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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 $O A+O B$ as $L$ varies, is
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The correct answer is:
4
Equation of line having slope ' $m$ ' passes through the point $(1,1)$ is
So, $A\left(\frac{m-1}{m}, 0\right)$ and $B(0,1-m)$
Now, $\quad O A+O B=\left(1-\frac{1}{m}\right)+(1-m)=2-\left(m+\frac{1}{m}\right)$
$\because \mathrm{m}$ is negative, so minimum value of
$$
-\left(m+\frac{1}{m}\right)=2
$$
So, minimum value of $O A+O B=2+2=4$
So, $A\left(\frac{m-1}{m}, 0\right)$ and $B(0,1-m)$
Now, $\quad O A+O B=\left(1-\frac{1}{m}\right)+(1-m)=2-\left(m+\frac{1}{m}\right)$
$\because \mathrm{m}$ is negative, so minimum value of
$$
-\left(m+\frac{1}{m}\right)=2
$$
So, minimum value of $O A+O B=2+2=4$
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