Dear student, The equation of the reflected ray is (y+1)=m(x+2) or, mx−y+2m−1=0 (1)Since it touches the circle x2+y2=1 ∴ length of ⊥ from (0,0) on (1)= radius =1 ⇒∣m(0)−0+2m−1∣1+m2=1⇒2m−11+m2=±1⇒(2m−1)2=(1+m)2⇒3m2−4m=0⇒m=0,4 3 ∴ Equation of the reflected ray is (y+1)= 43(x+2) or 4x−3y+5=0 Let α be the angle between the reflected the reflected ray and the line y=−1Then, tanα=43-0 1+43×0=±43∴ Slope of the incident ray =−43 (seeing the option this had been decided)Hence, equation of the incident ray is (y+1)=−43(x+2)i e , 3(y+1)=−4(x+2)⇒4x+3y+11=0 Regards

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Dear student,

T h e e q u a t i o n o f t h e r e f l e c t e d r a y i s (y + 1) = m (x + 2) o r, m x - y + 2 m - 1 = 0 . . . (1) S i n c e i t t o u c h e s t h e c i r c l e x^{2} + y^{2} = 1 . ∴ l e n g t h o f ⊥ f r o m (0, 0) o n (1) = r a d i u s = 1 \Rightarrow \frac{∣ m (0) - 0 + 2 m - 1 ∣}{\sqrt{1 + m^{2}}} = 1 \Rightarrow \frac{2 m - 1}{\sqrt{1 + m^{2}}} = \pm 1 \Rightarrow {(2 m - 1)}^{2} = {(1 + m)}^{2} \Rightarrow 3 m^{2} - 4 m = 0 \Rightarrow m = 0, \frac{4}{3} . ∴ E q u a t i o n o f t h e r e f l e c t e d r a y i s (y + 1) = \frac{4}{3} (x + 2) o r 4 x - 3 y + 5 = 0 . L e t α b e t h e a n g l e b e t w e e n t h e r e f l e c t e d t h e r e f l e c t e d r a y a n d t h e l i n e y = - 1 T h e n, \tan α = \frac{|\frac{4}{3} - 0| ​}{1 + \frac{4}{3} \times 0} = \pm \frac{4}{3} ∴ S l o p e o f t h e i n c i d e n t r a y = - \frac{4}{3} (s e e i n g t h e o p t i o n t h i s h a d b e e n d e c i d e d) H e n c e, e q u a t i o n o f t h e i n c i d e n t r a y i s (y + 1) = - \frac{4}{3} (x + 2) i . e ., 3 (y + 1) = - 4 (x + 2) \Rightarrow 4 x + 3 y + 11 = 0 . R e g a r d s

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