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Raziya Syed Grade: 12
        

A straight line through the point (2,2) intersects the line 31/2x +y =0 and 31/2x -y=0 at the point A and B. The equation to the line AB so that the triangle OAB is equailateral is..?

7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										Dear Raziya

given equation of line is y =v3x and y =-v3x

point at a distance r on the line y =v3x is ( r cos60 , r sin 60) =(r/2 , rv3/2)

point at a distance r on the line y =-v3 x is ( r cos120 ,r sin 120)=(-r/2 , rv3/2)

equation of line passing through these two point is
y = rv3/2

but this line also passes through ( 2,2)

so rv3 /2 =2
r = 4/v3

so equation of line is y =2

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Badiuddin
7 years ago
AskiitianExpert Pramod-IIT-R
47 Points
										

Dear Raziya
given equation of line is y =v3x and y =-v3x point at a distance r on the line y =v3x is ( r cos60 , r sin 60) =(r/2 , rv3/2) point at a distance r on the line y =-v3 x is ( r cos120 ,r sin 120)=(-r/2 , rv3/2) equation of line passing through these two point is y = rv3/2 but this line also passes through ( 2,2) so rv3 /2 =2 r = 4/v3 so equation of line is y =2


Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed
solution very  quickly.
We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

All the best.
Regards,
Askiitians Experts
Pramod Kumar
IITR Alumni

7 years ago
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