Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x -axis. (i) (ii) y – 2 = 0 (iii) x – y = 4
(i)
The equation of line is x − 3 y + 8 = 0 .
The normal form of general equation of line A x + B y + C = 0 is given by
x cos ω + y sin ω = p (1)
Here cos ω = A A 2 +B 2 , sin ω = 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 ω denotes the angle made by the perpendicular with the positive direction of the x axis.
Compare the equation of line with the form A x + B y + C = 0 .
A = 1 , B = − 3 A 2 +B 2 = 1 + ( 3 ) 2 = 1 + 3 = 4 = 2
Rearrange the terms of the equation.
− x + 3 y = 8
Divide the above equation on both sides by the term A 2 +B 2 ,
− x 2 + 3 2 y = 8 2 − 1 2 x + 3 2 y = 4 (2)
Compare the equation (2) with the equation (1).
cos ω = − 1 2 , sin ω = 3 2 , p = 4 .
Solve the above expression to obtain the value of ω .
cos ω = − 1 2 sin ω = 3 2 cos ω = − cos 60 ° sin ω = sin 60 ° cos ω = cos ( 180 ° − 120 ° ) sin ω = sin ( 180 ° − 120 ° ) cos ω = cos 120 ° sin ω = sin 120 ° ∵ sin ( 180 − θ ) = sin θ cos ( 180 − θ ) = − cos θ
Further simplify the above equation.
ω = 120 °
Substitute the value of ω and p in equation (1).
x cos 120 ° + y sin 120 ° = 4
Thus, the normal form of the line x − 3 y + 8 = 0 is x cos 120 ° + y sin 120 ° = 4 . The perpendicular distance of the line from the origin is 4 and the angle made by the perpendicular with the positive direction of the x axis is 120 ° .
(ii)
The equation of line is y − 2 = 0 .
Compare the equation of line with the form A x + B y + C = 0 .
A = 0 , B = 1 A 2 +B 2 = 0 + 1 2 = 1 = 1
Rearrange the terms of the equation.
0 ⋅ x + y = 2
Divide the above equation on both sides by the term A 2 +B 2 ,
0 1 ⋅ x + y = 2 1 0 ⋅ x + y = 2 (3)
Compare the equation (3) with the equation (1).
cos ω = 0 , sin ω = 1 , p = 2 .
Solve the above expression to obtain the value of ω .
cos ω = 0 sin ω = 1 cos ω = cos 90 ° sin ω = sin 90 °
Further simplify the above equation.
ω = 90 °
Substitute the value of ω and p in equation (1).
x cos 90 ° + y sin 90 ° = 2
Thus, the normal form of the line y − 2 = 0 is x cos 90 ° + y sin 90 ° = 2 . The perpendicular distance of the line from the origin is 2 and the angle made by the perpendicular with the positive direction of the x axis is 90 ° .
(iii)
The equation of line is x − y = 4 or x − y − 4 = 0 .
Compare the equation of line with the form A x + B y + C = 0 .
A = 1 , B = − 1 A 2 +B 2 = 1 + ( − 1 ) 2 = 1 + 1 = 2
Divide the equation of line on both sides by the term A 2 +B 2 ,
1 2 x − 1 2 y = 4 2 1 2 x + ( − 1 2 ) y = 2 2 ⋅ 2 2 1 2 x + ( − 1 2 ) y = 2 2 (4)
Compare the equation (4) with the equation (1).
cos ω = 1 2 , sin ω = − 1 2 , p = 2 2
Solve the above expression to obtain the value of ω .
cos ω = 1 2 sin ω = − 1 2 cos ω = cos 45 ° sin ω = − sin 45 ° cos ω = cos ( 360 ° − 315 ° ) sin ω = sin ( 360 ° − 315 ° ) cos ω = cos 315 ° sin ω = sin 315 °
Further simplify the above equation.
ω = 315 °
Substitute the value of ω and p in equation (1).
x cos 315 ° + y sin 315 ° = 2 2
Thus, the normal form of the line x − y = 4 is x cos 315 ° + y sin 315 ° = 2 2 . The perpendicular distance of the line from the origin is 2 2 and the angle made by the perpendicular with the positive direction of the x axis is 315 ° .
(i)
The equation of line is
The normal form of general equation of line
Here
The value
Compare the equation of line with the form
Rearrange the terms of the equation.
Divide the above equation on both sides by the term
Compare the equation (2) with the equation (1).
Solve the above expression to obtain the value of
Further simplify the above equation.
Substitute the value of
Thus, the normal form of the line
(ii)
The equation of line is
Compare the equation of line with the form
Rearrange the terms of the equation.
Divide the above equation on both sides by the term
Compare the equation (3) with the equation (1).
Solve the above expression to obtain the value of
Further simplify the above equation.
Substitute the value of
Thus, the normal form of the line
(iii)
The equation of line is
Compare the equation of line with the form
Divide the equation of line on both sides by the term
Compare the equation (4) with the equation (1).
Solve the above expression to obtain the value of
Further simplify the above equation.
Substitute the value of
Thus, the normal form of the line
