
Let the incident ray strike x-axis at the point A and the coordinate is (x,0).
Now from the figure
tanθ = (3 - 0)/(5-x)
=> tanθ = = 3/(5 - x) .............1
Again slope of the incident ray
tan(π - θ) = (0-2)/(x - 1)
=> - tanθ = -2/(x - 1)
=> tanθ = 2/(x - 1) .........2
from equation 1 and 2,
3/(5 - x) = 2/(x - 1)
=> 3(x-1) = 2(5-x)
=> 3x - 3 = 10 - 2x
=> 3x + 2x = 10 + 3
=> 5x = 13
=> x = 13/5
So the required point is (13/5, 0)