If a/(b+c)=b/(c+a)=c/(a+b)=k then value of k is Option 1: ± 1/2 Option 2: 1/2 or -1 Option 3: -1 Option 4: 1/2

Answer: 1/2 or -1. Solution: Given: a/(b+c) = b/(c+a) = c/(a+b) = k Let's say: a = k(b+c) ...(1) b = k(c+a) ...(2) c = k(a+b) ...(3) From (1): a/k = b + c From (2): b/k = c + a From (3): c/k = a + b Adding all equations: (a+b+c)/k = 2(a+b+c) Therefore: 1/k = 2 k = 1/2 If we subtract (2) from (1), we get : a-b = k(b-a) (a-b)=-k(a-b) 1=-k k=-1

Algebra

Linear Equation

Hard

If $\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=k$ then value of $k$ is

$\pm \frac{1}{2}$

$\frac{1}{2}$ or $-1$

$-1$

$\frac{1}{2}$

✅ Correct Option: 2

Given: $\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = k$

Let's say:

$a = k(b+c)$ ...(1)

$b = k(c+a)$ ...(2)

$c = k(a+b)$ ...(3)

From (1): $\frac{a}{k} = b + c$

From (2): $\frac{b}{k} = c + a$

From (3): $\frac{c}{k} = a + b$

Adding all equations:

$\frac{a+b+c}{k} = 2(a+b+c)$

Therefore:

$\frac{1}{k} = 2$

$k = \frac{1}{2}$

If we subtract (2) from (1), we get :

$a-b = k(b-a)$

$(a-b)=-k(a-b)$

$1=-k$

$k=-1$

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