Question

if x/b-c=y/c-a=z/a-b, prove that x+y+z=0​

Answer 1

Answer:

the answer is 0

Step-by-step explanation:

let x/b-c=y/c-a=z/a-b=k

so x=k(b-c)=kb-kc

similarly y=kc-ka

             z=ka-kb

so x+y+z=kb-kc+kc-ka+ka-kb=0

                                                   (answer)

Answer 2

To prove x + y + z = 0

Given data:

$\dfrac{x}{b-c}=\dfrac{y}{c-a}=\dfrac{z}{a-b}$

To prove:

x + y + z = 0

Step-by-step explanation:

Let, $\dfrac{x}{b-c}=\dfrac{y}{c-a}=\dfrac{z}{a-b}=k$ (say), where k is non-zero

Then x = k (b - c),

y = k (c - a) and

z = k (a - b)

Now, x + y + z

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

= kb - kc + kc - ka + ka - kb

= 0

So, x + y + z = 0 (Proved)

Note:

We can approach another way,

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

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

= k × 0

= 0

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