Question

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

Answer 1

Simplification of given data
There is a point $A$ on $x-\text{axis}$ on which ray reflects. A ray passing through $P(1,2)$ reflects on point $A$
On reflection, the ray passes through point $Q(5,3)$
Let coordinates of point $A$ be $(k,0)$
Let $\angle QAX=\theta$
Now,
$MA$ is normal
$\angle MAX={90}^{\circ }$
$\theta +\angle MAQ={90}^{\circ }$
$\angle MAQ={90}^{\circ }-\theta$
Also, $\angle MAP=\angle MAQ=90-\theta$
(Angle of incidence = Angle of reflection)
Now,
$\angle PAX=\angle MAP+\angle MAQ+\angle QAX$
$=(90-\theta )+(90-\theta )+\theta =180-\theta$

Slope of line $PA\&QA$
Line $PA$,
Slope of line that passes through points $(1,2)\&(k,0)$ is
Slope of $PA=\frac{0-2}{k-1}=\frac{-2}{k-1}$
But $PA$ makes angle $180-\theta$ With positive $x-\text{axis}$
Slope of $PA=\tan (180-\theta )=-\tan \theta$
So, $-\tan \theta =\frac{-2}{k-1}$
$\tan \theta =\frac{2}{k-1}$ $....\left(i\right)$

Line $QA$,
Slope of line $QA$ passing through points $(5,3)\&(k,0)$ is
Slope of $QA=\frac{0-3}{k-5}=\frac{-3}{k-5}$
But $QA$ makes angle $\theta$ with positive $x-\text{axis}$
Slope of $QA=\tan \theta$
So, $\tan \theta =\frac{-3}{k-5}$ $...\left(ii\right)$
From $\left(i\right)$ & $\left(ii\right)$
$\frac{2}{k-1}=\frac{-3}{k-5}$
$2(k-5)=-3(k-1)$
$2k-10=-3k+3$
$2k+3k=3+10$
$5k=13$
$k=\frac{13}{5}$
$\therefore A=(\frac{13}{5},0)$