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


The answer I got was


After simplifying further I got


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*}


Your expression was wrong. When you simplify it, you should get


$= (1/2) \times\left( x_1(a(1-a)y_1+ax_1(a^2-1)y_1+a^2x_1(1-a)y_1\right)$

$=(1/2) \times\left( x_1y_1(a(1-a)+a(a^2-1)+a^2(1-a)\right) = 0$ which is good.


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