Characteristic for non-vertical "straight lines" is that they correspond with functions that can be prescribed by $x\mapsto ax+b$ where $a,b$ are fixed real numbers.
Based on equation $y=ax+b$ we can find an expression for $x$ in $y$ under the extra condition that $a\neq0$: $$x=\frac1{a}(y-b)$$By interchanging $x$ and $y$ we find the inverse function is:$$y=\frac1{a}(x-b)$$ This tells us that such linear functions have an inverse if $a\neq0$. In case $a=0$ we are dealing with a constant function prescribed by $x\mapsto b$. This function is not injective hence has no inverse.
Best Answer
It’s just a bit more complicated than reflecting the graph about the line $y=x$, because the result may not be the graph of a function any more. Note that if you take your original picture and perform the reflection, there are two points in the new picture with $x$-coordinate $0$, and none with $x$-coordinate $1$.
The elaborate way of describing the situation is: consider a function $f\colon X\to Y$, $X$ being the domain of the function, and $Y$ being the target space. There is a left inverse $g\colon Y\to X$ satisfying $g(f(x))=x$ for all $x\in X$, if and only if the original $f$ was one-to-one. And there is a right inverse $h\colon Y\to X$ satisfying $f(h(y))=y$ for all $y\in Y$ if and only if $f$ was onto $Y$.
Examples with $X=Y=\mathbb R$: the exponential function $\exp(x)=e^x$ is one-to-one but not onto, and has this left inverse $g$: if $t>0$, $g(t)=\ln(t)$, while if $t\le0$, $g(t)=17$. The function $\tan^*$ cooked up from the tangent by setting $\tan^*(x)=\tan(x)$ if $x$ is not an odd multiple of $\pi/2$, and $\tan^*((k+1/2)\pi)=0$ is onto, and has the right inverse $h=\arctan$, the usual inverse tangent function. Notice that $g$ is not a right inverse of the exponential function any more than $h$ is a left inverse of the hoked-up tangent function.