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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

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

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,
```
7 years ago
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 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
```
7 years ago
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