Solution:

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3} & 1\end{array}\right|$
Since the points are collinear, the area they enclose is 0
$0=\frac{1}{2}\left|\begin{array}{lll}
\mathrm{a} & 0 & 1 \\
0 & \mathrm{~b} & 1 \\
1 & 1 & 1
\end{array}\right|$
Expanding with $C_{1}$
$\begin{array}{l}
\rightarrow 0=a(b-1)+(-b) \\
\rightarrow 0=a b-a-b \\
\rightarrow a+b=a b \\
\rightarrow \frac{a+b}{a b}=1 \\
\rightarrow \frac{1}{a}+\frac{1}{b}=1
\end{array}$
Hence proved.