The perpendicular distance between the line and origin is 5 units and the angle made by the perpendicular with the positive X-axis is 30°. Then, the equation of line is

Question

The perpendicular distance between the line and origin is 5 units and the angle made by the perpendicular with the positive X-axis is 30°. Then, the equation of line is (A) x+√3 y=10 (B) √3 x+y=10 (C) x-√3 y=10 (D) √3 x-y=10

Tardigrade Admin · Accepted Answer

Here, p=5 and α=30° <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b7f54358cf6c9579d7426cd6de6e2494-.png" /> So, equation of line in normal form is x cos α+y sin α =p ∴ x cos 30°+y sin 30° =5 ⇒ (√3/2) x+(1/2) y=5 ⇒ √3 x+y=10

Q. The perpendicular distance between the line and origin is 5 units and the angle made by the perpendicular with the positive $X$ -axis is $3 0^{\circ}$ . Then, the equation of line is

$x + 3 y = 10$

$3 x + y = 10$

$x - 3 y = 10$

$3 x - y = 10$

Here, $p = 5$ and $α = 3 0^{\circ}$

So, equation of line in normal form is
$x cos α + y sin α = p$
$∴ x cos 3 0^{\circ} + y sin 3 0^{\circ} = 5$
$\Rightarrow \frac{3}{2} x + \frac{1}{2} y = 5$
$\Rightarrow 3 x + y = 10$

Q. The perpendicular distance between the line and origin is 5 units and the angle made by the perpendicular with the positive X-axis is 30∘. Then, the equation of line is

x+3​y=10

3​x+y=10

x−3​y=10

3​x−y=10