0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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*}

$\endgroup$
1
$\begingroup$

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

$\,\,\,\,\,\,(1/2)\times\left(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right)$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.