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If x/b+c=y/c+a=z/a+b then show (b-c) x+(c-a) y+(a-b) z=0

If x/b+c=y/c+a=z/a+b then show (b-c) x+(c-a) y+(a-b) z=0

Grade:12th pass

3 Answers

Arun
25750 Points
6 years ago
Dear Aiswarya
 
I think you have done a midtake in the given condition.
 

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

There are three fractions.

Use the following properties of fractions.

1. We know that if both numerator and denominator of a fraction p/q are multiplied by the same term, it remains the same

p/q = px / qx = py / qy = pz/qz

-

2. a/b = c/d = e/f

Then adding all the numerators and denominators, the fraction doesn't change.

a/b = c/d = e/f = (a + c + e) / (b + d + f)

-

Using above two properties together

Multiply first fraction by (b - c)/(b - c), second by (c -a)/(c - a), third by (a - b)/(a - b) then adding them

Numerator = (b - c)x + (c -a)y + (a - b)z

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

Denominator = b2 - c2 - a(b - c) + c2 - a2- b(c - a) + a2 - b2 - c(a - b)

Denominator = - a(b - c)- b(c - a) - c(a - b)

Denominator = - ab + ac - bc + ab - ca +bc

Denominator = 0

Therefore

x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = [(b - c)x + (c -a)y + (a - b)z ] / 0 = k

(b - c)x + (c -a)y + (a - b)z = 0

Answer is zero

 

Regards

Arun (askIITians forum expert)

Hemant Patel
11 Points
6 years ago
Let x/b+c=y/c+a=z/a+b = k
then
x=(b+c)k        ...(1)
y=(c+a)k        ...(2)
z=(a+b)k        ...(3)
to prove
(b-c) x+(c-a) y+(a-b) z=0
L.H.S. = (b-c) x+(c-a) y+(a-b) z
= (b-c) (b+c) k + (c-a) (c+a)k + (a-b) (a+b)k                  by eq (1), (2) & (3)
= k [(b-c) (b+c)  + (c-a) (c+a) + (a-b) (a+b) ]
=k[b^2 – c^2 + c^2 – a^2 + a^2 – b^2]
= k [0]
= 0
=R.H.S.
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Here are three fractions.
Use the following properties of fractions.
1. We know that if both numerator and denominator of a fraction p/q are multiplied by the same term, it remains the same
p/q = px / qx = py / qy = pz/qz

2. a/b = c/d = e/f
Then adding all the numerators and denominators, the fraction doesn't change.
a/b = c/d = e/f = (a + c + e) / (b + d + f)
Using above two properties together
Multiply first fraction by(b - c)/(b - c), second by(c -a)/(c - a), third by(a - b)/(a - b)then adding them
Numerator = (b - c)x + (c -a)y + (a - b)z
Denominator = (b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)
Denominator = b2- c2- a(b - c) + c2- a2- b(c - a) + a2- b2- c(a - b)
Denominator = - a(b - c)- b(c - a) - c(a - b)
Denominator = - ab + ac - bc + ab - ca +bc
Denominator = 0
Therefore
x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = [(b - c)x + (c -a)y + (a - b)z ] / 0 = k
(b - c)x + (c -a)y + (a - b)z = 0
Answer is zero
Thanks and Regards

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