If x1,x2,x3 as well as y1,y2,y3 are in gp

I have got the answer of the following question but i have some doubt....

If $$x_1$$, $$x_2$$, and $$x_3$$ as well as $$y_1$$, $$y_2$$, and $$y_3$$ are in G.P. with same common ratio (not equal to one) then the points $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$

(a) lie on a straight line

(b) lie on an elipse

(c) lie on a circle

(d) are the vertices of a triangle.

I have solved this question by equating slopes and got the correct answer that is (a) but, I tried finding out the area of the triangle formed. That should have been zero but it wasn't zero.

Let the ratio be $$a$$.

$$x_1 = \frac{x_3}{a^2}$$

and similarly found the values of $$x_2$$ in terms of $$x_3$$ and values of $$y_1$$ and $$y_2$$ in terms of $$y_3$$. Then I substituted the values in formula

$$\frac{1}{2}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$

The answer I got was

$$\frac{(x_3)(y_3)(1-a)}{2a^2}$$

After simplifying further I got

$$\frac{(x_2)(y_2)(1-a)}{2}$$

For this expression to be zero $$a$$ should be equal to $$1$$ but it is not as the ration is not equal to one. $$x_2$$ and $$y_2$$ can not be zero as they are in G.P. So this proves that the three points are not concurrent.

I know I am making some mistake but I don't know where. Please help.

The area of the triangle is \begin{align*}\require{color} &\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\\ &=\frac{x_1(y_1a-y_1a^2)+x_1a(y_1a^2-y_1)+x_1a^2(y_1-y_1a)}{2}\\ &=\frac{x_1y_1}2[(a-a^2)+a(a^2-1)+a^2(1-a)]\\ &=\frac{x_1y_1}2[{\color{green}a}{\color{red}-a^2}{\color{blue}+a^3}{\color{green}-a}{\color{red}+a^2}{\color{blue}-a^3}]\\ &=0 \end{align*}
$$\,\,\,\,\,\,(1/2)\times\left(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right)$$
$$= (1/2) \times\left( x_1(a(1-a)y_1+ax_1(a^2-1)y_1+a^2x_1(1-a)y_1\right)$$
$$=(1/2) \times\left( x_1y_1(a(1-a)+a(a^2-1)+a^2(1-a)\right) = 0$$ which is good.