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

AskiitiansExpert-IITD

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..