Question

A ray of light passing through the point $A(1,2)$ is reflected at a point $B$ on the $x-$axis line mirror and then passes through $(5,3)$. Then the equation of $AB$ is

• A. $5x+4y=13$
• B. $5x-4y=-3$
• C. $4x+5y=4$
• D. $4x-5y=-6$

Answer 1

The correct option is A $5x+4y=13$

Given that,

There is a point $B$ on the $x-$axis on which rays reflects

A ray passing through $A(1,2)$ reflects on point $B$.

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

We need to find equation of line $AB$.

Since

the point $B$ is on the $x-$axis

its $y=0$

Let the coordinates of point $B(k,0)$

Let the angle $\angle CBX=\theta$

$PB$ be the normal

$\angle PBX={90}^o$

$\angle XBC+\angle CBP={90}^o$

$\theta +\angle CBP={90}^o$

$\angle CBP={90}^o$

Also, $\angle ABP=\angle CBP={90}^o=\theta$

Now, $\angle ABX=\angle ABP+\angle PBC+\angle CBX$

$=({90}^o-\theta )+({90}^o-\theta )+\theta$

$={180}^o-20+Q$

$\angle ABX={180}^o-\theta$

Now, we find slope of line $AB$ and $CB$

We know that

Slope of line

$m=\frac{y_2-y_1}{x_2-x_1}$

Line $AB$

Slope of line $AB$ passing through the points $(1,2)$ and $(k,0)$

Slope of $AB=\frac{0-2}{k-1}$

let $m_1=\frac{-2}{k-1}$

now,

$\angle ABx={180}^o-\theta$

$m_1=\tan ({180}^o-\theta )$

$=-\tan \theta$

$m_1=-\tan Q=\frac{-2}{k-1}$

$m_1=\tan Q=\frac{2}{k-1}$

Line $CB$ Slope of line $CB$ passing through the points $(5,3)$ and $(k,0)$

Slope of $CB=\frac{0-3}{k-5}$

$m_2=\frac{-3}{k-5}$

But $m_2=\tan \theta =\frac{2}{k-1}$

According to question

$m_1=m_2$

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

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

$2k+3k=3+10$

$5k=13$

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

Then point $B(k,0)=B(\frac{13}{5},0)=(x_2,y_2)$

Equation of line $AB$

$y-y_1=\frac{y_2-y_1}{x_2-x_1}$

$y-2=\frac{0-2}{\frac{13}{5}-1}(x-1)$

$y-2=\frac{-2}{\frac{13-5}{5}}(x-1)$

$y-2=\frac{-10}{8}(x-1)$

$y-2=\frac{-5}{4}(x-1)$

$4y-8=-5x+5$

$5x+4y-13=0$

Hence, the equation of line $AB$.

This is the answer.