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Consider the equation, $$\frac{1}{y}-\frac{1}{x}=c$$ here $c$ is any constant.
So letting $c=0$, we get $$\frac{1}{y}-\frac{1}{x}=0 \implies \frac{1}{y}=\frac{1}{x}$$ Now this is a straight line but I'm not sure whether origin lies on this line that is does this line passes through origin or not since $x,y$ are in denominator? Does that change anything from the equation of line $y=x$?

Amrit Awasthi
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    Strictly speaking $(x,y)=(0,0)$ is not a solution to $\frac{1}{y}-\frac{1}{x}=c$ for any $c$. there are, however, solutions arbitrarily close to the origin which do satisfy this equation. – Graviton Feb 17 '22 at 06:07

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Your original equation can't be evaluated when $x=0$ or $y=0$ because it contains undefined expressions, so no such values are solutions. It would be wrong to say that $\frac10-\frac10=0$.

However, a natural "fix" is to multiply through by $xy$, yielding $x-y=cxy$. This equation has the same solutions as the original one in the original domain, plus potentially new solutions on the coordinate axes. Indeed, when $c=0$,the origin $x=y=0$ satisfies the new equation.

Karl
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