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the previous question dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

the previous question 
 
 
 
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Grade:12th pass

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
Dear student

What do you mean my previous question

Regards
Vikas TU
14149 Points
4 years ago
P(π/4) = aCosπ/4 + bSinπ/4 
= (sqrt2 , 1 )-- Cordinate of P 
S coordinate is (sqrt 2  0 )
S' coordinate is (- sqrt 2 , 0) 
Equation of PS = x-sqrt2 = 0 
Equation of PS' = 2*sqrt2 y = x+ sqrt2 
Angle bisector 
(x - sqrt 2 )/ 1 = (+ or -) (x - 2sqrt2 +2 )/sqrt(1+8)
Solve this you will get 
x + sqrt2 y = 2*sqrt2 
A is correct ans. 

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