Using determinants prove that the points (a, b), (a', b')

$\left|\begin{array}{ccc}a & b & 1 \\ a^{\prime} & b^{\prime} & 1 \\ a-a^{\prime} & b-b^{\prime} & 1\end{array}\right|$

$\Rightarrow \Delta=\left|\begin{array}{ccc}a & b & 1 \\ a^{\prime}-a & b^{\prime}-b & 0 \\ a-a^{\prime} & b-b^{\prime} & 1\end{array}\right| \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ ]

$\Rightarrow \Delta=\left|\begin{array}{ccc}a & b & 1 \\ a^{\prime}-a & b^{\prime}-b & 0 \\ -a^{\prime} & -b^{\prime} & 0\end{array}\right| \quad$ [Applying $R_{3} \rightarrow R_{3}-R_{1}$ ]

$\Rightarrow \Delta=\left|\begin{array}{cc}a^{\prime}-a & b^{\prime}-b \\ -a^{\prime} & -b^{\prime}\end{array}\right|$

$\Rightarrow \Delta=-b^{\prime}\left(a^{\prime}-a\right)+a^{\prime}\left(b^{\prime}-b\right)$

$=-b^{\prime} a^{\prime}+b^{\prime} a+a^{\prime} b^{\prime}-a^{\prime} b$

$=b^{\prime} a-a^{\prime} b$

If the points are collinear, then $\Delta=0$.

So, $a b^{\prime}-a^{\prime} b=0$

Thus, $a b^{\prime}=a^{\prime} b$