Finding the equation of a perpendicular bisector

Probability

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I am not sure if I am posting in the right thread, I am not sure if this is a geometry, calculus or algebra type of problem?

OK I am working on a circle problem, I have found the gradient of a straight line which is - 2/3

I have found the midpoints of this line, which are [4/2, - 2/2]

I am asked to find from this the equation of the perpendicular bisector AB

I am struggling to understand two things;

1) Using y - y1 = m(x - x1)

If y - (- 2/2 ) = - 2/3 x - 4/2

and then I bring over the the RHS ( - 2/2) which then becomes (2/2) as the denominators are already the same, i.e.

y = -2/3 x - 4/2 - 2/2

y = - 2/3x - 6/2

At this point the denominators are now different which can't be right?

If I do this, - 2/3 - 6/2 = - 18/6 + 4/6 = - 14/6 = - 7/3

So my equation looks like;

y = - 7/3x

Which I am quite confident is incorrect.

If y = - 7/3 (4/2) = - 4.67

I am not convinced.
 
I am not sure if I am posting in the right thread, I am not sure if this is a geometry, calculus or algebra type of problem?

OK I am working on a circle problem, I have found the gradient of a straight line which is - 2/3

I have found the midpoints of this line, which are [4/2, - 2/2]

I am asked to find from this the equation of the perpendicular bisector AB

I am struggling to understand two things;

1) Using y - y1 = m(x - x1)

If y - (- 2/2 ) = - 2/3 x - 4/2

and then I bring over the the RHS ( - 2/2) which then becomes (2/2) as the denominators are already the same, i.e.

y = -2/3 x - 4/2 - 2/2

y = - 2/3x - 6/2

At this point the denominators are now different which can't be right?

If I do this, - 2/3 - 6/2 = - 18/6 + 4/6 = - 14/6 = - 7/3

So my equation looks like;

y = - 7/3x

Which I am quite confident is incorrect.

If y = - 7/3 (4/2) = - 4.67

I am not convinced.

How is this different from what you had posted in:

http://www.freemathhelp.com/forum/threads/78036-Finding-a-circle

You got a camera-ready answer there! What more do you want?
 
How is this different from what you had posted in:

http://www.freemathhelp.com/forum/threads/78036-Finding-a-circle

You got a camera-ready answer there! What more do you want?

It is as you say a perfect answer I fully agree. There are some areas of it that I either need to revise up on or ask further questions, but what looks to be an immediate difference is the technique used to find the answer. The methods used I can't see how things like the equations of bisectors are calculated?

This is why I posted again asking for advise with regards the methods used to find the equations, using the gradient and midpoints.

Kind regards

Probability:smile:
 
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