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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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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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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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) – 2019-01-11