Find the values of θ and p , if the equation is the normal form of the line .
The equation of the line is 3 x + y + 2 = 0 .
The normal form of the line is given by,
x cos θ + y sin θ = p (1)
Here, cos θ = A A 2 +B 2 , sin θ = B A 2 +B 2 , p = C A 2 +B 2 .
The value p denotes the perpendicular distance of the line from the origin and θ denotes the angle made by the perpendicular with the positive direction of the x axis.
Compare the equation of line with the standard form A x + B y + C = 0 .
A = 3 , B = 1 A 2 +B 2 = ( 3 ) 2 + 1 2 = 1 + 3 = 4 = 2
Rearrange the terms of the equation.
3 x + y = − 2 − 3 x − y = 2 (2)
Divide equation (2) on both the sides by the term A 2 +B 2 ,
− 3 2 x − 1 2 y = 2 2 − 3 2 x − 1 2 y = 1 (3)
Compare equation (3) with equation (1).
cos θ = − 3 2 , sin θ = − 1 2 , p = 1
Solve the above expression to obtain the value of ω .
cos θ = − 3 2 , sin θ = − 1 2 cos θ = − cos 30 ° sin θ = − sin 30 ° cos θ = cos ( 180 ° + 30 ° ) sin θ = sin ( 180 ° + 30 ° ) cos θ = cos 210 ° sin θ = sin 210 ° ∵ sin ( 180 + θ ) = − sin θ cos ( 180 + θ ) = − cos θ
Further simplify the above equation.
θ = 210 °
Thus, the respective values of θ and p are 210 ° and 1 .
The equation of the line is
The normal form of the line is given by,
Here,
The value
Compare the equation of line with the standard form
Rearrange the terms of the equation.
Divide equation (2) on both the sides by the term
Compare equation (3) with equation (1).
Solve the above expression to obtain the value of
Further simplify the above equation.
Thus, the respective values of
