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If the vertices \mathrm{P} and \mathrm{Q} of a triangle \mathrm{PQR} are given by \mathrm{(2,5)\, and \, (4,-11)}  respectively, and the point R moves along the line \mathrm{\mathrm{N}: 9 \mathrm{x}+7 \mathrm{y}+4=0}, then the locus of the centroid of the triangle \mathrm{\mathrm{PQR}}is a straight line parallel to

Option: 1

\mathrm{PQ}


Option: 2

\mathrm{QR}


Option: 3

\mathrm{RP}


Option: 4

\mathrm{N}


Answers (1)

best_answer

\mathrm{R}(\mathrm{x}, \mathrm{y})$ lies on $9 \mathrm{x}+7 \mathrm{y}+4=0

\mathrm{\Rightarrow \mathrm{R}\left(\mathrm{a}, \frac{(4+9 \mathrm{a})}{7}\right), centroid\: of\: \Delta \mathrm{PQR}=(\mathrm{h}, \mathrm{k})}
\mathrm{\mathrm{h}=\left(\frac{2+4+\mathrm{a}}{3}\right)=\frac{6+\mathrm{a}}{3}\quad \cdots (1)}

\mathrm{\mathrm{k}=\frac{5-11-\frac{(4+9 \mathrm{a})}{7}}{3}=\frac{-46-9 \mathrm{a}}{7 \times 3}\quad \cdots(2)}

from (1) & (2) we get

\mathrm{equating\: x\: \: 3 h-6=\frac{-(21 k-46)}{9} \Rightarrow 27 h+21 k-54+46=0}
\mathrm{or\: locus \: is \: 9 \mathrm{x}+7 \mathrm{y}-8 / 3=0}

\mathrm{\text{this line is }\amalg to \mathrm{N}}

Posted by

jitender.kumar

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