P and Q are points on the line joining A,(-2, | vedclass

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(A) Given that $P$ and $Q$ divide the line segment $AB$ into three equal parts such that $AP = PQ = QB$.This implies that $P$ and $Q$ are the points o

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$\left( \frac{1}{2}, 3 \right)$

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(A) Given that $P$ and $Q$ divide the line segment $AB$ into three equal parts such that $AP = PQ = QB$.This implies that $P$ and $Q$ are the points of trisection of the line segment $AB$.The mid-point of $PQ$ is the same as the mid-point of $AB$ because $P$ and $Q$ are symmetric with respect to the center of the segment $AB$.The coordinates of the mid-point of $AB$ are given by the formula $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.Substituting the coordinates $A(-2, 5)$ and $B(3, 1)$:Mid-point $= \left( \frac{-2 + 3}{2}, \frac{5 + 1}{2} \right) = \left( \frac{1}{2}, \frac{6}{2} \right) = \left( \frac{1}{2}, 3 \right)$.

$P$ and $Q$ are points on the line joining $A,(-2, 5)$ and $B,(3, 1)$ such that $AP = PQ = QB$. Then the mid-point of $PQ$ is

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