 # May i know what is the formula for orthocentre and circumcentre? AKASH GOYAL AskiitiansExpert-IITD
420 Points
12 years ago

Dear Apurva

Although 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  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

All the best.

AKASH GOYAL

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