IIT JEE , AIEEE Forum - Mathematics , Algebra

sharmadon

if (a-b)² + (b-c)²+(c-a)² = (a + b - 2c)²+ (a + c - 2b)²+ (c+ b - 2a)² with a,b,c real and non zero then if a/b = k, what will b/c be?

1. 2k

2. k + k²

3. k² - k

4. 1

computer001

if u havnt noticed yet chk out ans 4 ur probability ques...

sandeepramesh

solve and get a = b = c and hence k=1 and so b/c = k

hsbhatt

(a-b)² + (c-b)² = (a+c-2b)² + 2(a-b)(b-c) etc.

Hence 2

(a+c-2b)² + 2(a-b)(b-c) =

(a+c-2b)²

or

(a+c-2b)²+ 2(a-b)(b-c) = 0

or 2

(a-b)²= 0

Hence a=b=c

sandeepramesh

thank you sir for verifying my soln, i had a doubt abt that bcos he didnt specify a value for k n treated it as generic