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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 |
#2
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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.
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#3
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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! |
#4
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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. |
#5
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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?
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#6
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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. |
#7
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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 |
#8
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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}. |
#9
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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). |
#10
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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. |
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