Question

please answer this question

Answer 1

The normal form of a line is

xcosα + ysinα = p

Given equation is 3x + y -4 =0

⇒ 3x + y = 4

⇒ (3x + y)/(√(3²+1²)) = 4/(√(3²+1²))

⇒x(3/√10) + y(1/√10) = 4/√10

So cosα = 3/√10 ,sinα = 1/√10 and p = 4/√10

⇒(α,p)=(arctan(1/3),4/√10)

(pic credits:http://www.webmath.com/gline.html)