Maths Question! (1 Viewer)

[OriginalCopy]

Hi,

I've tried solving this question a few times, however I can't seem to do and I think the points that have been provided are incorrect. Here is the question:

The points A(-3,4) B(-5,-2) and C(2,6) form the vertices of a triangle. Prove that the perpendicular bisectors of the vertices are concurrent.

Could someone please provide the solution?

Thanks a lot for the help!

 

[Drongoski]

Hi,

I've tried solving this question a few times, however I can't seem to do and I think the points that have been provided are incorrect. Here is the question:

The points A(-3,4) B(-5,-2) and C(2,6) form the vertices of a triangle. Prove that the perpendicular bisectors of the vertices are concurrent.

Could someone please provide the solution?

Thanks a lot for the help!

Doesn't make sense. You cannot bisect a vertex!

But the perpendicular bisectors of the 3 sides of any proper triangle on a plane are concurrent. For this specific case, if P, Q and R are the mid-points of the 3 sides BC, CA and AB: you can find their mid-points P, Q and R, the gradients of BC, CA and AB, and therefore of their perpendiculars. Then you can find the equations of these 3 perpendiculars. Find where 2 perps intersect; show this point of intersection lies on the 3rd perpendicular.

 

[OriginalCopy]

Thanks! I did try to do that method, however the solution I got showed that the perpendicular bisectors were not concurrent.

 

[Drongoski]

Thanks! I did try to do that method, however the solution I got showed that the perpendicular bisectors were not concurrent.

The 3 mid-points are:


and the gradients of the perpendiculars at P, Q and R are resp: -7/8, -5/2 and -1/3

The equations of the 3 perpendiculars are (not very nice numbers):

P: y = -7x/8 + 11/16
Q: y = -5x/2 + 15/4
R: y = -x/3 - 1/3

The first 2 intersect at: (49/26, -0.96153 ...)
You now use x = 49/26 in the 3rd equation; should get y = -0.96153 ...

.: the 3 perpendicular are concurrent.

But this concurrency holds for all plane triangles; not just this particular case.