Stupid question about straight lines - need help
 

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...)

I tried to answer this by first finding the gradient of the straight line, then getting the perpendicular gradient of the straight line.
After that I got the midpoint of the straight line and got the equation for the perpendicular line.
From this I found out what A and B were and calculated the distance between them.

The exact numbers I got were:
gradient of straight line=2
gradient of perpendicular line= -0.5
midpoint of straight line= (4,4)
equation for perpendicular line: y=-0.5X+4
A=(8,0) B= (0,4)

The distance between A and B that I got was 4x"square root 5" which means, that I didn't prove this.

What did I get wrong?
Plz help me out, this really annoys me, I've gone through it about 500 times but I just don't see what I did wrong??

Thanks, guys

Re: Stupid question about straight lines - need help
 

Hey. Check your equation for the perpendicular line and your results for A and B. the other stuff looks fine.

Re: Stupid question about straight lines - need help
 

[ QUOTE ]
Hey. Check your equation for the perpendicular line and your results for A and B. the other stuff looks fine.

[/ QUOTE ]
Hey, thanks! for the equation I now got -o.5X+6, I forgot to double the one number. I must've looked at it about 20 times and just not realized!
Thanks again!

Re: Stupid question about straight lines - need help
 

Point (4,4) should be on the line y=-0.5X+4, but it is not
4=! -2+4.

You've made an aritmetical error somewhere. otherwise your solution is fine.

Re: Stupid question about straight lines - need help
 

Am I a huge, huge math nit for wanting it to be the perpendicular bisector of a line segment?

Re: Stupid question about straight lines - need help
 

[ 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.

Re: Stupid question about straight lines - need help
 

[ 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 ]

can you maybe elaborate on that, I don't understand what your meaning here?
thx

Re: Stupid question about straight lines - need help
 

[ 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}.

Re: Stupid question about straight lines - need help
 

[ QUOTE ]



can you maybe elaborate on that, I don't understand what your meaning here?
thx

[/ QUOTE ]

sorry, thought the question was distance between points
(3,2) and (5,6).

Re: Stupid question about straight lines - need help
 

[ QUOTE ]
Am I a huge, huge math nit for wanting it to be the perpendicular bisector of a line segment?

[/ 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.

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;)

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.

http://i32.photobucket.com/albums/d2...xyz/temppb.jpg

The black line joins the two points. The red line is
the perpendicular line bisecting the black at 4,4.

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'.

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.