A) \[p+q+r=pqr\]
B) \[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}=pr+rq\]
C) \[{{p}^{3}}+{{q}^{3}}+{{r}^{3}}=3pqr\]
D) None of these
Correct Answer: C
Solution :
[c] \[\therefore px+qy+r=0\] \[qx+ry+p=0\] \[\And rx+py+q=0\]be concurrent. \[\Rightarrow \left( p+q+r \right)\left\{ \left( r-p \right)\left( q-r \right)-{{\left( p-q \right)}^{2}} \right\}=0\] \[\Rightarrow \left( p+q+r \right)\left\{ rq-pq-{{r}^{2}}+pr-{{p}^{2}}-{{q}^{2}}+2pq \right\}=0\]\[\Rightarrow (p+q+r)({{p}^{2}}+{{q}^{2}}-{{r}^{2}}-pq-qr-rp)=0\] \[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}=3pqr\]You need to login to perform this action.
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