1. Given points

$A(-2,3),\quad B(1,9),\quad C(3,8)$

We need the equation of a line $L$ such that:

• it passes through the circumcentre of $\triangle ABC$,
• it bisects line segment $BC$ (so it passes through the midpoint of $BC$),
• it cuts the $y$-axis at $\left(0,\frac{\alpha}{2}\right)$.

So first we find:

1. the circumcentre of $\triangle ABC$,
2. the midpoint of $BC$,
3. the equation of the line through these two points.

2. Midpoint of $BC$

$B(1,9),\quad C(3,8)$

Midpoint of $BC$ is

$M=\left(\frac{1+3}{2},\frac{9+8}{2}\right)=\left(2,\frac{17}{2}\right).$

3. Circumcentre of triangle $ABC$

The circumcentre is the intersection point of perpendicular bisectors.

Perpendicular bisector of $AB$

Points: $A(-2,3),\quad B(1,9)$

Midpoint of $AB$: $\left(\frac{-2+1}{2},\frac{3+9}{2}\right)=\left(-\frac12,6\right).$

Slope of $AB$: $m_{AB}=\frac{9-3}{1-(-2)}=\frac{6}{3}=2.$

So perpendicular slope is $m=-\frac12.$

Equation of perpendicular bisector of $AB$: $y-6=-\frac12\left(x+\frac12\right).$

Multiply by 4: $4y-24=-2x-1$ $2x+4y-23=0.$

Perpendicular bisector of $AC$

Points: $A(-2,3),\quad C(3,8)$

Midpoint of $AC$: $\left(\frac{-2+3}{2},\frac{3+8}{2}\right)=\left(\frac12,\frac{11}{2}\right).$

Slope of $AC$: $m_{AC}=\frac{8-3}{3-(-2)}=\frac55=1.$

So perpendicular slope is $-1.$

Equation of perpendicular bisector of $AC$: $y-\frac{11}{2}=-(x-\frac12).$

$y-\frac{11}{2}=-x+\frac12$ $y=-x+6$ $x+y-6=0.$

Intersection of the two bisectors

Solve $2x+4y-23=0$ $x+y-6=0$

From the second equation, $x=6-y.$

Substitute into the first: $2(6-y)+4y-23=0$ $12-2y+4y-23=0$ $2y-11=0$ $y=\frac{11}{2}.$

Then $x=6-\frac{11}{2}=\frac12.$

Hence circumcentre is $O\left(\frac12,\frac{11}{2}\right).$

4. Equation of line $L$ through circumcentre and midpoint of $BC$

The line $L$ passes through $O\left(\frac12,\frac{11}{2}\right),\quad M\left(2,\frac{17}{2}\right).$

Slope of $OM$: $m=\frac{\frac{17}{2}-\frac{11}{2}}{2-\frac12}=\frac{3}{\frac32}=2.$

So equation of the line is $y-\frac{11}{2}=2\left(x-\frac12\right).$

$y-\frac{11}{2}=2x-1$ $y=2x+\frac92.$

5. Find the $y$-intercept

On the $y$-axis, $x=0$. Then $y=2(0)+\frac92=\frac92.$

Given that the line intersects the $y$-axis at $\left(0,\frac{\alpha}{2}\right),$ we compare:

$\frac{\alpha}{2}=\frac92.$

Therefore, $\alpha=9.$

6. Final answer

$\boxed{9}$

This matches the stored correct answer.