$\text{Show } \{(x,y):(x-y)(xy-1)=0\}$ is path connected 
Def.(path-connected)
A set $S⊆\mathbb{R}^n$ is path-connected if, for every pair of points $x$ and $y$ in $S$,
$∃$ continuous $γ:[0,1]→S$, such that $γ(0)=x$, and $γ(1)=y$.

$\text{Show } \{(x,y):(x-y)(xy-1)=0\}$ is path connected

My thoughts
Let $p_1,\dots,p_n\in\mathbb{R}^2$
If want the function 'walk though' the path of $p_1-\dots- p_n$ in $\mathbb{R}^2$ ,$\color{lightgrey}{\text{(I think this also works in $\mathbb{R}^n$)}}$
just define the $\gamma:[0,1]\to\mathbb{R}^2$ as the following would works
$$\gamma(t)= \left\{\begin{array}{l}
p_1\left(1-nt\right)+p_2\left(nt\right),t\in[0,\frac{1}{n}]
\\p_2\left(1-nt\right)+p_3\left(nt\right),t\in(\frac{1}{n},\frac{2}{n}]
\\\vdots
\\p_{n-2}\left(1-nt\right)+p_{n-1}\left(nt\right),t\in(\frac{n-2}{n},\frac{n-1}{n}]
\\p_{n-1}\left(1-nt\right)+p_n\left(nt\right),t\in(\frac{n-1}{n},1]\end{array}\right.$$
But this only walk though straight lines, and If we want to let $\gamma(t)$ go though curves, it seems doesn't make much sense to let $n\to\infty$
How do I define $\gamma$ in this case$?$
Thanks for your help.
 A: HINT. A good first step would be to plot $(x-y)(xy-1)=0$. You should get two sets of curves, one of which you may want to think of as two pieces. You should be able to parametrize each (part) of these curves easily by writing them as a function of a single variable, say $t$. Each of these curves are path connected. Two of these curves has a point in common with another one. Each curve is path connected (why?), then (being very careful) use a theorem you should know about path connected spaces having a point in common...
A: Hints: if you can connect $P$ and $Q$ as well as $Q$ and $R$ by paths then you can connect $P$ and $R$.
If $xy=1$ and $x >0$ then you can connect $(x,y)$ and $(1,1)$ by $\gamma (t)=(tx+(1-t)x,\frac 1 {tx+(1-t)x})$. Similarly if $xy=1$ and $x <0$ then you can connect $(x,y)$ and $(-1,-1)$. It is easy to see that any point on  the line $x=y$ can be connected to $(1,1)$ by a path. [ In particular $(-1,-1)$ and $(1,1)$ can be connected]. Put these together to finish the proof. 
