Graph for Coulomb force vs $1/r^2$ We know that when there is a relation like x inversely proportional to y^2, the graph would be like this:

But in this question"Plot a graph showing variation of Coulomb force(F) versus $1/r^2$, where $r$ is distance between two charges of each pair of charges($1\,\mu C$, $2\,\mu C$) and ($1\,\mu C$, $-3\,\mu C$). Interpret the graphs obtained.".. The answer in my book is given in a strange way like this:

I don't think this is a wrong answer as I have checked it in many books. So, why did they do it like that (that graph in answer looks like x directly proportional to y)? 
 A: Coulomb force is inversely proportional to the distance squared:
$$F = \frac{k}{r^2}$$
So if you plot $F$ versus $r$, as $r \rightarrow \infty$, $F \rightarrow 0$ (like the first image).
In the second image, $F$ is drawn versus $\frac{1}{r^2}$. As $r \rightarrow \infty$, $r^{-2} \rightarrow 0$ and $F \rightarrow 0$. "But why there are 
linear", you ask. 
Because if you plot $y$ versus $x$ according to the equation $y = mx$, you'll have a line with slope $m$. So if you plot $F$ versus $r^{-2}$, you'll have a line with slope $k$.
$$
\begin{align}
 y &= mx \\
&\updownarrow \\
 F &= k r^{-2}
\end{align}
$$
A: A graph between $x$ and $y$ is different from that of $x$ and $(1/y)$. 
Simply let $x$ and $y$ be related as follows:
$$y=k/x$$
where $k$ is a constant. You can see that $x$ is indirectly proportional to $y$. Now cross multiply and turn the equation into $(1/y)=(x/k)$ and consider $1/y$ as one variable $z$ (say). As a result the equation becomes $z=x/k$ where you can clearly see that $z$ is directly (not indirectly) proportional to $x$. So it is a straight line because $z$ is nothing but $1/y$.
