Equation of Reflected Ray

Let L1 ≡ a1x + b1y + c1 = 0 be the incident ray in the line mirror L2 ≡ a2x + b2y + c2 = 0.

Let L3 be the reflected ray from the line L2. Clearly L2 will be one of the bisectors of the angles between L1 and L3

Since L3 passes through A, so L3 ≡ L1 + λL2 = 0.

Let (h, k) be a point on L2 Then,

$ \displaystyle \frac{| a_1 h + b_1 k + c_1 | }{\sqrt{a_1^2 + b_1^2}} = \frac{|a_1 h + b_1 k + c_1 + \lambda (a_2 h + b_2 k + c_2)|}{\sqrt{( a_1 + \lambda a_2 )^2 + (b_1 + \lambda b_2 )^2}}$

Since (h, k) lies on L2 , a2h + b2k + c2 = 0

=> a12 + a22λ2 + 2a1a2λ + b12 + b22λ2 + 2b1b2λ = a12 + b12

=> λ = 0 or

$ \displaystyle \lambda = \frac{-2(a_1 a_2 + b_1 b_2)}{a_2^2 + b_2^2} $

But λ = 0 gives L3 = L1 . Hence

L3 ≡ $ \displaystyle L_1 + \frac{-2(a_1 a_2 + b_1 b_2)}{a_2^2 + b_2^2}L_2 = 0 $

Note: Some times the reflected ray L3 is also called the mirror image of L1 in L2

Also Read :

→ Co-ordinate Geometry
→ Area of a Triangle
→ Locus : Co-ordinate Geometry
→ Equations of Straight Line in Different Forms
→ Angle between Two Straight Lines
→ Bisectors of the angles b/w two lines
→ Family of Lines , Concurrency of Straight Lines
→ Pair of Straight Lines
→ Solved Problems : Straight Lines

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