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Question

Area of triangle whose vertices are (a,a2),(b,b2),(c,c2) is 1, and area of another triangle whose vertices are (p,p2),(q,q2) and (r,r2) is 4 then the value of ∣ ∣ ∣(1+ap)2(1+bp)2(1+cp)2(1+aq)2(1+bq)2(1+cq)2(1+ar)2(1+br)2(1+cr)2∣ ∣ ∣ is


Your Answer
A
4
Your Answer
B
8
Your Answer
C
16
Correct Answer
D
32

Solution

The correct option is D 32
Let
Δ=∣ ∣ ∣(1+ap)2(1+bp)2(1+cp)2(1+aq)2(1+bq)2(1+cq)2(1+ar)2(1+br)2(1+cr)2∣ ∣ ∣

Δ=∣ ∣ ∣(1+2ap+a2p2)(1+2bp+b2p2)(1+2cp+c2p2)(1+2aq+a2q2)(1+2bq+b2q2)(1+2cq+c2q2)(1+2ar+a2r2)(1+2br+b2r2)(1+2cr+c2r2)∣ ∣ ∣

Then,

Δ=∣ ∣ ∣12aa212bb212cc2∣ ∣ ∣∣ ∣111pqrp2q2r2∣ ∣
=2×(2Δ1)×(2Δ2)=2×(2×1)×(2×4)=32

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