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Area of triangle through (0,0) (2,0) (0,2) =2*2/2=1

Half of it = 1

Condition: x*x/2=1 gives the first line:

x/sqrt(2)+y/sqrt(2)=1

Condition: 1/2(2-x)(2-x)=1 gives horizontal and vertical lines:

y=2-sqrt(2)

x=2-sqrt(2)

Inserting these into the first line equation does not give the right hand side, meaning they do not intersect in a single point.

Hope this helps.

Note: To avoid any yelling of the type: "Why do you help scholars in their homework, they should do it themselves" I chose the subject nickname.

Posts: 4,027

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Joined: Aug 2005

Hi John;

thank for asking me this question. And I guess sqrt(2) just missed the triangle area; he wrote:

Quote:

Area of triangle through (0,0) (2,0) (0,2) =2*2/2=1

It is a visible typo, not an error. Right after that he writes

Quote:

half of it = 1

If I am not wrong, the problem has already been solved. And **sqrt(2)**, I for one believe that any help is welcome when someone asks for it. By doing what you did, you show that you know how to do and there's no problem sharing it. I'd never be yelling at you. On the other hand, if it is a test and someone is cheating... ;^)

Best regards!

Luiz (Brazil)

*Edited: 4 May 2005, 2:46 p.m. *