Answer
$P(m)=(\frac{1}{m^2},\frac{1}{m})$
Work Step by Step
$\mathrm{See\: the\:figure\:below}$. Let's say $\ P(a,b)\ $ is any point on the graph of $\ f(x)=\sqrt{x}\ $. If we input $\ x\ $ into the function $\ f(x)=\sqrt{x}\ $, we get the corresponding value of $\ y\ $ to be $\ \sqrt{x}\ $. And, since $\ P\ $ is on the graph of the function $\ \sqrt{x}\ $, we have a relation for its coordinates.
$P=(x,\sqrt{x})$
The slope of the line joining the point $\ P\ $ and the origin is $\ m=\frac{\sqrt{x}}{x}.$
After rearranging the equation, we have:
$\ m=\frac{\sqrt{x}}{x}$
$\Rightarrow\ m=\frac{1}{\sqrt{x}}$
$\Rightarrow\ \sqrt{x}=\frac{1}{m}$
$\Rightarrow\ x=\frac{1}{m^2}$
So, the coordinates of $\ P\ $ as a function of the slope $\ m\ $ can be written as:
$P(m)=(\frac{1}{m^2},\frac{1}{m})$ where $\ m\ $ represents the slope of the line.
