A line is drawn through the point (1, 2) to m | vedclass

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(A) The equation of line $PQ$ passing through $(1, 2)$ is $y - 2 = m(x - 1)$.The $x$-intercept (Point $P$) is $(1 - \frac{2}{m}, 0)$ and the $y$-inter

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(A) The equation of line $PQ$ passing through $(1, 2)$ is $y - 2 = m(x - 1)$.The $x$-intercept (Point $P$) is $(1 - \frac{2}{m}, 0)$ and the $y$-intercept (Point $Q$) is $(0, 2 - m)$.Since the area must be positive,we consider the magnitude of the intercepts. For the triangle to exist in the first quadrant,$m$ must be negative. Let $m = -k$ where $k > 0$.The intercepts are $P = (1 + \frac{2}{k}, 0)$ and $Q = (0, 2 + k)$.Area $A = \frac{1}{2} 	imes (1 + \frac{2}{k}) 	imes (2 + k) = \frac{1}{2} (2 + k + \frac{4}{k} + 2) = \frac{1}{2} (4 + k + \frac{4}{k}) = 2 + \frac{k}{2} + \frac{2}{k}$.By $AM$-$GM$ inequality,$\frac{k}{2} + \frac{2}{k} \ge 2 \sqrt{\frac{k}{2} 	imes \frac{2}{k}} = 2$.Equality holds when $\frac{k}{2} = \frac{2}{k} \implies k^2 = 4 \implies k = 2$.Since $m = -k$,the slope $m = -2$.

$A$ line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a $\triangle OPQ$,where $O$ is the origin. If the area of $\triangle OPQ$ is least,then the slope of the line $PQ$ is

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