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Question

Find the equation of the straight line which passes through the origin and making angle 60 with the line x+3y+33=0

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Solution

Given line is x+3y+33=0
y=(13)x3
Slope of (1) = 13
Let slope of the required line be m Also between there lines is given to be 60
tan60=∣ ∣ ∣m(1/3)1+m(1/3)∣ ∣ ∣3=3m+13m3m+13m=±3
3m+13m=3
3m+1=33m
m=13
Using y = mx + c the equation of the required line is y=13x+0
i.e. x3y=0 ( This passes through origin so c = 0)
3m+13m=3
3m+1=3+3m
m is not defined
The slope of the required line is not defined Thus the required line is a vertical line This line is to pass through the origin
The equation of the required line is x = 0

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