F.4 MATHS

(a) Mid - point of OA is M(4,0).
Location of the centroid is on MB and is 1/3 the length of MB from M.
So its x - coordinate is [2(4) + 6(1)]/3 = (8 + 6)/3 = 14/3.
Its y - coordinate is [ 2(0) + 6(1)]/3 = 2
So centroid is (14/3, 2).
(b)
The orthocenter is right below B, so x - coordinate is 6.
Let its co-ordinates be P(6, y).
PA is perpendicular to OB.
Slope of PA = (y - 0)/(6 - 8) = -y/2
Slope of OB = (6 - 0)/(6 - 0) = 1
so (-y/2)(1) = - 1
y = 2
Orthocenter is therefore ( 6,2).
(c)
The circumcenter is right above mid- point of OA, so its x - coordinate is 4.
Let it be R(4, r)
Mid -point of AB is H( 7, 3).
RH is perpendicular to AB
Slope of RH = ( r - 3)/(4 - 7) = (3 - r)/3.
Slope of AB = (6 - 0)/(6 - 8) = 6/(-2) = - 4
So (3 - r)(-4)/3 = - 1
12 - 4r = 3
r = (12 - 3)/4 = 9/4
So circumcenter is ( 4, 9/4).
(d) Radius of circumcenter = OR = sqrt [ 4^2 + (9/4)^2] = [sqrt (256 + 81)]/4
= (sqrt 337)/4.