starting at the origin a beam of light hits a mirror (in the form of a line) at the point A (4,8) and reflected line passes through the point B(8,12 ).compute the slope of the mirror
Slope of the incident ray = (8-0)/(4-0) = 2 ( angle A)
Slope of the reflected ray = (12-8)/(8-4) = 1( angle B)
Now let us find the line that bisects the 2 rays
Tan (A+B) = (tan A + tan B)/(1- Tan A tan B) = (1+2)/(1-2) = - 3
Let C be the angle bisector between A and B
A + B = 2C say m is slope of the bisector
So we get 2m/(1-m^2) = - 3
Or 2m = -3 + 3m^2
Or 3m^2 – 2m – 3 = 0
=> m = (1/6) [2 + √(4 + 36)] ... [m positive taken to the correct slope
=> m = (1/3) [1 + √(10)].
this is the slope of the mirror
Slope of the reflected ray = (12-8)/(8-4) = 1( angle B)
Now let us find the line that bisects the 2 rays
Tan (A+B) = (tan A + tan B)/(1- Tan A tan B) = (1+2)/(1-2) = - 3
Let C be the angle bisector between A and B
A + B = 2C say m is slope of the bisector
So we get 2m/(1-m^2) = - 3
Or 2m = -3 + 3m^2
Or 3m^2 – 2m – 3 = 0
=> m = (1/6) [2 + √(4 + 36)] ... [m positive taken to the correct slope
=> m = (1/3) [1 + √(10)].
this is the slope of the mirror