Question

A ray of light is sent along the line `x-2y-3=0`. Upon reaching the line`3x-2y-5=0`, the ray is reflected from it. If the equation of the line containing reflected ray is `ax-2y = b`, then find the value of`(a+b)`?

Answer 1

`a+b = 60`

Explanation:

We are given 3 equations:

`x-2y-3=0" [1]"`
`3x-2y-5=0" [2]"`
`ax-2y = b" [3]"`

Find the slope of the line corresponding to equation [1]

`-2y = -x+3`

`y = 1/2x-3/2`

`m_1 = 1/2`

The angle that it forms with the x axis is `theta_1 = tan^-1(1/2)`

Find the slope of the line corresponding to equation [2]:

`-2y=-3x+5`

`y=3/2x-5/2`

`m_2=3/2`

The angle that it forms with the x axis is `theta_2 = tan^-1(3/2)`

The angle from line [1] to line [2] is the angle of incidence:

`theta_i = tan^-1(3/2)-tan^-1(1/2)" [4]"`

Find the slope of the line corresponding to equation [3]:

`-2y = -ax+ b`

`y = a/2x-b/2`

`m_3 = a/2`

The angle that it forms with the x axis is `theta_3 = tan^-1(a/2)`

The angle from line [2] to line [3] is the angle of reflection:

`theta_r = tan^-1(a/2) - tan^-1(3/2)" [5]"`

Because the angle of incidence equals the angle of reflection we can set the right side of equation [4] equal to the right side of equation 5:

`tan^-1(a/2) - tan^-1(3/2) = tan^-1(3/2)-tan^-1(1/2)`

`tan^-1(a/2) = 2tan^-1(3/2)-tan^-1(1/2)`

`a = 2tan(2tan^-1(3/2)-tan^-1(1/2))`

`a = 29`

Substitute into equation [3]:

`29x-2y = b" [3.1]"`

Find the point of intersection of lines [1] and [2]:

`x-2y-3=0" [1]"`
`3x-2y-5=0" [2]"`

Subtract [1] from [2]:

`2x -2 = 0`

`x = 1`

Substitute 1 for x into equation [1]:

`1-2y-3=0`

#-2y -2 = 0

`y = -1`

The point is `(1, -1)`

Equation [3.1] must contain the same point:

`29(1) - 2(-1) = b`

`b = 31`

`a+b = 60`