frac{\mathrm{\alpha }}{2}-\frac{11}{2}=2\left(0-\frac{1}{2}\right)\Rightarrow \mathrm{\alpha }-11=4\left(-\frac{1}{2}\right)\\ \Rightarrow & \mathrm{\alpha }=11-2=9\\ \Rightarrow & \mathrm{\alpha }=9\end{array}$