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

Option: 1

PQ


Option: 2

QR


Option: 3

RP


Option: 4

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 \ldots{(1)}
\mathrm{\mathrm{k}=\frac{5-11-\frac{(4+9 \mathrm{a})}{7}}{3}=\frac{-46-9 \mathrm{a}}{7 \times 3}\quad \ldots(2)}

from (1) & (2) we get

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

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chirag

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