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JIPMAT 2022

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:

\newline

a = k(b+c)

...(1)

\newline

b = k(c+a)

...(2)

\newline

c = k(a+b)

...(3)

From (1):

\frac{a}{k} = b + c

\newline

From (2):

\frac{b}{k} = c + a

\newline

From (3):

\frac{c}{k} = a + b

Adding all equations:

\newline

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

Therefore:

\newline

\frac{1}{k} = 2

k = \frac{1}{2}

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

a-b = k(b-a)

\newline

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

\newline

1=-k

\newline

k=-1

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