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

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(C) $1$. Given $P = (5, 3)$. Since $PQ$ is parallel to the $x-$ axis,the $y-$ coordinate of $Q$ is the same as $P$,so $Q = (x_Q, 3)$.$2$. $Q$ lies on

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$2x - 5y = 0$

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(C) $1$. Given $P = (5, 3)$. Since $PQ$ is parallel to the $x-$ axis,the $y-$ coordinate of $Q$ is the same as $P$,so $Q = (x_Q, 3)$.$2$. $Q$ lies on the line $RQ: x - 2y = 2$. Substituting $y = 3$,we get $x - 2(3) = 2$,which implies $x - 6 = 2$,so $x = 8$. Thus,$Q = (8, 3)$.$3$. $R$ lies on the $x-$ axis,so its $y-$ coordinate is $0$. Let $R = (x_R, 0)$. Since $R$ lies on $x - 2y = 2$,we have $x_R - 2(0) = 2$,so $x_R = 2$. Thus,$R = (2, 0)$.$4$. The centroid $G$ of $\Delta PQR$ is given by $(\frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3}) = (\frac{5 + 8 + 2}{3}, \frac{3 + 3 + 0}{3}) = (5, 2)$.$5$. We check which line equation is satisfied by $(5, 2)$:- For $A: 5 - 2(2) + 1 = 5 - 4 + 1 = 2 
eq 0$.- For $B: 2(5) + 2 - 9 = 10 + 2 - 9 = 3 
eq 0$.- For $C: 2(5) - 5(2) = 10 - 10 = 0$. This is satisfied.- For $D: 5(5) - 2(2) = 25 - 4 = 21 
eq 0$.$6$. Therefore,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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