Q.

Normal form of  x+3y=2

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a

xcosπ2+ysinπ6=2

b

xcosπ3+ysinπ3=2

c

xcosπ3+ysinπ3=1

d

xcosπ6+ysinπ6=1

answer is C.

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Detailed Solution

given line is      x+3y=2          divide with a2+b2=1+3=2                 12(x)+32y=22=1                xcosπ3+ysinπ3=1                ⇒xcosα+ysinα=p                here α=π3 ; p = 1
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Normal form of  x+3y=2