Question

A straight line L through the point (3,-2) is inclined
at an angle of 60° to the line √3x + y =1. If L also
intersects the x-axis, then the equation of L is

(a) y + √3x + 2 – 3√3 = 0
(b) y-√3x +2 +3√3 = 0
(c) √3y - x +3+2√3=0
(d) √3y + x - 3+2√3 = 0
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Answer 1

Answer:

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can't do it by ur own..

Answer 2

Answer:

(B) y -√3x +2+3√3 = 0

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Solution :-

Given line (√3x+y ) = 1

let slope the line is m1

y = -√3x+1

m1= dy/dx = -√3 + 0 = -√3

Let the slope of the line L is m2

We know that , to find angle between lines

tan¢ = | m2-m1/1+m1*m2 |

•°•tan60° =| m2+√3/ 1+ (-√3×m2)

+-√3 = m2 + √3 / 1+ (-√3 ×m2)

Taking as tan60° = +√3______ case(1)

=> √3(1-√3m2) = m2 + √3

=>√3 -3m2 = m2 + √3

=>4m2 = 0

=>m2 = 0

taking as tan60° = -√3 _______case (2)

=>-√3 ( 1-√3m2) = m2 + √3

=> 2m2 = 2√3

=> m2 = √3

Equation of line with slope m and passing through (x1,y1) is y-y1 =

m(x-x1)

•°• Equation of line with slope √3 and passing through (3,-2)

=>y -(-2) = √3(x-3)

=> y +2 = √3x-3√3*

=> y - √3x +2 +3√3 Answer

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