$\begin {array} {1 1} (A)\;\bigg( \large\frac{5}{13}, 0 \bigg) & \quad (B)\;\bigg( -\large\frac{5}{13}, 0 \bigg) \\ (C)\;\bigg( \large\frac{13}{5}, 0 \bigg) & \quad (D)\;\bigg( -\large\frac{13}{5}, 0 \bigg) \end {array}$

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- Slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $ m=\large\frac{y_2-y_1}{x_2-x_1}$
- Also slope $m=\tan \theta$ where $\theta$ is the angle of inclination.

Let the coordinates of the point A be ( a,0) .

Let AL be the perpendicular to $x$ - axis.

By Law of reflection we know,

angle of incidence = angle of reflection.

Hence $ < BAL = < CAL = \phi$

Let $ < CAX = \theta$

$ < OAB = 180^{\circ} - ( \theta + 2\phi)$

$ \qquad \qquad = 180^{\circ} - [ \theta + 2( 90^{\circ}- \theta )$

$( \because \phi = 90 - \theta)$

$ \therefore < OAB = 180^{\circ} - \theta - 180^{\circ} + 2\theta$

$ = \theta \Rightarrow < BAX = 180^{\circ} - \theta$

Slope of the line $AC, m = \large\frac{3-0}{5-a}$

also $m = \tan \theta$

$ \Rightarrow \tan \theta = \large\frac{3}{5-a}$-------(1)

Slope of the line $AB = \large\frac{2-0}{1-a}$

$ \therefore \tan ( 180^{\circ} - \theta ) = \large\frac{2}{1-a}$

But $ \tan ( 180^{\circ}-\theta)=-\tan \theta$

$ \therefore -\tan \theta = \large\frac{2}{1-a}$

$ \therefore \tan \theta = \large\frac{2}{a-1}$------(2)

equally (1) and (2) we get,

$ \large\frac{3}{5-a}$$ = \large\frac{2}{a-1}$

$ \Rightarrow 3(a-1)=2(5-a)$

$ 3a-3=10-2a$

$ \Rightarrow 5a=13$

$ \therefore a = \large\frac{13}{5}$

Hence the coordinate of A are $ \bigg( \large\frac{13}{5}$$, 0 \bigg)$

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