1. Interpret the geometry

The reflecting line is $x=1,$ which is a vertical line.

The ray comes from point $P=(2,2\sqrt{3})$ and is incident on the line at point $A$ with angle $30^\circ$.

For a vertical mirror, the normal is horizontal. So an angle of incidence $30^\circ$ means the incident ray makes angle $30^\circ$ with the horizontal.

Hence the slope of the incident ray is $m=-\tan 30^\circ=-\frac{1}{\sqrt{3}},$ since the ray goes downward from $(2,2\sqrt3)$ to the mirror.

2. Find the point of incidence $A$ on $x=1$

Equation of incident ray through $(2,2\sqrt3)$: $y-2\sqrt3=-\frac{1}{\sqrt3}(x-2).$

At the mirror, $x=1$. So $y-2\sqrt3=-\frac{1}{\sqrt3}(1-2)=\frac{1}{\sqrt3}.$ Thus $y=2\sqrt3+\frac{1}{\sqrt3}=\frac{6+1}{\sqrt3}=\frac{7}{\sqrt3}=\frac{7\sqrt3}{3}.$

So, $A=\left(1,\frac{7\sqrt3}{3}\right).$

3. Find the reflected ray and point $B$ on the x-axis

Reflection from a vertical line reverses the sign of slope. So reflected ray has slope $m=+\frac{1}{\sqrt3}.$

Equation of reflected ray through $A$: $y-\frac{7\sqrt3}{3}=\frac{1}{\sqrt3}(x-1).$

Point $B$ lies on the $x$-axis, so $y=0$: $0-\frac{7\sqrt3}{3}=\frac{1}{\sqrt3}(x-1).$ Multiply by $\sqrt3$: $-7=x-1,$ so $x=-6.$

Hence $B=(-6,0).$

4. Equation of line $AB$

Using point $A=(1,7\sqrt3/3)$ and slope $1/\sqrt3$: $y-\frac{7\sqrt3}{3}=\frac{1}{\sqrt3}(x-1).$

Rewrite: $y=\frac{x-1}{\sqrt3}+\frac{7\sqrt3}{3}.$ Since $\frac{7\sqrt3}{3}=\frac{7}{\sqrt3},$ we get $y=\frac{x-1+7}{\sqrt3}=\frac{x+6}{\sqrt3}.$

So line $AB$ is $y=\frac{x+6}{\sqrt3}.$

5. Check which option lies on this line

We test each option in $y=\frac{x+6}{\sqrt3}.$

• A: $(3,-\sqrt3)$ $\frac{3+6}{\sqrt3}=\frac{9}{\sqrt3}=3\sqrt3 \ne -\sqrt3$ Not on the line.

• B: $(4,-\sqrt3)$ $\frac{4+6}{\sqrt3}=\frac{10}{\sqrt3} \ne -\sqrt3$ Not on the line.

• C: $\left(4,-\frac{\sqrt3}{2}\right)$ $\frac{10}{\sqrt3} \ne -\frac{\sqrt3}{2}$ Not on the line.

• D: $\left(3,-\frac{1}{\sqrt3}\right)$ $\frac{9}{\sqrt3}=3\sqrt3 \ne -\frac{1}{\sqrt3}$ Not on the line.

So none of the options matches under this interpretation.

6. Check the standard exam convention

In many such problems, “incident at angle $30^\circ$ on the line” is interpreted as angle with the line itself, not with the normal.

Since the mirror is vertical, the incident ray then makes angle $30^\circ$ with the vertical, i.e. angle $60^\circ$ with the horizontal. Thus incident slope is $m=-\tan 60^\circ=-\sqrt3.$

Incident ray through $(2,2\sqrt3)$: $y-2\sqrt3=-\sqrt3(x-2).$ At $x=1$, $y-2\sqrt3=-\sqrt3(1-2)=\sqrt3,$ so $y=3\sqrt3.$ Thus $A=(1,3\sqrt3).$

After reflection from vertical line, slope becomes $+\sqrt3$. Equation of reflected ray: $y-3\sqrt3=\sqrt3(x-1).$ To meet x-axis, set $y=0$: $-3\sqrt3=\sqrt3(x-1) \implies x-1=-3 \implies x=-2.$ So $B=(-2,0).$

Now line $AB$ has equation $y-3\sqrt3=\sqrt3(x-1),$ that is $y=\sqrt3 x+2\sqrt3.$

Test options:

• A: $(3,-\sqrt3)$ $\sqrt3(3)+2\sqrt3=5\sqrt3\ne -\sqrt3$
• B: $(4,-\sqrt3)$ $6\sqrt3\ne -\sqrt3$
• C: $\left(4,-\frac{\sqrt3}{2}\right)$ $6\sqrt3\ne -\frac{\sqrt3}{2}$
• D: $\left(3,-\frac{1}{\sqrt3}\right)$ $5\sqrt3\ne -\frac{1}{\sqrt3}$

Again, none matches.

7. Conclusion

Using the physically correct reflection interpretation, the line $AB$ is found, but none of the given options lies on it. Therefore the stored answer A does not agree with the derived result.

Most likely, there is an error in the question/options or in the stored key.