Reduce the equation √3x+y-8=0 into normal form. Find the length of perpendicular from the origin to the given line and angle made by the perpendicular with x-axis.

Question

Reduce the equation √3x+y-8=0 into normal form. Find the length of perpendicular from the origin to the given line and angle made by the perpendicular with x-axis. (A) 8, 120° (B) 6, 120° (C) 4, 30° (D) 5, 15°

Tardigrade Admin · Accepted Answer

Given equation is √3x+y-8=0 ldots(i) Dividing (i) by √(√3)2+(1)2=2, we get (√3/2)x+(1/2) y=4 or cos 30°(x)+sin 30°(y)=4 ldots(ii) Comparing (ii) with x cos ω+y sin ω=p, we get p=4 and ω=30°

Q. Reduce the equation $3 x + y - 8 = 0$ into normal form. Find the length of perpendicular from the origin to the given line and angle made by the perpendicular with $x$ -axis.

$8$ , $12 0^{\circ}$

$6$ , $12 0^{\circ}$

$4$ , $3 0^{\circ}$

$5, 1 5^{\circ}$

Given equation is $3 x + y - 8 = 0 \dots (i)$
Dividing $(i)$ by $(3)^{2} + (1)^{2} = 2$ , we get
$\frac{3}{2} x + \frac{1}{2} y = 4$ or $cos 3 0^{\circ} (x) + s in 3 0^{\circ} (y) = 4 \dots (ii)$
Comparing $(ii)$ with $x cos ω + y s in ω = p$ , we get $p = 4$ and $ω = 3 0^{\circ}$

Q. Reduce the equation 3​x+y−8=0 into normal form. Find the length of perpendicular from the origin to the given line and angle made by the perpendicular with x-axis.

8, 120∘

6, 120∘

4, 30∘

5,15∘