orthocenter, circumcenter

**Archie Meade** · January 2nd 2010, 04:31 PM

Keep working on it,

it's 1.30 am here, so it's my naptime.

Remember the following though.

When you write the equation of a line, the x and y values are the co-ordinates of all of the points on the line.
A different x will go with a different y.

When you have 2 line equations, as in these cases,
once you solve the simultaneous equations,
you find the only point on both lines, because only those x and y co-ordinates are on both lines.

This is why you need to first write the equations of the lines that the orthocentre and circumcentre are on, before you solve the simultaneous equations.

**Archie Meade** · January 2nd 2010, 04:40 PM

Ok,

two perpendicular bisectors, already written are....

$y-32=\frac{4}{11}(x-78)$

$y-24=\frac{70}{6}(x-46)$

Hence...

$11y-352=4(x-78)$

$11y-352=4x-312$

$11y=4x+40$

Also...

$6y-144=70x-3290$

$6y=70x-3146$

Since $y=\frac{4x+40}{11}$

then $6(\frac{4x+40}{11})=70x-3146$

Now, work out this x and use it to find y.

**donut166** · January 2nd 2010, 05:19 PM

this is still about the circumcenter, right?
if it is, thanks.

**donut166** · January 2nd 2010, 05:55 PM

does x equal 46.71?

**Archie Meade** · January 2nd 2010, 06:12 PM

It sure does,
now you will have no bother finding y.

If you go back over all this and write it out neatly for yourself,
with the diagrams of the triangles and the lines,
understanding the geometry, point co-ordinate geometry,
and how the geometry is expressed in algebra,
you will have learned quite a lot through your efforts,
and it will be quite memorable for you.

Good job!

**donut166** · January 2nd 2010, 06:16 PM

can u finish the problem for me?
i'm confused on the y.

**Archie Meade** · January 2nd 2010, 06:38 PM

Maybe, you've worn yourself out for the day!
Ok, but promise me you'll do it all again from scratch soon???
If you do, you will know why I asked.

You are probably confused about the y because you don't know which equation to put it in.

You will get the same y if you put the x co-ordinate of the circle centre into either perpendicular bisector....

The circle centre x co-ordinate is 46.7

Therefore $11y=4(46.71)+40=186.84+40=226.84$

so $y=\frac{226.84}{11}=20.62$

Or $6y=70(46.71)-3146=3269.7-3146=123.7$

so $y=\frac{123.7}{6}=20.6$

**donut166** · January 2nd 2010, 06:43 PM

ok thanks

**donut166** · January 2nd 2010, 06:46 PM

no wait 20.6-24 equals 3.4
while 20.6-32= -11.3
O_O

**Grandad** · January 2nd 2010, 10:38 PM

Hello everyone

I think it's time we drew this to a close. Here's what I got for the circumcentre, continuing from the point that I and red_dog had left at.

Solving the equations simultaneously, it's not difficult, but the arithmetic is tedious:

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

$\Rightarrow x^2-164x+82^2+y^2-42y+21^2=x^2-24x+12^2+y^2-54y+27^2$

$\Rightarrow 12y-140x+6724+441-144-729=0$

$\Rightarrow 12y-140x+6292=0$

$\Rightarrow 3y-35x+1573=0$ (1)

And:

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

$\Rightarrow x^2-24x+144+y^2-54y+729=x^2-148x+5476+y^2-86y+1849$

$\Rightarrow 32y+124x-6452=0$

$\Rightarrow 8y+31x-1613=0$ (2)

Multiply (1) by $8$ and (2) by $3$ :