If A(2ca,cb),B(ca,0) and C(1+ca,1b) are three points, then
A
(AB)2+(BC)2−(CA)2=2c
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B
(AB)2+(BC)2(CA)2=c2+1(c−1)2
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C
(AB)2+(BC)2−(CA)2=2c(a2+b2)a2b2
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D
(AB)2−(BC)2−(CA)2=2b
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Solution
The correct options are B(AB)2+(BC)2(CA)2=c2+1(c−1)2 C(AB)2+(BC)2−(CA)2=2c(a2+b2)a2b2 As A≡(2ca,cb),B≡(ca,0) and C≡(1+ca,1b) AB=√(ca−2ca)2+(0−cb)2=√c2a2+c2b2BC=√(1+ca−ca)2+(1b−0)2=√1a2+1b2AC=√(1+ca−2ca)2+(1b−cb)2=√(1a−ca)2+(1b−cb)2 Hence, (AB)2+(BC)2(AC)2=c2a2+c2b2+1a2+1b2(1a−ca)2+(1b−cb)2=c2+1(c−1)2(AB)2+(BC)2−(AC)2=c2a2+c2b2+1a2+1b2−(1a−ca)2−(1b−cb)2=2c(a2+b2)a2b2