If the centroid and circumcenter of a triangle are (3,3),(6,2),then the orthocenter is?
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Let the Orthocenter be H(x,y). Let Centroid be G(3,3). Let the circumcenter be O(6,2).
Then we have 2 * OG = GH or GH = 2 * GO (to remember better)
GH^2 = 4 GO^2
so (x - 3)^2 + (y-3)^2 = 4 [ 3^2 + 1^2] = 40
x^2 + y^2 - 6x - 6 y - 22 = 0 --- (1)
The circumcenter O, centroid G and orthocenter H are always collinear. So find the slope of the Euler line :
(y-3)/(x-3) = (2-3)/(6-3) = -1/3
3 y - 9 = 3 - x
x = 3 (4 - y) --- (2)
Substituting for y in (1) we get:
y^2 - 6 y + 9 (4 - y)^2 - 18 (4 - y) - 22 = 0
y^2 - 6 y + 5 = 0
=> y = 5 or 1 => so x = -3 or 9
So Orthocenter can be : (-3, 5) or (9, 1)
There are two triangles possible.
Then we have 2 * OG = GH or GH = 2 * GO (to remember better)
GH^2 = 4 GO^2
so (x - 3)^2 + (y-3)^2 = 4 [ 3^2 + 1^2] = 40
x^2 + y^2 - 6x - 6 y - 22 = 0 --- (1)
The circumcenter O, centroid G and orthocenter H are always collinear. So find the slope of the Euler line :
(y-3)/(x-3) = (2-3)/(6-3) = -1/3
3 y - 9 = 3 - x
x = 3 (4 - y) --- (2)
Substituting for y in (1) we get:
y^2 - 6 y + 9 (4 - y)^2 - 18 (4 - y) - 22 = 0
y^2 - 6 y + 5 = 0
=> y = 5 or 1 => so x = -3 or 9
So Orthocenter can be : (-3, 5) or (9, 1)
There are two triangles possible.
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