# orthocenter, circumcenter

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• Jan 1st 2010, 07:16 PM
donut166
orthocenter, circumcenter
I have a triangle with these coordinates:
X= 82,21
Y=12,27
Z=74,43
I need to find the orthocenter and the circumcenter of this triangle.
• Jan 1st 2010, 11:32 PM
red_dog
$X(82,21), \ Y(12,27), \ Z(74,43)$

For the circumcenter:

Let $C(x,y)$ be the circumcenter. Then $CX=CY=CZ$ which is equivalent to $CX^2=CY^2=CZ^2$.

$(x-82)^2+(y-21)^2=(x-12)^2+(y-27)^2=(x-74)^2+(y-43)^2$

and you have to solve the system.

For the ortocenter you have to find the equations of two heights of the triangle and then to solve the system of two equation.

The equation for the height passing through X:

Find the slope of the line YZ: $m_{YZ}=\frac{27-43}{12-74}=\frac{8}{31}$. Then the slope of the height is $-\frac{31}{8}$.

The equation of the height is $y-21=-\frac{31}{8}(x-82)$

In the same way find the equation of the height pasiing through Y, then solve the system.
• Jan 2nd 2010, 05:23 AM
donut166
x and y solve
do i solve the final equations u wrote?
these equations:

http://www.mathhelpforum.com/math-he...5e2039b2-1.gif
(Surprised)
• Jan 2nd 2010, 05:55 AM
Hello donut166
Quote:

Originally Posted by donut166

Read carefully what red_dog wrote. To find the circumcentre, you'll need to solve:
$(x-82)^2+(y-21)^2=(x-12)^2+(y-27)^2=(x-74)^2+(y-43)^2$
which you'll do by taking, say, the first two expressions:
$(x-82)^2+(y-21)^2=(x-12)^2+(y-27)^2$
expanding the brackets, and simplifying to obtain a linear equation in $x$ and $y$ (i.e. there won't be any $x^2$ or $y^2$ terms).

Then take the second and third expressions:
$(x-12)^2+(y-27)^2=(x-74)^2+(y-43)^2$
and do the same.

Then solve the resulting pair of simultaneous equations.

For the orthocentre, red_dog showed you how to find the equation of one of the heights (altitudes) of the triangle. You will need to find a second one in the same way, and then solve this pair of simultaneous equations.

Can you do it now?

• Jan 2nd 2010, 08:32 AM
donut166
simplication ok
so i simplify this equation:http://www.mathhelpforum.com/math-he...5e2039b2-1.gif

bit by bit?
two by two?

• Jan 2nd 2010, 09:30 AM
donut166
and finding slope
which slope should i find?
XZ?
YZ?
XY?
• Jan 2nd 2010, 09:58 AM
earboth
Quote:

Originally Posted by red_dog
$X(82,21), \ Y(12,27), \ Z(74,43)$

...

For the ortocenter you have to find the equations of two heights of the triangle and then to solve the system of two equation.

The equation for the height passing through X:

Find the slope of the line YZ: $m_{YZ}=\frac{27-43}{12-74}=\frac{8}{31}$. Then the slope of the height is $-\frac{31}{8}$.

The equation of the height is $y-21=-\frac{31}{8}(x-82)$

In the same way find the equation of the height pasiing through Y, then solve the system.

Quote:

Originally Posted by donut166
which slope should i find?
XZ?
YZ?
XY?

1. Do what red-dog suggested: Find the height passing through Y. This height must be perpendicular to the slope of $\overline{XZ}$.

2. You should come out with a slope $m = \frac4{11}$. The height passes through Y( 12, 27). Thus the equation of the height is:

$h_Y: y-27 = \frac4{11} (x-12)$

$h_X: y-21=-\frac{31}{8}(x-82)$ (This equation was posted by red-dog!)

3. Solve this system of simultaneous equations for x and y, the coordinates of the orthocenter.
• Jan 2nd 2010, 11:02 AM
donut166
x=-8?
i Got x= -8.
• Jan 2nd 2010, 11:12 AM
donut166
x and y?
x and y are the coordinates of the orthocenter, right?
i got x= -8.
some1 check this plz.(Speechless)
• Jan 2nd 2010, 11:56 AM
Hi Donut166,

you've drifted off track.
The orthocentre x co-ordinate must be between 12 and 82.
The orthocentre y co-ordinate must be between 21 and 43.

A simple way to find the orthocentre is as follows...

The orthocentre is the point of intersection of the perpendicular lines to the 3 triangle sides that contain the opposite vertex.

Find the slope of any 2 of the 3 sides.
Write the perpendicular slopes.
Write the line equations of the perpendicular lines containing the point opposite the triangle side.

Therefore, you could write the slope of the line XY...
write the perpendicular slope...
write the equation of the line with the perpendicular slope that contains Z.

write the slope of XZ....
write the perpendicular slope..
write the equation of the line with this perpendicular slope that contains Y.

Solve the resulting 2 simultaneous equations to find the orthocentre.
Pairs of linear simultaneous equations find the point of intersection of 2 lines.

Study the .pdf diagram to help.
• Jan 2nd 2010, 11:59 AM
donut166
any number?
so the x coordinate can be any number (i mean i can pick out 70..) between 12 and 82?
and the y coordinate can also be any number? (examble 39)
• Jan 2nd 2010, 12:02 PM
donut166
and circumcenter
will the x and y number be in any sections of numbers?
example: 6-90
• Jan 2nd 2010, 12:26 PM
No Donut166,
they are very specific values of x and y.

You must discover them through finding points of intersection.

You will need to study the diagrams of orthocentre and circumcentre first.
Then you must understand how solving simultaneous equations finds
specific points of intersection.
You must also understand how to write the line slopes and line equations given points.

You are given 3 points.
From these you can write slopes and equations,
but you must understand the geometry.

Take your time and don't rush,
you are too eager to get the answer!
be patient, understand the geometry first.
• Jan 2nd 2010, 12:28 PM
donut166
and is this
• Jan 2nd 2010, 12:50 PM
Yes, you must solve this for x and y.

Here's a start for you, but you must practice from beginning to end.

What that equation is saying is....

The distance from all 3 points to the circle centre is the circle radius.

$(x-82)(x-82)+(y-21)(y-21)=x^2-82x-82x+82^2+y^2-21y-21y+21^2$

$=x^2+y^2-164x-42y++6560+441=x^2+y^2-164x-42y+7001.$

Multiply out the others, then all 3 expansions must be equal.
Try that.

There is another way to find the circumcentre, but see how you manage that anyway.
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