Consider the coordinates of point A as (a, 0)

Construct a line (AL) which is perpendicular to the x-axis

Here the angle of incidence is equal to angle of reflection

∠BAL = ∠CAL = Φ

∠CAX = θ

It can be written as

∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]

On further calculation

= 180° – θ – 180° + 2θ

= θ

So we get

∠BAX = 180° – θ

Slope of line = \( \dfrac{3-0}{5-a}\)

tanθ = \( \dfrac{3}{5-a}\) ..............(1)

Slope of line AB = \( \dfrac{2-0}{1-a}\)

tanθ =(180° – θ) = \( \dfrac{2}{1-a}\)

- tanθ = \( \dfrac{2}{1-a}\)

- tanθ = \( \dfrac{2}{a-1}\)

From equations (1) and (2) ,

\( \dfrac{3}{5-a}\)= \( \dfrac{2}{a-1}\)

By cross multiplication,

3a -3 =10-2a

= \( \dfrac{13}{5}\)

Therefore, the coordinates of point A are \( ( \dfrac{13}{5},0)\)

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