Question

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 $A$.

Answer 1

There is a point A on x-axis on which ray reflects

REF. IMAGE 1

A ray passing through $P(1,2)$ reflect on point A

On reflection, the ray passes through point $Q(5,3)$

We need to find coordinate of A

Since point A is on the a-axis, its y-coordinate is 0

Let coordinates of point A be $(k,0)$

So, we need to find value of k

REF. IMAGE 2

We need to find angle with positive x-axis of both lines

Line $QA$ makes angle $\angle QAX$ with positive x-axis

Line PA makes $\angle PAX$ with positive x-axis

Let $\angle QAX=\theta$

Now,

$MA$ is normal

$\angle MAX={90}^{\circ }$

$\theta +\angle MAQ={90}^{\circ }$

$\angle MAQ={90}^{\circ }-\theta$ (Angle of incidence = Angle of reflection)

Also, $\angle MAP=\angle MAQ=90-\theta$

Now,

REF. IMAGE 3

$\angle PAX=\angle MAP+\angle MAQ+\angle QAX$

$=(90-\theta )+(90-\theta )+\theta =180-\theta$

Now, we find slope of line $PA$ & $QA$

We know that slope of line that passes through points $(x_1,y_1)$ & $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$

Line $PA$

Slope of line $PA$ passing 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-axis

Slope of $PA=\tan (180-\theta )$

$=-\tan \theta$

So, $-\tan \theta =\frac{-2}{k-1}$

$\tan \theta =\frac{2}{k-1}$ ___(1)

Line $QA$

Slope of line $QA$ passing through points $(5,3)$ &$(k,0)$ is

Slope pf $QA=\frac{0-3}{k-5}$

$=\frac{-3}{k-5}$

But $QA$ makes angle $\theta$ with positive x-axis

Slope of $QA=\tan \theta$

So, $\tan \theta =\frac{-3}{k-5}$ ___(2)

From (1) & (2)

$\frac{2}{k-1}=\frac{-3}{k-5}$

$2(k-5)=-3(k-1)$

$2k-10=-3k+3$

$2k+3k=3+10$

$5k=13$

$k=\frac{13}{5}$

Hence point $A(\frac{13}{5},0)$