(a) The point B on AB,2x+y=0 may be chosen as (α,−2α) and the point C on AC;x−y=3 be taken as (β,β−3).
If P(2,3) is centroid, i.e. (13Σx1,13Σy1) then
2=1+α+β,3=−2−2α+β−33
∴α+β=5,−2α+β=14
∴α=−4,β=8
∴B is (−3,6) and C is (8,5).
Both these points lie on BC,x+py=q
∴−3+6p=q and 8+5p=q
Solving p=11,q=63
∴p+q=74 ∴(a)⟹(iv)
(b) If P(2,3) is orthocentre, then
CP⊥BC and BP⊥CA . Apply m1m2=−1
∴2β−12=β−2⟹β=10,2α+3=α−2
⟹α=−5
∴B is (−5,10) and C is (10,7). Both these points lie on BC.
∴−5+10p=q and 10+7p=q
∴p=5,q=45⟹p+q=50
∴(b)⟹(ii)
(c) If P(2,3) is circumcentre, then PA=PB=PC
26=(α−2)2+(2α+3)2=(β−2)2+(β−6)2
(5α+13)(α−1)=0⟹α=−135 as α=1 will give (1,−2)
Which is point A not B. Similarly y(β−7)(β−1)=0⟹β=7 as β=1 will give (1,−2) which is point A not C.
∴B is (−135,265) and C is (7,4).
Both these points lie on x+py=q
∴−135+265p=q and 7+4p=q
Solving p=8,q=39 ∴p+q=47
∴(c)⟹(i)