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Question

In the adjoining figure, the sides AB,BC and CA of the triangle ABC are 2x+y=0,x+py=q and xy=3 respectively. The point P inside the triangle is (2,3), then match the entries of Column I and Column II.
Column I - P(2,3)
Column II - Value of p+q

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Solution

(a) The point B on AB,2x+y=0 may be chosen as (α,2α) and the point C on AC;xy=3 be taken as (β,β3).
If P(2,3) is centroid, i.e. (13Σx1,13Σy1) then
2=1+α+β,3=22α+β33
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
CPBC and BPCA . Apply m1m2=1
2β12=β2β=10,2α+3=α2
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=45p+q=50
(b)(ii)
(c) If P(2,3) is circumcentre, then PA=PB=PC
26=(α2)2+(2α+3)2=(β2)2+(β6)2
(5α+13)(α1)=0α=135 as α=1 will give (1,2)
Which is point A not B. Similarly y(β7)(β1)=0β=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)

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