Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

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

Let assume that coordinates of B be (a, b).

As, B lies on x + y = 0

It implies, a + b = 0

It means, b = - a

So, Coordinates of B be ( a, - a ).

Now, Slope of x + y = 0 is - 1.

As, BD is perpendicular to x + y = 0.

So, slope of BD = 1.

Now,

$\rm :\longmapsto\:Slope \:of \: AB = \dfrac{ - a - 3}{a - 2} = \dfrac{a + 3}{2 - a}$

Now,

$\rm :\longmapsto\:Slope \:of \: BC = \dfrac{ - a - 3}{a - 5} = \dfrac{a + 3}{5 - a}$

Since, BD is normal

So,

$\rm :\longmapsto\:i = r$

$\rm :\longmapsto\:tani = tanr$

$\rm :\longmapsto\:\bigg |\dfrac{1 - \dfrac{a + 3}{2 - a}}{1 + 1 \times \dfrac{a + 3}{2 - a}} \bigg| = \bigg |\dfrac{1 - \dfrac{a + 3}{5 - a}}{1 + 1 \times \dfrac{a + 3}{5 - a}} \bigg|$

$\rm :\longmapsto\:\bigg |\dfrac{2 - a - a - 3}{2 - a + a + 3} \bigg| = \bigg |\dfrac{5 - a - a - 3}{5 - a + a + 3} \bigg|$

$\rm :\longmapsto\:\bigg |\dfrac{ - 2a - 1}{5} \bigg| = \bigg |\dfrac{2(1 - a)}{8} \bigg|$

$\rm :\longmapsto\:\bigg |\dfrac{ 2a + 1}{5} \bigg| = \bigg |\dfrac{1 - a}{4} \bigg|$

$\rm \implies\:\dfrac{2a + 1}{5} = \dfrac{1 - a}{4} \: or \: \dfrac{2a + 1}{5} = \dfrac{a - 1}{4}$

$\rm \: 8a + 4 = 5 - 5a \: \: or \: \: 8a + 4 = 5a - 5$

$\rm \: 8a + 5a = 5 -4 \: \: or \: \: 8a - 5a = - 4 - 5$

$\rm \: 13a = 1 \: \: or \: \:3a = - 9$

$\bf\implies \:a = \dfrac{1}{13} \: \: or \: \: a = - 3$

Hence, Coordinates of B are

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:-\begin{cases} &\sf{( - 3, \: 3)} \\ \\ &\sf{\bigg(\dfrac{1}{13}, \: - \dfrac{1}{13} \bigg) } \end{cases}\end{gathered}\end{gathered}$