Solved

# finding triangle circumcentre with equations for the triangle sides

Posted on 2010-01-12
Medium Priority
340 Views
Hi

I'm trying to find the triangle circumcentre from the equations for the triangle sides. I don't think I'm doing it right as I can't see how to finish the question off.

To get the circumcentre I need the point of intersection of 2 perpendicular bisectors. Given the equation of the side, the gradient of the bisector is -1/m but i don;t know how to get a point on this line to then work out the equation for the line using y - y1 = m(x - x1). I could say that the midpoint of the side is ((x1 + x2)/2, (y1 + y2)/2) and this point is on the perpendicular bisector but i don't know how to put this knowledge into y - y1 = m(x - x1). There's too many different x and y variables.

Once i have got the equation for one bisector i can get the others and then solve these equations simulataneously to get the midpoint.

The equations of the triangle are
y = 3x
y + 3x = 0
3y -x +12 = 0

Many thanks
0
Question by:andieje
[X]
###### Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

• Help others & share knowledge
• Earn cash & points
• 4
• 4
• 2
• +2

Author Comment

ID: 26297619
perhaps i need the coordinates of the corners first?
0

Author Comment

ID: 26297728
I can't get anything right today.

Are the corners A (0,0), B = (1.5, 4.5), C= (-1.2,3.6)
and midpoints AB = (0.75, 2.25), AC (-0.6, 1.8), BC(0.15, 4.05)

thanks
0

LVL 27

Expert Comment

ID: 26297762
Something is wrong here
y = 3x
y + 3x = 0
3y -x +12 = 0
The first two lines are parallel to each other, hence the third cannot make a triangle.
0

Author Comment

ID: 26297770
no they aren't!
0

LVL 27

Expert Comment

ID: 26297839
y = 3x
y = -3x
you are right
0

LVL 27

Expert Comment

ID: 26297896
Note that the perpendicular bisector of line 2 is parallel to line 3
0

LVL 15

Accepted Solution

yuk99 earned 2000 total points
ID: 26297900
Solve systems of all equations pairwise and you will get corners coordinates. For example, 1st and 2nd give (0,0).
0

LVL 27

Expert Comment

ID: 26297912
and perp bi of line 3 is parallel to line 2
0

Author Comment

ID: 26297915
Hi, that's what I did. Got the corners and then the bisectors. Thanks
0

LVL 3

Expert Comment

ID: 26297947
I answered this one for someone earlier, but the question was slightly different.  I think the solution is the same.

Use this after figuring out your corners.

http://www.experts-exchange.com/Other/Math_Science/Q_25045192.html?cid=1131#a26286230
0

LVL 15

Expert Comment

ID: 26297954
Are you sure you accepted the correct answer? It was just a hint. Have you find the circumcentre?
0

LVL 32

Expert Comment

ID: 26298163
andieje,

here's a no-point site of interest - has all sorts of visual aids:
http://www.mathopenref.com/trianglecircumcenter.html
http://www.mathopenref.com/constcircumcircle.html

0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

How to Win a Jar of Candy Corn: A Scientific Approach! I love mathematics. If you love mathematics also, you may enjoy this tip on how to use math to win your own jar of candy corn and to impress your friends. As I said, I love math, but I guâ€¦
When we purchase storage, we typically are advertised storage of 500GB, 1TB, 2TB and so on. However, when you actually install it into your computer, your 500GB HDD will actually show up as 465GB. Why? It has to do with the way people and computersâ€¦
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaacâ€¦
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templaâ€¦
###### Suggested Courses
Course of the Month10 days, 19 hours left to enroll