Please solve q 33?

Question

1151 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T17:33:10

The answer is $"option (b)"$

Let

$x/(b+c-a)=k$

$y/(c+a-b)=k$

$z/(a+b-c)=k$

Therefore,

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

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

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

So,

$x(b-c)+y(c-a)+z(a-b)$

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

$=k(b^2-c^2-a(b-c)+c^2-a^2-b(c-a)+a^2-b^2-c(a-b))$

$=0$

The answer is $"option (b)"$

$"Question (31)"$

Let

$a/(b+c)=k$ , $=>$ , $a=(b+c)k$

$b/(c+a)=k$ , $=>$ , $b=(a+c)k$

$c/(a+b)=k$ , $=>$ , $c=(a+b)k$

Therefore,

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

$(a+b+c)=(bk+ck)+(ak+ck)+(ak+bk)$

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

$k=1/2$

Similarly if $k'=1/k$

$k'=2$

The answer is $"option (d)"$