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Question

For what value of PQ does A (a,a), B (-a,-a) and C (Pa,Qa) form the vertices of an equilateral triangle?


A

-1

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B

1

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C

2

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D

-2

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Solution

The correct option is A

-1


Let A = (-a,-a), B = (a,a) and C = (Pa,Qa)
Since ABC is an equilateral triangle
AB = BC
AB2=BC2
(a(a))2+(a(a))2=(Paa)2+(Qaa)2
(2a)2+(2a)2=a2(P22P+1+Q22Q+1)
8a2=a2(P2+Q22P2Q+2)
8=(P2+Q22P2Q+2)
P2+Q22(P+Q)=6 ---------- (1)
AB = AC
AB2=AC2
(a(a))2+(a(a))2=(Pa(a))2+(Qa(a))2
(2a)2+(2a)2=a2(P2+2P+1+Q2+2Q+1)
8a2=a2(P2+Q2+2P+2Q+2)
8=(P2+Q2+2P+2Q+2)
P2+Q2+2(P+Q)=6 ---------- (2)
Subtracting (2) from (1), we get
4(P+Q)=0
P+Q=0
P=-Q
Or, PQ=1


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