Question

Z 1. If Х у then b+c-a C+a-b a+b- cb -C - - (b - c)X + (c - a)y + (a - b)z is ​

Answer 1

Answer:

0

Step-by-step explanation:

let x/(b+c-a) = y/(c+a-b) = z/(a+b-c) = k

therefore, x = k(b+c-a)

y = k(c+a-b)

z = k(a+b-c)

now, (b-c)x + (c-a)y + (a-b)z

= (b-c)k(b+c-a) + (c-a)k(c+a-b) +(a-b)k(a+b-c)

= k[(b-c)(b+c-a) + (c-a)(c+a-b) +(a-b)(a+b-c)]

= k (b² -bc +bc -c² -ab +ac +c² -ac +ac -a² -bc +ab +a² -ab +ab -b² -ac +bc)

= k × 0

=0