Let P(2, 3),Q(6, 0),and R(alpha, beta) be thr | vedclass

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(A) The expression $|PR + QR| + |PR - QR|$ represents the sum of the distances from $R$ to $P$ and $Q$ plus the absolute difference of those distances

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(A) The expression $|PR + QR| + |PR - QR|$ represents the sum of the distances from $R$ to $P$ and $Q$ plus the absolute difference of those distances.By the triangle inequality,$|PR + QR| \ge |PQ|$.Also,$|PR - QR| \ge 0$,and the minimum value is $0$ when $R$ lies on the line segment $PQ$.Thus,the minimum value of the entire expression is $|PQ|$,which occurs when $R$ lies on the line segment $PQ$.However,for the expression to be minimized specifically as a function of $R$,$R$ must lie on the line segment $PQ$.Given the standard interpretation of such geometric minimization problems,$R$ is the midpoint of $PQ$.$R = \left( \frac{2+6}{2}, \frac{3+0}{2} \right) = (4, 1.5)$.So,$\alpha = 4$ and $\beta = 1.5$.Then,$\alpha - 2\beta = 4 - 2(1.5) = 4 - 3 = 1$.

Let $P(2, 3)$,$Q(6, 0)$,and $R(\alpha, \beta)$ be three points in the $x-y$ plane such that $|PR + QR| + |PR - QR|$ is minimum. Then the value of $(\alpha - 2\beta)$ is:

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