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The equation of a straight line is given by:

y = mx + c

where m is the slope and c is the intercept

let m_{1} be the slope of the perpendicular line and m_{2} be the slope of the straight line

m_{1} = tan(60) = √3

since m_{1} and m_{2} are perpendicular, m_{2} = -1/m_{1}

m_{2} = -1/√3

Thus, y = -1/√3 x + c

multiply through by √3

y√3 = - x + c√3

x + y√3 - c√3 = 0 ------------- eqn(1)

d = |(ax_{1} + by_{1} + c)/(√a^{2} + b^{2})|

x_{1}, y_{1} = 0

d = distance from the origin to the perpendicular length = 2

from eqn(1),

a = 1 [coefficient of x],

b = √3 [coefficient of y]

c = - c√3 [constant]

2 = |(c)/(√a^{2} + b^{2})|

2 = |(-c√3)/(√1^{2} + √3^{2})|

2 = |(-c√3)/(√1 + 3)|

2 = |(-c√3)/(√4)|

2 = |(-c√3)/2|

2 = (c√3)/2

c√3 = 4

substitute into eqn(1),

x + y√3 - 4 = 0

x + √3 y = 4

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