There is a question:

 If (ay-bx)/p = (cx-az)/q = (bz-cy)/r, then prove that x/a = y/b = z/c.

I want to know whether the question is correct? If so what is the solution.

The question is not correct . The question is (ay - bx) / c = (cx - az) / b = (bz - cy) / a  instead of  (ay-bx)/p = (cx-az)/q = (bz-cy)/r 

Solution :-

(ay - bx) / c = (cx - az) / b = (bz - cy) / a  ----->  c * (ay - bx) / c * c  =  b * (cx - az) / b * b = a * (bz - cy) / a * a

------>  (cay - cbx) / c² = (bcx - baz) / b² = (abz - cay) / a² 

------> (cay - cbx + bcx - baz + abz - cay) / a² + b² + c²  =   0 / a² + b² +  c²  =  0 

------>  (cay - cbx) / c² = (bcx - baz) / b² = (abz - cay) / a² =  0

So , (cay - cbx) / c² = 0 ---->  cay - cbx = 0 ---->  cay = cbx  ----> ay = bx  ----> y / b = x / a _--------  (eqn 1)

and ,  (bcx - baz) / b² = 0  ---->  bcx - baz = 0  ----->  bcx = baz  ---->  cx = az -----> x / a = z / c  _--------  (eqn 2)

From 1 and 2 , we get ,

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

Hi Satyajit, Thanks for the solution. Can you please let me know why  ------>  (cay - cbx) / c² = (bcx - baz) / b² = (abz - cay) / a² =  0

(cay - cbx) / c ²  = (bcx - baz) / b ² = (abz - cay) / a ²  =  0 because :):):):)

(cay - cbx) / c ²  = (bcx - baz) / b ² = (abz - cay) / a ²   and on solving it , we get  :-

(cay - cbx + bcx - baz + abz - cay) / a² + b² + c²  =   0 / a² + b² +  c²  =  0 .

Hence (cay - cbx) / c ²  = (bcx - baz) / b ² = (abz - cay) / a ²   =  0 .

Hope it helps !!!!!!  Satyajit

i have a question :-  Sir , we have been said in school to not use this method for solving because it is not acceptable in ICSE .

Then how do we do this ?

 Satyajit, you have used the method: If a/b=c/d=e/f, then each ratio = sum of anticedents/sum of consequence.

Whereas the above solution has been done by the method by Sir is called 'k method'. This method is very much acceptable. (Please see Concise Mathematics by Selina Publishers). 

Prakashik