# orthocenter, circumcenter

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• Jan 2nd 2010, 01:00 PM
donut166
to archie
can u tell me the other way to find the circumcenter?
and what expansion?
• Jan 2nd 2010, 01:16 PM
By expansion, I mean multiplying out $(x-12)^2+(y-27)^2$
and the third part of that equality,
as you need to go through those steps, to find the x and y co-ordinates of the circumcentre.

The other way is to write the equations of two of the three perpendicular bisectors of the sides of the triangle.
Then you find the point of intersection of these 2 lines.

you need to work with the geometric diagram of the circumcentre that i uploaded or some other suitable one.

You need to do this carefully step by step, as you are learning this.

So, just reply saying which method you want to continue with first.
• Jan 2nd 2010, 01:27 PM
donut166
to archie...agian
the second way sounds easier
• Jan 2nd 2010, 01:46 PM
Ok,

you've got to be able to follow the geometry.

All 3 points are on the circumference of a circle.
The 3 points are the triangle vertices.

Join any of the two points together and also to the circle centre.
You now have an isosceles triangle.
If you split an isosceles triangle in two, you have two identical right-angled triangles.

This ok so far?

Ok,
1.we find the midpoint of a triangle side.
2.We write the slope of the triangle side.
3.Then write the perpendicular slope.
4.Next write the line equation with that perpendicular slope and the midpoint.
5.You now have the equation of the perpendicular bisector.
6.The circle centre lies on this line.

Do the same steps for a second side.
You now have the equations of two lines.
These two lines intersect at the circumcentre.
Solve the simultaneous equations to find the co-ordinates of the centre.

Let me get you started......

midpoint of XY is $(\frac{82+12}{2},\frac{21+27}{2})=(47,24)$

The slope of XY is $\frac{21-27}{82-12}=-\frac{6}{70}$

The slope of the perpendicular bisector is $\frac{70}{6}$

The equation of this perpendicular bisector is $y-24=\frac{70}{6}(x-47)$

Do the same procedure for another side...
You will now have 2 equations.
The solution of the simultaneous equations is the circumcentre.
• Jan 2nd 2010, 02:00 PM
donut166
lol help plz
can u write out the other equation?
• Jan 2nd 2010, 02:08 PM
You really need to practice.
First, you should try the steps I've shown for the first equation.
Getting the answer is not what is important.

Knowing how to perform the steps to get there is!

You have the equation of the perpendicular bisector of XY.

What is the first step for the side XZ ?

The slope of XZ or the midpoint ?

It does not matter!!

You need both in order to write the equation of the perpendicular bisector of XZ.

Try to work out the slope of XZ and find it's midpoint.

Will you try that first ?
• Jan 2nd 2010, 02:52 PM
X is $(82,21)=(x_1,y_1)$
Z is $(74,43)=(x_2,y_2)$

The slope of XZ is $\frac{y_2-y_1}{x_2-x_1}=\frac{43-21}{74-82}$

$=\frac{22}{-8}=-\frac{11}{4}$

The midpoint of XZ is $\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$

$=\frac{82+74}{2},\frac{21+43}{2}=(78,32)=(x_m,y_m) .$

The slope of the perpendicular bisector is $\frac{4}{11}$

The perpendicular bisector has slope $\frac{4}{11}$
and contains (78,32)

so it's equation is $y-y_m=\frac{4}{11}(x-x_m),\ so\ y-32=\frac{4}{11}(x-78)$
• Jan 2nd 2010, 02:52 PM
donut166
i got the orthocenter!
is it:

x=74.5
y=49.75?
• Jan 2nd 2010, 03:09 PM
Maybe I'm wrong?
• Jan 2nd 2010, 03:17 PM
you're right!
• Jan 2nd 2010, 03:36 PM
donut166
omg
holy crap i am???
• Jan 2nd 2010, 03:42 PM
Yep!

Don't round off so much, though,
x=74.579
y=49.76

Now congratulate yourself and try the circumcentre whenever you're ready!
• Jan 2nd 2010, 03:47 PM
donut166
fdsfs
ok the circumcenter is really killing me.(Angry)
• Jan 2nd 2010, 03:55 PM
You just need to build on what you did with the orthocentre.

You must choose 2 sides and get the midpoints first.
Get the slopes and then write the perpendicular slopes.

Write the equations of these lines that have the perpendicular slopes and contain the midpoints of the two triangle sides.