Browse Questions

# A ray of light passing through the point (1, 2) reflects on the $x$ - axis at point $A$ and the reflected ray passes through the point (5, 3). Find the coordinates of $A$

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

Toolbox:
• 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)$