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