Question

a straight line passing through origin and inclined at 60° to the line √3x+y=1 is

Answer 1

Given :

Line , L1 = √3x+y=1

Point , O = Origin ( 0 , 0)

To Find : Equation of line passing through origin and inclined at 60° with L1.

Formula Used :

$\\ y \: = \: mx \: + \: c$

&

$\\ \tan( \alpha ) \: = | \frac{m2 \: - \: m1}{1 \: + \: m2m1} |$

&

$\\ (y \: - \: y1) \: = \: m( \: x \: - \: x1 \: )$

DETAILED SOLUTION IS IN THE ADJOINING ATTACHMENT.

_________________________

ANSWER : The equation of the required line is

y = √3 x and y = 0 i.e., X-axis.

Answer 2

Answer:

y = 0 or y = √3x

Step-by-step explanation:

Hi,

Given equation of line is √3x + y = 1

Writing in slope-intercept form , we get

y = -√3x + 1

Hence, the slope of  the line is -√3

We know that if Ф is the inclination of line with positive x -axis,

slope is given by tanФ,

so tanФ = -√3

⇒ Ф = - 60°

Now, given that line makes an angle of 60° with the given line,

hence its inclination would be either (Ф + 60) or (Ф - 60)

= -60 +60 or -60 - 60

= 0° or -120°

Hence the slope of the required line would be either tan0 = 0 or

tan(-120) = -√3

As , we know any equation of line passing through origin and

having slope 'm' will be of the form y = mx,

Hence the equation of the required line would be either

y = 0 or y = √3x

Hope, it helps !