Question

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

Answer 1

Answer:

y = √3x (or) y = 0

Step-by-step explanation:

Given that the line passes through the origin. So, its coordinates will be O(0,0)

Slope of the given line: √3x + y = 1 i.e m₁ = -√3.

Let the slope of the required line which makes 60° with above line is m.

∴ tan 60° = |-√3 - m/1 - √3m|

⇒ √3 = |-√3 - m/1 - √3m|

⇒ -√3 - m = √3 - 3m (or) -√3 - m = -√3 + 3m

⇒ m = √3 (or) m = 0

Given that line is passing through (0,0).

Hence, the equation for the required line is:

⇒ y + 0 = √3(x - 0)  (or) y + 0 = 0(x - 0)

⇒ y = √3x       (or)  y = 0

Hope it helps!

Answer 2

Answer:

y = 0

y = √3 x

Step-by-step explanation:

Given :

√3 x + y = 1

The line passes through the origin and is inclined 60° with the line √3 x + y = 1

y + √3 x = 1

⇒ y = 1 - √3 x

Comparing with y = m x + c :

⇒ m = -√3

Let the slope of the line be M

m = | ( -√3 - M ) / ( 1 - √3 M ) |

⇒ tan 60 = | ( -√3 - M ) / ( 1 - √3 M ) |

⇒ √3 = | ( -√3 - M ) / ( 1 - √3 M ) |

EITHER

⇒ √3 =  ( -√3 - M ) / ( 1 - √3 M )

⇒ √3 - 3 M = -√3 - M

⇒ 2 M = 2√3

⇒ M = √3

OR

-√3 = ( -√3 - M ) / ( 1 - √3 M )

⇒ -√3 + 3 M = - √3 - M

⇒ 4 M = 0

⇒ M = 0

When M = √3

y - y1 = √3 ( x - x1 )

⇒ y - 0 = √3 x - 0

⇒ y = √3 x

When M = 0

y - y1 = 0 ( x - x1 )

⇒ y - 0 = 0

⇒ y = 0

Hence these are the two equations .