Answer
$x=\dfrac{4}{2 \tan \theta+1}; y= \dfrac{4 \tan \theta}{2 \tan \theta+1}$; $0\leq \theta \leq \dfrac{\pi}{2}$
Work Step by Step
The point-slope equation of the line joining $(0,2)$ and $(4,0)$ is defined as:
$\dfrac{y-0}{x-4}=\dfrac{2-0}{0-4}$
$y=-\dfrac{x}{2}+2$
or, $x \tan \theta=-\dfrac{x}{2}+2$
$\implies x=\dfrac{4}{2 \tan \theta+1}$
Now, $y=-\dfrac{x}{2}+2=-(\dfrac{1}{2})(\dfrac{4}{2 \tan \theta+1})+2$
$y= \dfrac{4 \tan \theta}{2 \tan \theta+1}$
Hence, $x=\dfrac{4}{2 \tan \theta+1}; y= \dfrac{4 \tan \theta}{2 \tan \theta+1}$; $0\leq \theta \leq \dfrac{\pi}{2}$
