Question

A ray of light passing through the point $A(2,3)$ reflected at a point B on line $x+y=0$ and then passes through $(5,3)$. Then the coordinates of B are

• A. $(\frac{1}{3},-\frac{1}{3})$
• B. $(\frac{2}{5},-\frac{2}{5})$
• C. $(\frac{1}{13},-\frac{1}{13})$
• D. $Noneofthese$

Answer 1

Answer: (A)

$(\frac{1}{3},-\frac{1}{3})$

According to the law of reflection $\angle i=\angle r$

Given line $x=-y$, $slope\left(m_1\right)=-1$

let the point of reflection on line $x=-y$ at $(x_1,y_1)$ , But $y_1=-x_1$

Now equation of line($L_1$) passing through $(2,3)$ and $(x_1,-x_1)$ is

$\Rightarrow m=\frac{y_2-y_1}{x_2-x_1}=\frac{-x_1-3}{x_1-2}$

$\Rightarrow y-y_1=m(x-x_1)$

$\Rightarrow y-3=\frac{-x_1-3}{x_1-2}(x-2)$ and slope$\left(m_2\right)=\frac{-x_1-3}{x_1-2}$

Angle b/w line any 2 line is $\varphi =ta{n}^{-1}\frac{(m_1-m_2)}{1+m_1{m}_2}$

Hence Angle b/w line ($x=-y$) and $L_1$ is

${\varphi }_1=ta{n}^{-1}[\frac{-1+\frac{x_1+3}{x_1-2}}{1+\frac{x_1+3}{x_1-2}}]$

similarly for line ($L_2)$ which passing through $(5,3)$ and $(x_1,-x_1)$

Angle b/w line ($x=-y$) and $L_2$ is

${\varphi }_1=ta{n}^{-1}[\frac{-1+\frac{x_1+3}{x_1-5}}{1+\frac{x_1+3}{x_1-5}}]$

Now using law of reflection $\angle i=\angle r$

$tan{\varphi }_1=tan{\varphi }_2$

On solving ,we will get $x_1=\frac{1}{3}$

hence point of reflection is $(\frac{1}{3},\frac{-1}{3})$