Question

If (x+y)/(3a-b)=(y+z)/(3b-c)=(z+x)/(3c-a) then prove that (x+y+z)/(a+b+c)=(ax+by+cz)/(a^2+b^2+c^2).

Answer 1

Please see the attachment

Answer 2

Answer:

Step-by-step explanation:

( x + y ) / ( 3a - b ) .........( 1 )

( y + z ) / ( 3b - c )  .........( 2 )

( z + x ) / ( 3c - a ) .........( 3 )

Add equation 1, 2 and 3

( 2x + 2y + 2z ) / (3a - b + 3b - c + 3c - a )

=  ( 2x + 2y + 2z ) / ( 2a + 2b + 2c )

= [ 2 ( x + y + z ) ] / [ 2 ( a + b + c )

= ( x+ y + z ) / ( a + b +c )

= ( x / a ) = ( y / b ) = ( z / c )

⇒ ax / a²  =  by / b² = cz / c²

= ( ax + by + cz ) / ( a² + b² + c² )

It proves ( x+ y + z ) / ( a + b +c ) = ( ax + by + cz ) / ( a² + b² + c² )