Question

given
a/q-r= b/r-p=y/p-q
prove
a+b+y=0
pa+qb+ry=0​

Answer 1

ANSWER:

Given:

• a/(q-r) = b(r-p) = y(p-q)

To Prove:

• a + b + y = 0
• pa + qb + ry = 0

Solution:

We are given that,

$\implies\dfrac{a}{q-r}=\dfrac{b}{r-p}=\dfrac{y}{p-q}$

Let us assume, that these all are equal to a constant k. So,

$\implies\dfrac{a}{q-r}=\dfrac{b}{r-p}=\dfrac{y}{p-q}=k$

So,

$\implies\dfrac{a}{q-r}=k$

$\implies a=k(q-r)- - - - (1)$

$\implies\dfrac{b}{r-p}=k$

$\implies b=k(r-p)- - - - (2)$

$\implies\dfrac{y}{p-q}=k$

$\implies y=k(p-q)- - - - (3)$

Now, for the first proof, we will add (1), (2) and (3).

So,

$\implies a+b+y=k(q-r)+k(r-p)+k(p-q)$

$\implies a+b+y=k(q-r+r-p+p-q)$

$\implies a+b+y=k[(q\!\!\!/-q\!\!\!/)+(r\!\!\!/-r\!\!\!/)+(p\!\!\!/-p\!\!\!/)]$

$\implies a+b+y=k(0)$

$\implies\bf a+b+y=0$

HENCE PROVED!!

Now, for the second proof, we'll multiply (1) with p, (2) with q & (3) with r.

So,

$\hookrightarrow pa=pk(q-r)$

$\implies pa=k(pq-pr)- - - - (4)$

$\hookrightarrow qb=qk(r-p)$

$\implies qb=k(qr-qp)- - - - (5)$

$\hookrightarrow ry=rk(p-q)$

$\implies ry=k(rp-rq)- - - - (6)$

Adding (4), (5) and (6),

$\implies pa+qb+ry=k(pq-pr)+k(qr-qp)+k(rp-rq)$

$\implies pa+qb+ry=k(pq-pr+qr-qp+rp-rq)$

$\implies pa+qb+ry=k(pq-pr+qr-pq+pr-qr)$

$\implies pa+qb+ry=k[(pq\!\!\!\!\!/\:-pq\!\!\!\!\!/\:)+(qr\!\!\!\!\!/\:-qr\!\!\!\!\!/\:)+(rp\!\!\!\!\!/\:-rp\!\!\!\!\!/\:)]$

$\implies pa+qb+ry=k(0)$

$\implies\bf pa+qb+ry=0$

HENCE PROVED!!