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(D) The equation of a line with slope $m$ passing through $(1,1)$ is $y - 1 = m(x - 1)$.Since the line cuts the positive coordinate axes at $A$ and $B

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$4$

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(D) The equation of a line with slope $m$ passing through $(1,1)$ is $y - 1 = m(x - 1)$.Since the line cuts the positive coordinate axes at $A$ and $B$,we set $y=0$ to find $A$ and $x=0$ to find $B$.For $A$,$0 - 1 = m(x - 1) \implies x - 1 = -\frac{1}{m} \implies x = 1 - \frac{1}{m}$. Thus,$A = (1 - \frac{1}{m}, 0)$.For $B$,$y - 1 = m(0 - 1) \implies y - 1 = -m \implies y = 1 - m$. Thus,$B = (0, 1 - m)$.Since $A$ and $B$ are on positive axes,$1 - \frac{1}{m} > 0$ and $1 - m > 0$. Since $m  0$.Then $OA = 1 + \frac{1}{k}$ and $OB = 1 + k$.$OA + OB = 2 + k + \frac{1}{k}$.By $AM$-$GM$ inequality,$k + \frac{1}{k} \ge 2\sqrt{k \cdot \frac{1}{k}} = 2$.The minimum value is $2 + 2 = 4$.

$A$ straight line $L$ with negative slope passes through the point $(1,1)$ and cuts the positive coordinate axes at the points $A$ and $B$. If $O$ is the origin,then the minimum value of $OA + OB$ as $L$ varies,is

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