Explanation

(a)
Since, $x_{1},x_{2},x_{3}$ and $y_{1},\ y_{2},y_{3}$ are in GP with the same common ratio.
$\therefore\ x_{2} = rx_{1},\ x_{3} = r^{2}x_{1},\ y_{2} = ry_{1},\ y_{3} = r^{2}y_{1}$
Area of triangle
$= \frac{1}{2}\left| \begin{matrix}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{matrix} \right| = \frac{1}{2}\left| \begin{matrix}
x_{1} & y_{1} & 1 \\
rx_{1} & ry_{1} & 1 \\
r^{2}x_{1} & r^{2}y_{1} & 1
\end{matrix} \right|$
$= \frac{1}{2}x_{1}y_{1}\left| \begin{matrix}
1 & 1 & 1 \\
r & r & 1 \\
r^{2} & r^{2} & 1
\end{matrix} \right| = 0$ $(\because\$two columns are identical)
Hence, three points are in a straight line.