(i) Given equation is x − $\sqrt{3}y$ + 8 = 0 ⇒ x − $\sqrt{3}y$ = −8 ⇒ − x + $\sqrt{3}y$ = 8 ... (1) Also, $\sqrt{(\text{coeff. of } x^2)+(\text{coeff. of } y^2)}$ = $\sqrt{(-1)^2+(\sqrt{3})^2}$ = $\sqrt{1+3}$ = $\sqrt{4}$ = 2 Now dividing both the sides of (1) by 2, we get − $\frac{1}{2}x + \frac{\sqrt{3}}{2}y$ = 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 $\sqrt{(\text{coeff. of } x^2)+(\text{coeff. of } y^2)}$ = $\sqrt{(1)^2(-1)^2}$ = $\sqrt{1+1}$ = $\sqrt{2}$ Now dividing both the sides of (1) by $\sqrt{2}$ , we get $\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}}$ = $2\sqrt{2}$ ⇒ x cos 45° − y sin 45° = 2 $\sqrt{2}$ ⇒ x cos (360° - 45°) + y sin (360° - 45°) = $2\sqrt{2}$ ⇒ x cos 315° + y sin 315° = 2 $\sqrt{2}$ , is the required equation in normal form. Q The normal form is x cosω + y sinω = p So, p = 2 $\sqrt{2}$ 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°.