From the figure why is $\frac{y}{x}$ greater than $\frac{q}{p}$
Why is $\frac{y}{x}$ greater than $\frac{q}{p}$ -Figure
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1Think of the slopes of the lines passing through the points and the origin. – 2012-08-03
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0To add a little to David Mitra's comment, you can also think of the slope of the line in the middle. – 2012-08-03
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0@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 – 2012-08-03
2 Answers
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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0Which is the vertical line ? – 2012-08-03
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0@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.) – 2012-08-03
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1@MistyD I added a diagram, for what's it's worth. – 2012-08-03
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0A great way to put this pictorially! – 2012-08-03
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0Thanks for the image , that explains a lot – 2012-08-03
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0@DavidMitra After reviewing this answer it seems that we can only compare the two points P1 and P2 only when they have a common coordinate in this case the X-axis. Am I correct ? – 2012-08-04
$y/x =\arctan\theta\,\,,\,\theta=\,$ 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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0Can you rephrase it in terms of slopes and lines – 2012-08-03
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0Yes: $\,m:=\arctan y/x = \,$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 $\,\arctan q/p\,$ – 2012-08-03
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0The 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) – 2012-08-03