(i) Given equation is x − √3y + 8 = 0
⇒ x − √3y = −8
⇒ − x + √3y = 8 ... (1)
Also, √(coeff.of‌x2)+(coeff.of‌y2)
= √(−1)2+(√3)2 = √1+3 = √4 = 2
Now dividing both the sides of (1) by 2, we get −

1

2

x+

√3

2

y = 4
⇒ – cos 60°x + sin 60° y = 4.
⇒ {cos (180° – 60°)} x + {sin (180° – 60°)}y = 4
⇒ cos 120° x + sin 120° y = 4
∴ x cos 120° + y sin 120° = 4 is the required equation in normal form
Straight Lines 179
Since The normal form is x cosω + y sinω = p
So, ω = 120° and p = 4
∴ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.
(ii) Given equation is y – 2 = 0
⇒ y = 2 ⇒ 0 . x + 1 . y = 2
⇒ x cos 90° + y sin 90° = 2 is the required equation in normal form
Since The normal form is x cosω + y sinω = p So, ω = 90° and p = 2
∴ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.
(iii) Given equation is x – y = 4 ... (1)
Also √(coeff.of‌x2)+(coeff.of‌y2)
= √(1)2(−1)2 = √1+1 = √2
Now dividing both the sides of (1) by √2 , we get

x

√2

−

y

√2

= 2√2
⇒ x cos 45° − y sin 45° = 2 √2
⇒ x cos (360° - 45°) + y sin (360° - 45°) = 2√2
⇒ x cos 315° + y sin 315° = 2 √2 ,
is the required equation in normal form.
Q The normal form is x cosω + y sinω = p
So, p = 2 √2 and ω = 315°
∴ Distance of the line from the origin is 2 2 and the angle between
perpendicular and the positive x-axis is 315°.