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  #11  
Old 10-08-2007, 05:59 PM
runningmarvel runningmarvel is offline
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Default Re: Stupid question about straight lines - need help

Not at all. In fact, I had trouble understanding the problem at first, because it's unclear what is meant by a "bisector of a line," given that every point on a line can (arguably) be a bisector.

[/ QUOTE ]I just copied it down as it was in my mathsbook, which is obviously wrong. I'll keep that in mind though, next time my teacher asks
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  #12  
Old 10-08-2007, 07:58 PM
jogsxyz jogsxyz is offline
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Default Re: Stupid question about straight lines - need help

[ QUOTE ]

The intercepts are at (0,6) and (12,0). So the sides are 6 and 12, and the hypoteneuse is 6\sqrt{5}.

[/ QUOTE ]

I agree with this answer.



The black line joins the two points. The red line is
the perpendicular line bisecting the black at 4,4.
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  #13  
Old 10-08-2007, 11:18 PM
Senor Cardgage Senor Cardgage is offline
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Default Re: Stupid question about straight lines - need help

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Hey guys,
I have to do some homework about straight lines. So far I've managed everything without problems, but the I stumpled upon this question:

The perpendicular bisector of the straight line joining the points (3,2) and (5,6) meets the x axis at A and the y axis at B. Prove that the distance AB is equal to 6x"square root 5"(I don't know how to do the symbol...)



[/ QUOTE ]

Kinda hard to prove the distance is 6x"square root 5",
when the actual distance is 2x"square root 5".

It's a 1 X 2 X SR 5 triangle. Except here the sides
are 2 and 4, so the hypotenuse is 2 SR 5.

[/ QUOTE ]

The intercepts are at (0,6) and (12,0). So the sides are 6 and 12, and the hypoteneuse is 6\sqrt{5}.

[/ QUOTE ]

And I don't notice till days later that I misspelled hypotenuse. /wrist

So much for math nittin'.
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  #14  
Old 10-09-2007, 02:30 PM
jogsxyz jogsxyz is offline
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Join Date: Mar 2005
Posts: 1,167
Default Re: Stupid question about straight lines - need help

[ QUOTE ]
Not at all. In fact, I had trouble understanding the problem at first, because it's unclear what is meant by a "bisector of a line," given that every point on a line can (arguably) be a bisector.

[/ QUOTE ]I just copied it down as it was in my mathsbook, which is obviously wrong. I'll keep that in mind though, next time my teacher asks

[/ QUOTE ]

Not wrong. SenorCardgage posted the correct solution.
Your A and B points are wrong.
A = (0,6) not (0,4)
B = (12,0) not (8,0)

The bisector is the point which splits the line
into two equal halves. The point is (4,4).
You already listed that point in your solution.
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