Find the eqn. of the line which is at distance 5 from origin it makes an angle of 120o with +ve x-axis.
1It is also simple one.....
Just given exterior angle made by the line is 120 ...so interior will be 60
DIstance = 5
So x -intecept = 5/sin(60) = 10 √3
y-intercept = 5/sin(30) = 10
So equation of line
x/a + y/b = 1
x / 10√3 + y/10 = 1
x + y √3 = 10 √3
1@ Scintillating dev : Mine is coming to √3x+y = ±10 (same as that of sahil)
Given that angle it makes with +ve x-axis is 120°, hence slope= -√3.
Let the equation of the line be y=-√3x + c.
From the diagram, DAO = 60° and OD=5 units.
So, AO = 5sin60° = 10√3.
tan60°= OBOA
or, OB=10 units.
Therefore, c=10.
Now if the line passed through the 4th quadrant, through similar calculations, c= -10.
SO c=±10.
The equation becomes
y = -√3x ± 10
or √3x + y = ± 10........[Answer]
1Sorry, i forgot 2 check d answer before. however, it's given √3x+y-10=0
if we take the line in the 4th quad, the angle with the +ve X axis is 60°(not 120°), so c=-10 is not considered.
Anyways, thank you for the concept.
1Yes i didn't notice that....
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