# Why is $\frac{y}{x}$ greater than $\frac{q}{p}$ -Figure

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

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Think of the slopes of the lines passing through the points and the origin. – David Mitra Aug 3 '12 at 0:03
To add a little to David Mitra's comment, you can also think of the slope of the line in the middle. – Matt Groff Aug 3 '12 at 0:07
@DavidMitra so the slope of the line passing through (p,q) would be $\frac{q}{p}$ and slope of line passing through (x,y) and origin will be $\frac{y}{x}$ however i still dont get how $\frac{y}{x}$ is greater in value – MistyD Aug 3 '12 at 0:11

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.)

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Which is the vertical line ? – MistyD Aug 3 '12 at 0:31
@MistyD Just the vertical line that passes through the point $(p,q)$. $C$ is the point of intersection of this line with $\ell_1$. ($C$ is introduced so one can compute the slope of $\ell_1$ and $\ell_2$ using $p$ as the "run" for both.) – David Mitra Aug 3 '12 at 0:35
@MistyD I added a diagram, for what's it's worth. – David Mitra Aug 3 '12 at 0:56
A great way to put this pictorially! – Pedro Tamaroff Aug 3 '12 at 1:25
Thanks for the image , that explains a lot – MistyD Aug 3 '12 at 2:24

$y/x =\arctan\theta\,\,,\,\theta=\,$ 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...

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Can you rephrase it in terms of slopes and lines – MistyD Aug 3 '12 at 0:08
Yes: $\,m:=\arctan y/x = \,$slope of the line that passes through (x,y) and (0,0), i.e. of the line $\,y=mx\,$ , and the same with the other point and $\,\arctan q/p\,$ – DonAntonio Aug 3 '12 at 0:11
The slope of the line passing through (p,q) and origin would be $\frac{q}{p}$ and slope of line passing through (x,y) and origin will be $\frac{y}{x}$ however i still dont get how $\frac{y}{x}$ is greater in value (unless i know their values) – MistyD Aug 3 '12 at 0:12