If $x_1, x_2, x_3$ as well as $y_1, y_2, y_3$ are in geometric progression with the same common ratio, then…
If $x_1, x_2, x_3$ as well as $y_1, y_2, y_3$ are in geometric progression with the same common ratio, then the points $\left(x_1, y_1\right)$, $\left(x_2, y_2\right),\left(x_3, y_3\right)$ are
vertices of an equilateral triangle
vertices of a right angled triangle
vertices of a right angled isosceles triangle
collinear
Solution
It is given that $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in GP with the same common ratio.
Let $r$ be the common ratio.
$\therefore \quad x_1=x_1, x_2=x r \text { and } x_3=x r^2$
Similarly, $y_1=y, y_2=y r$ and $y_3=y r^2$
$\therefore \quad \text { Area of } \Delta=\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_1 & 1 \\
x_3 & y_1 & 1
\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}
x & y_1 & 1 \\
x r & y r & 1 \\
x r^2 & y r^2 & 1
\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 1 \\ r & r & 1 \\ r^2 & r^2 & 1\end{array}\right|=\frac{1}{2} \times 0=0$
$\left[\because \mathrm{C}_1\right.$ and $\mathrm{C}_2$ are identical]
$\therefore \quad$ The given points do not form any triangle. They are collinear.