If $x_1,x_2,x_3$  and $y_1,y_2,y_3$  are both in GP with the same common ratio, then the points $\left(x_1,y_1\right),\left(x_2,y_2\right)$  and $\left(x_3,y_3\right)$

If $x_1,x_2,x_3$  and $y_1,y_2,y_3$  are in GP.

Then let $x_2=r{x}_1,x_3=r^2{x}_1$

And $y_2=r{y}_1,y_3=r^2{y}_1$

With common ratio $r$ , then the points are

$\left(x_1,y_1\right),\left(r{x}_1,r{y}_1\right)$  and $\left(r^2{x}_1,r^2{y}_1\right)$ .

Now $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}{x}_1& y_1& 1\\ r{x}_1& r{y}_1& 1\\ r^2{x}_1& r^2{y}_1& 1\end{array}\right|$

$=x_1{y}_1\left|\begin{array}{ccc}1& 1& 1\\ r& r& 1\\ r^2& r^2& 1\end{array}\right|$

$=x_1{y}_1\left(0\right)=0$

(Since two columns are identical)

Thus these points lie on a straight line.

Alternate Solution

Let $x_1=a\Rightarrow x_2=ar$  and $x_3=a{r}^2$

And $y=b\Rightarrow y_2=br$  and $y_3=b{r}^2$

Let the points are $A\left(a,b\right),B\left(ar,br\right)$  and $C\left(a{r}^2,b{r}^2\right)$ .

Now slope of $AB=\frac{b\left(r-1\right)}{a\left(r-1\right)}=\frac{b}{a}$

And slope of $BC=\frac{b\left(r^2-r\right)}{a\left(r^2-r\right)}=\frac{b}{a}$

$\because$   slope of $AB=$ slope of $BC$

$\Rightarrow AB\parallel BC$

But $B$  is a common point.

$\therefore A,B$  and $C$  are collinear.

i.e., the points $\left(x_1,y_1\right),\left(x_2,y_2\right)$  and $\left(x_3,y_3\right)$  lie on a straight line.