NCERT Solutions
Class 11 Maths
Straight Lines
Ex.10.3 Q.3
Reduce the following equations into normal form. Find their perpendicular distances from the
origin and angle between perpendicular and the positive x-axis.
(1) x – √3y + 8 = 0
(2) y – 2 = 0
(3) x – y = 4
(1) The given equation is x – √3y + 8 = 0
It can be reduced as:
x – √3y = -8
=> -x + √3y = 8
On dividing both sides by √ {(-1)2 +(√3)2} = √ (1 + 3) = √4 = 2, we get
=> -
+ √3
= 
=> - + √3 = 4
=> (-
)x + (√3 ÷ 2) y = 4
=> x cos 120° + y sin 120° = 4 …………. (1)
Equation (1) is in the normal form.
On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get
ω = 120° and p = 4
Thus, the perpendicular distance of the line from the origin is 4,
while the angle between the perpendicular and the positive x-axis is 120°.
(2) The given equation is y – 2 = 0.
It can be reduced as 0.x + 1.y = 2
On dividing both sides by √ {02 + 12} = 1, we get
0 × x + 1 × y = 2
=> x cos 90° + y sin 90° = 2 ………... (1)
Equation (1) is in the normal form.
On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get
ω = 90° and p = 2
Thus, the perpendicular distance of the line from the origin is 2,
while the angle between the perpendicular and the positive x-axis is 90°.
(3) The given equation is x – y = 4.
It can be reduced as 1 × x + (–1) y = 4
On dividing both sides by √ {12 + (-1)2} = √2, we get
(1 ÷ √2) x + (-1 ÷ √2) = (4 ÷ √2)
=> x cos (2π – 
) + y sin (2π – ) = 2√2
=> x cos 315° + y sin 315° = 2√2 ………... (1)
Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line x cos ω + y sin ω = p, we get
ω = 315° and p = 2√2
Thus, the perpendicular distance of the line from the origin is 2√2, while the angle
between the perpendicular and the positive x-axis is 315°.