Find the values of θ and p, if the equation xcosθ+ysinθ=p is the normal form of the line √3x+y+2=0.
The equation of the given line is √3x+y+2=0
this equation can be reduced as −√3x−y=2
on dividing both sides by √(−√3)2+(−1)2=2,
we obtain −√32x−12y=22
⇒{−√32}x+{−12}y=1....(1)
On comparing equation (1) to xcosθ+ysinθ=p,
we obtain cosθ=−√32,sinθ=−12, and p=1
Since the value of sinθ and cosθ are both negative, θ is in the third quadrant
∴θ=π+π6=7π6
Thus, the respective values of θ and p are 7π6 and 1