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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..?
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 detailedsolution very quickly.We are all IITians and here to help you in your IIT JEE & AIEEE preparation.All the best. Regards,Askiitians ExpertsPramod KumarIITR Alumni
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