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Analytical Geometry

Cool goIITian

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8 Sep 2010 11:05:26 IST

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The points (0,8/3),(1,3)&(82,30) rthe verticesofa)noneb)obtusec)isoscelesd)acute

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The points (0,8/3),(1,3)&(82,30) rtheverticesofa)noneb)obtusec)isoscelesd)acute

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jay

Cool goIITian

Joined: 11 Nov 2008

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8 Sep 2010 11:15:27 IST

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When we find there determinant we get 0 these proves they are collinear but when we take there slops two at a time then they dont come out to be equal showing they are noncollinear so what should i do?

Yagyadutt Mishra

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Joined: 19 Feb 2009

Posts: 1984

8 Sep 2010 15:54:09 IST

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This is because they are not forming triangles....

slope when you take (1,3) and (0,8/3) comes equal to 1/3

when take ( 82,30) and (1,3) comes equal to 1/3

means all three are collinear with 1/3 at starting point

More explanation:-

First of all

Don't panic after seeing big values of co-ordinate....because they are arranged in such manner..that you can easily solve it....many term will be canceled out ...

A ( 0,8/3 ) B(1,3) C ( 82,30)

Slope of AB = (3-8/3)/(1-0) = 1/3

Slope of CB = (30-3)/(82-1) = 27/81 = 1/3

And it is not possible that two side in a triangle has same slope.....so obviously this co-ordinate is not forming any triangle,,,:D

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