Question : If $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$, find the value of $(pa+qb+rc)$.
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
Correct Answer: 0
Solution :
Given: $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$
Let $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}=k$
$\frac{a}{q-r}=k$
Multiplying by $p$, we get:
$pa=k(pq-pr)$ -------------------(1)
Similarly, $qb=k(qr-qp)$ ------------------(2)
and $rc=k(rp-rq)$ --------------------------(3)
Adding equations (1), (2), and (3), we get,
$⇒pa+qb+rc=k(pq-pr+qr-qp+rp-rq)$
$\therefore pa+qb+rc=0$
Hence, the correct answer is 0.
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