HP Forum Archive 15

[ Return to Index | Top of Index ]

Vieira, Luiz C. (Brazil) please help Message #1 Posted by John Littlefield on 4 May 2005, 2:46 a.m. I need help w/ this homework problem, can you help me? Draw a triangle with vertices (0,0),(0,2)and(2,0), with lines parallel to each side that each divide the triangle into two equal parts. Determine whether or not these lines are concurrent (all meet in one spot).

Re: Vieira, Luiz C. (Brazil) please help Message #2 Posted by sqrt(2) on 4 May 2005, 6:51 a.m., in response to message #1 by John Littlefield 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.

(edited) Sorry I'm late... d8^( Message #3 Posted by Vieira, Luiz C. (Brazil) on 4 May 2005, 10:51 a.m., in response to message #1 by John Littlefield 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.

...cheating Message #4 Posted by Michael F. Coyle on 4 May 2005, 10:41 p.m., in response to message #3 by Vieira, Luiz C. (Brazil) Hi Luiz! Quote: On the other hand, if it is a test and someone is cheating... ;^) ...Then he deserves full credit for his resourcefullness! (And a harder test next time.) - Michael

Re: ...cheating Message #5 Posted by Vieira, Luiz C. (Brazil) on 4 May 2005, 11:42 p.m., in response to message #4 by Michael F. Coyle Now you see why I did not conclude the thinking... ;^)

Thank you Luiz C... Message #6 Posted by sqrt(2) on 5 May 2005, 4:02 a.m., in response to message #3 by Vieira, Luiz C. (Brazil) ... for the correction of the typo. It should have read: 2 ENTER 2 x 2 / I have a problem using the equals sign;)

[ Return to Index | Top of Index ]

Go back to the main exhibit hall