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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

Ok we got this question in maths.

Find the midpoint M, of the line passing through points A and B of circle:

if the line has equation:

Noone could get it, well I could of but I didn't want to use the conventional method, I wanted to use a method noone else would think of because that the kinda of guy I am. Plus the teacher didn't give us enougth time before she excited and confidently told us how to do it.

The way she told us how to do it was with some simultaneous stuff, I wasn't listening I was working on my method.

I found that the midpoint of an interval under the circumstances above is the point in which the perpundicular line to the interval specifed passes through the center of the circle.

So by finding the perpundicular distance I found the hypotinuse of the triangle whos base and height will give the coordinates of M in relation to C if C is the center of the circle.

Now this is all works in theory. But I couldn't get it to work on my page. I knew one of the angles of the triangle because I knew the gradient of the line. Using atan(-1/m) of the line given. Using Sin Cos Tan (I refuse to use the childish word SOHCAHTOA) I can calculate the coordinates of M. But when I did so, I recieved irrational answer. But the answers my teacher found were very rational.

Heres my working out: (Or should i say not-working out, hahaha ... bad joke )

It's obviously very wrong. Can someone help me out?

*Last edited by Zmurf (2006-05-11 18:59:47)*

*"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Do you mean x² + y² = 8? Otherwise, it isn't a circle.

Assuming that you do, the question is to find the mid point of AB, correct?

Assuming that as well, what I would do is just find A, then find B, then find the midpoint between the two by:

No?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

I think yes. Now the question is to find A and B:

*Last edited by krassi_holmz (2006-05-12 03:42:15)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

How's that graphic?

I'm starting to make good Mathematica graphics:)

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Looks great, except I would find a way to left (or right) justify the points so the numbers don't go through the lines.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

Yes, I know that way, I wanna know why it won't work when I use perpundicular distance.

*"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

I don't know where the mistake is, but there obviously is a mistake.

For Ricky: I left the points unjustified to see who will notice this and post some note.

Now I see you're a good moderator.

IPBLE: Increasing Performance By Lowering Expectations.

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