The length of the perpendicular from the origin to a line is 7 and line makes an angle of 150° with the positive direction of y -axis then the equation of the line is

Question

The length of the perpendicular from the origin to a line is 7 and line makes an angle of 150° with the positive direction of y -axis then the equation of the line is (A) 4 x + 5 y = 7 (B) -x + 3 y = 2 (C) √3 x - y =10 √2 (D) √3 x+y=14

Tardigrade Admin · Accepted Answer

Here p = 7 and α = 30° <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/fb1dba4c53d5f398811de5a2d085d306-.png" /> ∴ Equation of the required line is x cos 30°+y sin 30°=7 or × (√3/2)+y × (1/2)=7 or √3 x+y=14

Q. The length of the perpendicular from the origin to a line is $7$ and line makes an angle of $15 0^{\circ}$ with the positive direction of $y$ -axis then the equation of the line is

4 x + 5 y = 7

-x + 3 y = 2

$3 x - y = 102$

$3 x + y = 14$

Here $p = 7$ and $α = 3 0^{\circ}$

$∴$ Equation of the required line is
$x cos 3 0^{\circ} + y sin 3 0^{\circ} = 7$
or $\times \frac{3}{2} + y \times \frac{1}{2} = 7$ or
$3 x + y = 14$

Q. The length of the perpendicular from the origin to a line is 7 and line makes an angle of 150∘ with the positive direction of y -axis then the equation of the line is

4 x + 5 y = 7

-x + 3 y = 2

3​x−y=102​

3​x+y=14

Here p=7 and α=30∘ ∴ Equation of the required line is xcos30∘+ysin30∘=7 or ×23​​+y×21​=7 or 3​x+y=14

Q. The length of the perpendicular from the origin to a line is $7$ and line makes an angle of $15 0^{\circ}$ with the positive direction of $y$ -axis then the equation of the line is

$3 x - y = 102$

$3 x + y = 14$

Here $p = 7$ and $α = 3 0^{\circ}$

$∴$ Equation of the required line is
$x cos 3 0^{\circ} + y sin 3 0^{\circ} = 7$
or $\times \frac{3}{2} + y \times \frac{1}{2} = 7$ or
$3 x + y = 14$