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. The equation of the reflected ray is
A
13x−y=14
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B
17x−2y=19
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C
29x−2y=31
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D
x+y=11
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Solution
The correct option is C29x−2y=31 Incident ray : x−2y−3=0
Line mirror : 3x−2y−5=0
Solving these two equation , P(1,−1)
Let A′(α,β) be the image of A(3,0)
w.r.t 3x−2y−5=0 ⇒α−33=β−0−2=−2(9−0−5)5⇒A′=(1513,1613) ∴ Equation of PA′(reflected ray) is 29x−2y−31=0