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# 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..? Badiuddin askIITians.ismu Expert
147 Points
11 years ago
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 preparation. All the best. Regards, Askiitians Experts Badiuddin
11 years ago

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.

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Regards,