What are the formulae for finding the orthocentre and circumcentre in co-ordinate geometry?

Question

What are the formulae for finding the orthocentre and circumcentre in co-ordinate geometry?

Answer 1

it looks cooler, and it is correct to write co-ordinate as coördinate

also in coöperate

anytime both o´s are pronounced when written in succession.

Answer 2

I'll tell you the formula I've worked out for the circumcentre, but in my opinion it's easier just to find the equation of the perpendicular bisectors of two sides (mid point formula, gradient formula inverted with sign changed, then use point gradient formula), and solve these simultaneous equations.  That's what I did using points (x1, y1), (x2, y2), (x3, y3).

Calculate the determinants

D = | x1 .. x2 .. x3 |

,.,,, | y1 .. y2 .. y3 |

..... | 1 ..... 1 .... 1 |

D1 = same as above, except that the x values are squared

D2 = same as above, except that the y values are squared

Then the circumcentre is given by

x = [D1 - (y1 - y2)(y2 - y3)(y3 - y1)] / 2D

y = [D2 - (x1 - x2)(x2 - x3)(x3 - x1)] / 2D

For the orthocentre, it's not very hard to find the equation of the line through A perpendicular to BC, the line through B perpendicular to AC, and solve these simultaneous equations.  I haven't yet worked out a formula for the result of that.  Will edit this if I do.  Email me if you'd like an example.

OK, I have a formula for the orthocentre.  Calculate the determinants

D3 = | x2x3 .. x3x1 .. x1x2 |

....... | .. y1 ..... y2 ........y3 |

....... | ..1 ......... 1 ........ 1 .|

D4 the same but with y2y3 etc in the top row.  Then

x = -(D3 + D4)/D

Now calculate D5 and D6, the same as these but with x replacing y in the middle row.  Then

y = (D5 + D6)/D

I may have made a slip somewhere here -- better to use the method I outlined above!