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(D) Let $m$ be the slope of the line $PQ$. The equation of the line is $y - 2 = m(x - 1)$.The line meets the $X$-axis at $P$ where $y=0$,so $0 - 2 = m

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$-2$

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(D) Let $m$ be the slope of the line $PQ$. The equation of the line is $y - 2 = m(x - 1)$.The line meets the $X$-axis at $P$ where $y=0$,so $0 - 2 = m(x - 1) \implies x - 1 = -\frac{2}{m} \implies x = 1 - \frac{2}{m}$. Thus,$P = (1 - \frac{2}{m}, 0)$.The line meets the $Y$-axis at $Q$ where $x=0$,so $y - 2 = m(0 - 1) \implies y = 2 - m$. Thus,$Q = (0, 2 - m)$.The area $A$ of $	riangle OPQ = \frac{1}{2} |OP| |OQ| = \frac{1}{2} |(1 - \frac{2}{m})(2 - m)|$.Since the line must form a triangle in the first quadrant for a minimum area,$1 - \frac{2}{m}  0$,implying $m $A = \frac{1}{2} |2 - m - \frac{4}{m} + 2| = \frac{1}{2} |4 - (m + \frac{4}{m})|$.Since $m  0$. Then $A = \frac{1}{2} |4 - (-k - \frac{4}{k})| = \frac{1}{2} (4 + k + \frac{4}{k})$.To minimize $A$,we minimize $k + \frac{4}{k}$. By $AM$-$GM$ inequality,$k + \frac{4}{k} \geq 2\sqrt{k \cdot \frac{4}{k}} = 4$.Equality holds when $k = \frac{4}{k} \implies k^2 = 4 \implies k = 2$.Since $m = -k$,the slope $m = -2$.

$A$ line passing through the point $(1, 2)$ meets the axes at $P$ and $Q$ such that it forms a triangle $OPQ$,where $O$ is the origin. If the area of triangle $OPQ$ is minimum,then the slope of the line $PQ$ is:

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