Question

Help me find the circumcenter of the triangle with the coordinates:
B(2,2)
P(4,10)
H(12,2)

Answer 1

@ganeshie8

Answer 2

(G+H)/2 = (1+6, 0+0)/2 = (7/2, 0)
Because it's along the x-axis, (slope, m=0) its perpendiculars all have equations, x = c, which are all parallel to the y-axis (and with slope = -1/m = ∞). Here, it's
x = 7/2

Taking GJ next, its midpoint is
(G+J)/2 = (1+3, 0-6)/2 = (2, -3)
Slope of GJ is given by m = (-6-0)/(3-1) = -6/2 = -3
Slope of the perp is a = -1/m = 1/3
So we're looking for a line of slope = 1/3 passing through (2, -3):
y = x/3 + b --> -3 = 2/3 + b
b = -11/3
y = x/3 - 11/3

Together with our first equation, x = 7/2, gives
y = 7/6 - 11/3 = -15/6 = -5/2
So (x,y) = (7/2, -5/2)

The third P.B., of HJ, its midpoint is
(H+J)/2 = (9/2, -3)
Slope of HJ = (0-(-6))/(6-3) = 6/3 = 2
Slope of P.B. of HJ = -1/2; P.B.passes through (9/2, -3):
y = -x/2 + b
-3 = -9/4 + b; b = -3/4
y = -x/2 - 3/4

To check that the point we already found lies on this line, we just plug the point's coordinates in:
-5/2 =? -(7/2)/2 - 3/4 = -7/4 - 3/4 = /10/4 = -5/2
Success!

Answer 3

My triangle doesn't have a G.. I'm confused @zobobozo

Answer 4

you're here @annipuppi ?

Answer 5

Now I am @ganeshie8

Answer 6

hey

Answer 7

draw it out

Answer 8

Yeah I've done that but I don't get it still.