From the figure why is $\frac{y}{x}$ greater than $\frac{q}{p}$

From the figure why is $\frac{y}{x}$ greater than $\frac{q}{p}$

Let $\ell_1$ be the line through the origin $O$ and the point $P_1=(x,y)$. Let $\ell_2$ be the line through $O$ and the point $P_2=(p,q)$. Let $B=(0,p)$ and let $C=(p,c_2)$ be the point of intersection of the vertical line through $P_2$ and the line $\ell_1$.
Clearly (cough), $C$ lies above the pictured blue line, so $c_2>q$.
The slope of $\ell_1$ is ${y\over x} = {c_2\over p}$. The slope of $\ell_2$ is $q\over p$. Since $c_2>q$, we have ${c_2\over p}>{q\over p}$; and thus ${y\over x}>{q\over p}$.
(I think you can argue intuitively: the "steeper" the line, the greater its slope.)

$y/x =\arctan\theta\,\,,\,\theta=\,$ the angle between the line through the origin and $\,(x,y)\,$ and the positive direction of the $\,x-\,$ axis, and $\,q/p\,$ is the arctangent of the line throught the origin and $\,(p,q)\,$ and the positive direction of the $\,x-\,$ axis. Since the former angle is clearly greater than the latter we're done (you know the inverse trigonometric functions, right?).
The above, of course, can also be rephrased in terms of slopes of straight lines...