Given three points P, Q, R with P(5, 3) and R | vedclass

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(D) The equation of line $RQ$ is $x - 2y = 2$. Since $R$ lies on the $x-$ axis,its $y-$ coordinate is $0$. Substituting $y = 0$ in the equation of $RQ

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$2x - 5y = 0$

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(D) The equation of line $RQ$ is $x - 2y = 2$. Since $R$ lies on the $x-$ axis,its $y-$ coordinate is $0$. Substituting $y = 0$ in the equation of $RQ$,we get $x - 2(0) = 2$,so $x = 2$. Thus,$R = (2, 0)$. Since $PQ$ is parallel to the $x-$ axis,the $y-$ coordinate of $Q$ is the same as that of $P$,which is $3$. Substituting $y = 3$ in the equation of $RQ$,we get $x - 2(3) = 2$,which implies $x - 6 = 2$,so $x = 8$. Thus,$Q = (8, 3)$. The centroid $G$ of $\Delta PQR$ with vertices $P(5, 3)$,$Q(8, 3)$,and $R(2, 0)$ is given by: $G = \left( \frac{5 + 8 + 2}{3}, \frac{3 + 3 + 0}{3} \right) = \left( \frac{15}{3}, \frac{6}{3} \right) = (5, 2)$. Now,we check which line equation is satisfied by the point $(5, 2)$: For option $D$: $2x - 5y = 2(5) - 5(2) = 10 - 10 = 0$. Thus,the centroid lies on the line $2x - 5y = 0$.

Given three points $P, Q, R$ with $P(5, 3)$ and $R$ lies on the $x-$ axis. If the equation of $RQ$ is $x - 2y = 2$ and $PQ$ is parallel to the $x-$ axis,then the centroid of $\Delta PQR$ lies on the line

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