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Q.

The normal form of line 3x−y+3=0  is

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a

xcosπ6+ysinπ6

b

xcos1500+ysin1500=3

c

x+y=3

d

xcos1500+ysin1500=32

answer is D.

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Detailed Solution

−3x+y=3   make the constant positive and divide with a2+b2=3+1=2         −32x+12y=32     ⇒x cos α+y sin α =p , where cos α=-32, sin α=12⇒α belongs to second quadrant.⇒α=150°     xcos1500+ysin1500=32
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