Question

hello everybody ,
solve this question :
$\large\tt{if\:\frac{a+b}{xa+yb}=\frac{b+c}{xb+yc} =\frac{c+a}{xc+ya} \:where, x+y\neq 0, \: and\: a+b+c\neq 0}$
then find the vlaue of each of the ratios

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atlas 99, taken name, your helper adi

Answer 1

To Find :

the value of each of the ratios given

Solution :

let, each ratio be equal to 'k', so :

$\tt{\frac{a+b}{xa+yb} =\frac{b+c}{xb+yc} = \frac{c+a}{xc+ya} = k }$

$\implies \tt{\frac{a+b}{xa+yb} = k$

$\tt{\implies a+b = kxa+kyb ---eq1}$

$\implies \tt{\frac{b+c}{xb+yc} = k}$

$\implies \tt{b+c = kxb+kyc----eq2}$

$\tt{\implies \frac{c+a}{xc+ya} = k }$

$\tt{\implies c+a = kxc+kya-----eq3}$

now, adding the (eq1), (eq2) and (eq3), we get :

$\tt{\implies eq1+eq2+eq3}$

$\tt{\implies (a+b) +(b+c)+(c+a)=(kxa+kyb)+(kxb+kyc)+(kxc+kya)}$

$\tt{\implies 2a+2b+2c = k(xa+yb+xb+yc+xc+ya)}$

$\tt{\implies 2(a+b+c) = k[x(a+b+c)+y(a+b+c)]}$

$\tt{\implies 2(a+b+c) = k(x+y)(a+b+c)}$

$\large\tt{\implies k = \frac{2(a+b+c)}{(x+y)(a+b+c)}}$

$\tt{\implies k = \frac{2}{x+y}}$

hence, value of each ratio = k

or, value of each ratio =2/x+y

thanks for believing in me :]

Answer 2

☆ To Find :-

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• The value of each ratio

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☆ Solution :-

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Let each ratio be "r". Therefore :-

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$\boxed{ \frac{a + b}{xa + yb} = \frac{b + c}{xb + yc} = \frac{c + a}{xc + ya} = r}$

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$: \implies \: \frac{a + b}{xa + yb} = r \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \: a + b = r(xa + yb) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \: a + b = rxa + ryb... (eq. 1)$

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$: \implies \: \frac{b + c}{xb + yc} = r \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \: b + c= r(xb + yc) \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \: b + c = rxb + ryc..( eq. 2)$

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$: \implies \: \frac{c + a}{xc + ya} = r \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \: c + a= r(xc + ya) \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \: c + a = rxc + rya \:.. ( eq. 3)$

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Now, on adding these equations, we get,

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$\sf: \implies \: (a + b) + (b + c) + (c + a) = (rxa + ryb) + (rxb + ryc) + (rxc + rya) \\\sf : \implies \:2a + 2b + 2c = r[(xa + yb) + (xb + yc) + (xc + ya)] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies \: 2(a + b + c) = r (x + y) ( a + b + c ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\sf: \implies \: r = \frac{2 \cancel{(a + b + c)}}{(x + y) \cancel{(a + b + c)}} = \frac{2}{x + y} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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$\color{yellow} \bigstar \: \: \color{orange}{\boxed{r = \frac{2}{x + y}}}$

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Therefore, each ratio or "r" is $\frac{2}{x + y}$.

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I hope that helps! :)