If the vertices P and Q of a triangle PQR are | vedclass

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(D) Let the centroid of $\Delta PQR$ be $(h, k)$.Given $P = (2, 5)$ and $Q = (4, -11)$.Let $R = (x_0, y_0)$. Since $R$ lies on the line $N: 9x + 7y +

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(D) Let the centroid of $\Delta PQR$ be $(h, k)$.Given $P = (2, 5)$ and $Q = (4, -11)$.Let $R = (x_0, y_0)$. Since $R$ lies on the line $N: 9x + 7y + 4 = 0$,we have $9x_0 + 7y_0 + 4 = 0$.The centroid $(h, k)$ is given by:$h = \frac{2 + 4 + x_0}{3}$ $\Rightarrow 3h = 6 + x_0$ $\Rightarrow x_0 = 3h - 6$$k = \frac{5 - 11 + y_0}{3}$ $\Rightarrow 3k = -6 + y_0$ $\Rightarrow y_0 = 3k + 6$Substituting $x_0$ and $y_0$ into the equation of line $N$:$9(3h - 6) + 7(3k + 6) + 4 = 0$$27h - 54 + 21k + 42 + 4 = 0$$27h + 21k - 8 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $27x + 21y - 8 = 0$.This can be written as $3(9x + 7y) - 8 = 0$,or $9x + 7y - \frac{8}{3} = 0$.Since the coefficients of $x$ and $y$ are the same as in line $N$,the locus is parallel to line $N$.

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

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