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Step-by-Step Solution
Step 1: Identify the mirror line and the given data
• The mirror (reflecting surface) is the vertical line $x = 1$.
• A light ray originates at the point $(2,\,2\sqrt{3})$ and strikes $x=1$ at some point $A=(1,\,a)$.
• The ray is said to be incident “at an angle of $30^\circ$ on the line $x=1$.” In typical optics/geometry usage, “angle on the mirror” means the angle the ray makes with the mirror itself (not with the normal).
• Since $x=1$ is vertical, making an angle of $90^\circ$ with the horizontal, having a $30^\circ$ incidence with the vertical implies its angle with the horizontal (the usual axis for measuring slope) is $60^\circ$. Consequently, the slope of the incident ray must be $\pm\,\tan(60^\circ)=\pm\,\sqrt{3}.$
• Because the ray travels from $(2,\,2\sqrt{3})$ leftward to $x=1,$ we determine the correct sign of the slope by consistency with the final geometry (and reflection).
Step 2: Find the point of incidence A
Let $A=(1,\,a).$ The slope of the line from $(2,\,2\sqrt{3})$ to $(1,\,a)$ is
\[
m \;=\;\frac{\,a - 2\sqrt{3}\,}{\,1 - 2\,}\;=\;\frac{\,a - 2\sqrt{3}\,}{-1}\;=\;2\sqrt{3}-a.
\]
We want this slope $m$ to be $+\sqrt{3}$ or $-\sqrt{3}$ depending on how the angle of $60^\circ$ with the horizontal is oriented. Checking consistency leads to choosing
\[
2\sqrt{3}-a \;=\; +\sqrt{3}.
\]
Hence
\[
a \;=\; 2\sqrt{3} - \sqrt{3}\;=\;\sqrt{3}.
\]
Therefore, the point of incidence is
\[
A \;=\;(1,\;\sqrt{3}).
\]
Step 3: Determine the reflection path
Under reflection in a vertical line $x=1,$ the horizontal component of a ray’s direction reverses sign, which effectively flips the slope from $m$ to $-\,m.$ Since the incident slope at $A$ was $+\sqrt{3},$ the reflected slope becomes $-\sqrt{3}.$
Step 4: Find where the reflected ray meets the x-axis
Let the reflected ray meet the x-axis at $B=(b,0).$ The slope of the line from $A=(1,\,\sqrt{3})$ to the unknown $B=(b,\,0)$ is $-\sqrt{3}.$ The equation of the reflected line through $A$ is:
\[
y - \sqrt{3} \;=\; -\sqrt{3}\,\bigl(x-1\bigr).
\]
Setting $y=0$ (to find the intersection with the x-axis) gives:
\[
0 - \sqrt{3} \;=\; -\sqrt{3}\,\bigl(b-1\bigr)
\;\;\Longrightarrow\;\;
-\sqrt{3}
\;=\;
-\sqrt{3}\,(\,b-1\,)
\;\;\Longrightarrow\;\;
b-1
\;=\;1
\;\;\Longrightarrow\;\;
b=2.
\]
Hence $B=(2,\,0).$
Step 5: Check the line AB and verify the given point
The line $AB$ goes from $A=(1,\,\sqrt{3})$ to $B=(2,\,0).$ Its slope is
\[
\frac{\,0 - \sqrt{3}\,}{\,2-1\,} \;=\; -\sqrt{3}.
\]
The equation of line $AB$ can be written as:
\[
y - \sqrt{3}\;=\;-\sqrt{3}\,(\,x -1\,).
\]
Substitute $x=3$:
\[
y - \sqrt{3}
\;=\;
-\sqrt{3}\,\bigl(3-1\bigr)
\;=\;
-2\sqrt{3}
\;\;\Longrightarrow\;\;
y
\;=\;
\sqrt{3} \;-\;2\sqrt{3}
\;=\;
-\sqrt{3}.
\]
So the point $\bigl(3,\,-\sqrt{3}\bigr)$ lies on $AB,$ exactly matching the answer option given.
Conclusion
Therefore, the reflected ray meets the x-axis at $B=(2,\,0),$ and the line $AB$ indeed passes through the point $\boxed{\,(3,\,-\sqrt{3})}.$
