I got it, thanks Archie ( I round to the nearest thousandth)
Thanks a lot for your help.
You use the perpendicular slopes again.
Now, instead of using side midpoints, you use the co-ordinates of the opposite vertex instead (the point that is not on the triangle side).
Again, you only need 2 perpendiculars (altitudes).
The perpendiculars may not even be touching the triangle side if the triangle has an obtuse angle.
They will be parallel to the bisectors in the circumcentre case.
It's my naptime so i'll check tomorrow to see how you're getting on.
recheck your equations for the 2 perpendicular lines (altitudes)
that contain the opposite vertex.
At least one is wrong.
You say you have no idea how to continue with the algebra,
at least you are honest but it means you should study why the line equations
are needed, how you write them and how you find the point of intersection.
I calculate (-47.67, 350.2) for the orthocentre.
I will post the calculations in fraction form,
but you should practice until you are very clear about your algebraic calculations and how they relate to the geometry.
The 3 equations of the altitudes are...
Only 2 are required to find the orthocentre.
The x and y values of these equations are the co-ordinates of points on the line. If we let x be the same on all 3 lines and also let y be the same on all 3 lines, then we are referring to the orthocentre, the common point of intersection of the 3 altitudes.
Using the following 2 lines...
=orthocentre x co-ordinate.
Place this co-ordinate into any of the 3 altitudes to find the orthocentre y co-ordinate.
All 3 should give the exact same y.